# ABAW : Facial Expression Recognition in the wild

Dr. Darshan Gera
SSSIHL, Brindavan Campus
Bengaluru, Karnataka, India
<EMAIL_ADDRESS>
&Badveeti Naveen Siva Kumar
SSSIHL, Prasanthi Nilayam Campus
Sri Sathya Sai District, Andhra Pradesh, India
<EMAIL_ADDRESS>
Bobbili Veerendra Raj Kumar
SSSIHL, Prasanthi Nilayam Campus
Sri Sathya Sai District, Andhra Pradesh, India
<EMAIL_ADDRESS>
&Dr. S Balasubramanian
SSSIHL, Prasanthi Nilayam Campus
Sri Sathya Sai District, Andhra Pradesh, India
<EMAIL_ADDRESS>

###### Abstract

The fifth Affective Behavior Analysis in-the-wild (ABAW) competition has
multiple challenges such as Valence-Arousal Estimation Challenge, Expression
Classification Challenge, Action Unit Detection Challenge, Emotional Reaction
Intensity Estimation Challenge. In this paper we have dealt only expression
classification challenge using multiple approaches such as fully supervised,
semi-supervised and noisy label approach. Our approach using noise aware model
has performed better than baseline model by 10.46% and semi supervised model
has performed better than baseline model by 9.38% and the fully supervised
model has performed better than the baseline by 9.34%.

_K_ eywords Facial Expression Recognition $\cdot$ Aff-Wild2 $\cdot$ Semi-
supervised Learning $\cdot$ Noisy label approach $\cdot$ Complementary label

##  1 Introduction

Facial expression recognition (FER) is also a rapidly growing field of
research that has become increasingly important in recent years. The ability
to accurately detect and interpret human emotions based on facial expressions
has a wide range of potential applications, from improving human-computer
interaction to enhancing mental health diagnosis and treatment. To get models
that are independent of demographic features such as age, gender, region, we
need to train the models on real-in-wild datasets such as RAF-DB, Affectnet,
Aff-wild2 [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17]. These
in-wild datasets poses multiple challenges such as variation in illumination,
variation in poses, blur, etc. In this paper we have dealt with this problem
in multiple approaches such as fully supervised,semi-supervised and noisy
label approach.

##  2 Method

In this section, we present our solution to the Expression (Expr)
Classification Challenge at the 5th Affective Behavior Analysis in-the-wild
(ABAW) Competition.

###  2.1 Baseline

The white paper[11] released by organisers of ABAW competition provided as
baseline, a VGG16 network with fixed convolutional weights from a pre-trained
checkpoint on VGGFACE dataset for feature extraction and the last three fully
connected layers are trainable along with output layer equipped with softmax
activation function that gives the predictions on 8 expression classes.

###  2.2 Fully Supervised approach with finetuning

The approach taken in the baseline model of [11] is to use a pre-trained
checkpoint as a fixed feature extractor. We improve this by using resnet-18
[18] as basenet with pre-trained weights from resnet-18 model trained on MS-
Celeb [19] face recognition data-set as our initial point, which acts as
feature extractor, this is fine-tuned on the current task of facial expression
recognition. This basenet is followed by a dropout layer and fully connected
layer equipped with softmax activation learns to predict on the 8 expression
classes considered as part of this challenge.

To train this model, images were augmented using augmentations such as
horizontal flip, random crop and were resized to 224*224*3 before feeding them
to the model to obtain predictions. Cross entropy ${}^{\ref{eq:lossCE}}$ was
used as loss function for the baseline. We used Adam optimizer with learning
rate 0.0005 and weight decay as $e^{-4}$.

$L_{CE}=(-\sum^{8}_{c=1}\tilde{y}_{i=1}log(p^{c}(x_{i},\theta))))$ (1)

Here $\theta$ refers to the model parameters $p^{c}$ refers to prediction
probability of class c and $\tilde{y}_{c}$ refers to ground truth value of
that class on a sample.

###  2.3 Semi-Supervised learning with complementary labels

There are a total of 1089929 images in train set, but nearly $50\%$ of this
data, precisely 502970 images have invalid label(-1), therefore do not belong
to any of the eight classes considered as part of the challenge. Therefore in
total we have 586959 images as actual training images. To improve upon the
performance of the fully supervised approach we could use these invalid images
as unlabeled images and make use of semi-supervised learning which benefits
from unlabeled images. Motivated from a recent work in semi-supervised
learning named MutexMatch [20] which effectively uses unlabeled data to
improve overall performance of the model in limited label setting.

MutexMatch uses a fixed threshold to divide unlabeled samples as high
confident and low confident ones. Unlike other works in semi-supervised
learning which emphasise on utilizing the high confident samples in varied
varieties, MutexMatch uses the low confident samples to predict negative
labels which is a simpler goal by the means of a True-Negative
classifier(TNC).

Similar to general approaches to semi-supervised learning it uses True-
Positive classifier(TPC) to classify the images into considered set of
classes. It has a supervised loss $L_{sup}$ 1 on model predictions and ground
truth labels, which helps the network learn features and weights, to classify
appropriately. Once the model has learnt from the labeled data, this model is
used to predict pseudo-labels on the unlabeled data using the True-Positive
classifier. Fixed threshold is used to separate unlabeled samples into
confident and non-confident samples. Predictions on confident samples are used
as pseudo-labels to help the model utilize unlabeled data.

To learn pseudo-labels effectively a pseudo-label loss $L_{p}$ is introduced
which is the cross entropy${}^{\ref{eq:lossCE}}$ loss between predictions of
the model on weak and strong augmentations of high confident samples. Two more
losses $L_{sep}$ and $L_{n}$ involve the predictions from the complementary
labels on low confident samples on the True-Negative classifier. $L_{sep}$
loss is cross entropy${}^{\ref{eq:lossCE}}$ on TNC prediction of weak
augmentation and the class with lowest confidence on weak augmented image as
predicted by the TPC. $L_{n}$ ${}^{\ref{eq:lossLn}}$ is defined only on the
low confident samples, which is negative consistency loss between prediction
of TNC on weak and strong augmented versions of an image for top-k
complementary predicted classes.

####  2.3.1 Loss functions

$L_{p}=\frac{1}{\mu_{B}}\sum^{\mu_{B}}_{n=1}1(max(p^{w}_{n})\geq\tau)H(\tilde{p}^{w}_{n},\tilde{p}^{s}_{n}))$
(2)

Here $\mu_{B}$ is the number of unlabeled images in a batch, H is the cross
entropy function as in 1, $p^{w}_{n}$ is the TPC prediction on weak augmented
image, $\tilde{p}^{s}_{n}$ is the TPC prediction on strong augmented image,
$\tilde{p}^{w}_{n}$ is $argmax(p^{w}_{n})$.

$L_{n}=\frac{1}{\mu_{B}}\sum^{\mu_{B}}_{n=1}1(max(p^{w}_{n})<\tau)(-\frac{1}{k}\sum^{C}_{i=1}g_{n,(i)}r^{w}_{n,(i)}log(r^{s}_{n,(i)}))$
(3)

Here $\mu_{B}$ and $p^{w}_{n}$ are as defined above, $r^{w}_{n,(i)}$ is the
${i}^{th}$ probability component in the prediction of TNC on weak augmented
image, $r^{s}_{n,(i)}$ is the $i^{th}$ probability component in the prediction
of TNC on strong augmented image, $g_{n,(i)}$ is a mask which selects the
classes that a in top-k of largest probability components in the complementary
label prediction.

Inspired from [21, 22, 23] which use dynamic adaptive threshold 4 that caters
to inter and intra class differences that exist in facial expressions, we use
dynamic adaptive threshold to divide the unlabeled samples into confident and
non-confident ones. This class adaptive threshold is calculated by the model’s
performance on train set and is dynamically scaled up as the training
progresses.

$T^{c}=\frac{\beta*({\frac{1}{N^{s}}}\sum^{N^{s}}_{i=1}\delta^{c}_{i}*p_{i})}{1+\gamma^{-ep}}\hskip
10.00002pt,\hskip 10.00002pt\mathrm{where}$ (4)
$\delta^{c}_{i}=\left\\{\begin{array}[]{ll}1&\mathrm{if\hskip
10.00002pt}\tilde{y}_{i}=c,\\\ \\\ 0&\mathrm{otherwise}.\end{array}\right.$

The values $\beta=0.95$ and $\gamma=e$ are taken from [21]

###  2.4 Noise aware model

During the training of the fully supervised model, we observed that accuracy
and f1-score on an average was 98.7 and 98.2% respectively on the training
set, whereas average accuracy was 52.6 and the best f1-score was obtained to
be 32.35%.This shows that the model doesn’t generalize well on unseen data.
Generally, models do not generalize well due to the following reasons:

  * •

Capacity of the model : If the capacity of the model is low, then the model
underfits the data leading to poor generalization. This can be solved by using
a model with more capacity. But it can’t be the case since the accuracy and
f1-score on train is very high implying over fitting of model.

  * •

Distribution change : If there is a mismatch in the distribution of training
data and the unseen validation data, then the model performs poorly on unseen
data. But in general we treat the training data and unseen data are sampled
from same distribution.

  * •

Presence of noise : If there is noise in the labels of dataset and if the
model over-fits on the training data, then the model can’t generalize well.

Assuming that there is noise in the labels, To deal with the noisy label
problem, We propose a noise aware model.

This model has its backbone as resnet-18 followed by a fully connected layer
with softmax activation to predict the 8 classes in expression classification
task. We use pretrained weights from resnet18 model trained on affectnet
dataset. In this model, we use two different types of augmentations of the
image for consistency. MSE loss${}^{\ref{eq:lossMSE}}$ is used for consistency
and Weighted cross entropy loss${}^{\ref{eq:weightedCE}}$ is used for
supervised loss. The weights are learnt by the model as per in SCN [24] paper.
Not all the samples are sent to supervised loss. We take out the samples whose
prediction probabilities that fail to be greater than dynamic adaptive
threshold[25].

We use dynamic adaptive threshold in every epoch to tackle inter and intra
class differences in prediction probabilities of expression classification.
Dynamic adaptive threshold was calculated by taking the class vise mean of all
the prediction probabilities in every batch. For a given class $i$, we treat
the samples whose ground truth prediction probabilities that are greater than
the dynamic adaptive threshold for class $i$ as clean and those samples that
fails are treated as noisy samples. Only clean samples are used for supervised
loss and all the samples are used for consistency loss where we force the
consistency between the attention maps of the weak augmented image and strong
augmented image.

The attention maps[26] are calculated in the following way:

We first extract the feature maps from last but second layer from the backbone
and weights from the fully connected layer by multiplying the weights and
feature maps we obtained the attention maps for every class.

$L_{WCE}=\frac{1}{N}(-\sum^{N}_{i=1}log(\frac{e^{\alpha_{i}W^{T}_{y_{i}}}x_{i}}{\sum^{C}_{j=1}e^{\alpha_{i}W^{T}_{j}}x_{i}}))$
(5)

$L_{MSE}=-\frac{1}{NLHW}\sum^{N}_{i=1}\sum^{L}_{j=1}||AM_{ij}-AM^{{}^{\prime}}_{ij}||_{2}$
(6)

We used Adam optimizer with 0.0001 as learning rate on all the model
parameters.

####  2.4.1 Loss functions

We have used weighted cross entropy loss${}^{\ref{eq:weightedCE}}$ for
supervised loss where N is number of samples in batch C is number of classes
and $\alpha_{i}$ are the weights. The loss function that we have used for
consistency loss is MSE loss${}^{\ref{eq:lossMSE}}$ where N is number of
samples in the batch, L is number of feature maps, H and W are height and
width of feature maps, AM is attention maps on weak augmented image and
$AM^{{}^{\prime}}$ is attention maps on strong augmented flipped image.

###  2.5 Problem formulation

Let $D={\\{(x_{i},\tilde{y}_{i})\\}}^{N}_{i=1}$ be the dataset of N samples.
Here $x_{i}$ is $i^{th}$ image where $\tilde{y}_{i}$ represents expression
class ${y}^{Exp}_{i})$ of $i^{th}$ image. The backbone network is
parameterized by $\theta$ ( ResNet-18 [18] pre-trained on different datasets
like rafdb [27] and affectnet [28] for different approaches as backbone). We
denote $x_{w}$ as weak augmented image and $x_{s}$ as strong augmented image,
$P_{Exp}$ represent probability distribution predicted by Expression
classifier. Weak augmentations include random cropping with padding and
horizontal flipping of the input image. Strong augmentations includes weak
augmentations along with Randaugment [29].

##  3 Dataset

###  3.1 Dataset

s-AffWild2 [17] database is a static version of Aff-Wild2 database and
contains a total of 11,10,367 images for training and 4,53,535 images for
validation. Out of which 5,02,970 images are to be disregarded from training
set and 20,438 images are not present but the image paths are given. In
validation data 2,79,749 are valid and remaining all are with -1 as label
which we should disregard.

###  3.2 Implementation Details

The performance measure fo experssion classification problem is the average F1
score on all the 8 classes.The F1 score is a harmonic mean of the recall (i.e.
Number of positive class images correctly identified out of true positive
class) and precision (i.e. Number of positive class images correctly
identified out of positive predicted). The F1 score takes values in the range
[0, 1]. Here we need to present F1 score as percentage. The F1 score is
defined as:

$F_{1}={\frac{2*precision*recall}{precision+recall}}$ (7)

##  4 Results

We report our results on the official validation set from the ABAW 2023
Challenge [11] in Table 1 . Our best performance achieves overall score of
33.46% on validation set which is a significant improvement over baseline.

Table 1: Performance comparison on Aff-Wild2 validation set Method | Exp-F1 score
---|---
Baseline [11] | 23
Fully supervised model | 32.34
Semi-Supervised learning with complementary labels | 32.82
Noise aware model | 33.46

##  5 Conclusions

In this paper, we presented multiple methods. Firstly fully supervised model
gave F1 score that is greater than the given baseline model[11] by 9.34%. Then
to overcome the limitations of fully supervised model, we proposed semi
supervised learning with complementary labels and noise aware model that
perform better over baseline by a margin of 9.38% and 10.46% respectively.

## Acknowledgments

We dedicate this work to Our Guru and Guide Bhagawan Sri Sathya Sai Baba,
Divine Founder Chancellor of Sri Sathya Sai Institute of Higher Learning,
Prasanthi Nilayam, Andhra Pradesh, India.

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# Generating functions in Riesz spaces

Youssef Azouzi and Youssef Nasri
Research Laboratory of Algebra, Topology, Arithmetic, and Order
Department of Mathematics
COSAEF Faculty of Mathematical, Physical and Natural Sciences of Tunis
Tunis-El Manar University, 2092-El Manar, Tunisia The authors are members of
the GOSAEF research groupThis document is the results of the research project
funded by the National Science Foundation

###### Abstract

We introduce and study the concept of generating function for natural elements
in a Dedekind complete Riesz space equipped with a conditional expectatnion
operator. This allows to study discrete processes in free-measure setting. In
particular we improve a result obtained by Kuo, Vardy and Watson concerning
Poisson approximation.

Key words. Generating function, Dedekind complete Riesz space, Conditional
expectation, order convergence, unbounded order convergence.

##  1 Introduction

We introduce in this paper the notion of generating function for special
elements of a Riesz space called natural elements. Recently Ben Amor and
Omrani published a paper in which they treated the notion of moment generating
function in Riesz spaces; the main purpose of their paper is to get a
generalization of Chernoff Inequality in the setting of free-measure theory of
Riesz spaces [6]. Our appraoch here is more general and leads to a more
general Chernoff Inequality. Also we are able to give a generalization of a
result obtained by Kuo, Vardy, and Watson [15]. In these two papers the
authors studied only Bernoulli processes and sums of independent Bernoulli
processes which give Binomial processes. The main idea in our work here is to
develop a general approach to study discrete processes in Riesz spaces. For
this reason Functional Calculus via Daniell integral as they are developed in
[11] and [5] are more suitable than Functional Calculus as they are developed
in [7]. This allows to work in the universal completion $E^{u}$ of $E,$ and
even in its sup-completion $E^{s}$. When dealing with natural elements we can
use their distributions or their generating functions to define convergence in
distribution. This notion can be defined via order and unbounded order
convergence and it will be useful to note that these modes of convergence
agree for sequences if the space is universally complete; this result is due
to Grobler and can be found in [10]. The latter paper contains several
important results about unbounded order convergence.

The paper is organized as follows. Section 2 contains some preliminaries
needed in the following sections. We prove some results concerning band
projections and functional calculus developed using Daniell integral on Riesz
spaces following Grobler [11]. In Section 3 we develop some results on series
in Riesz spaces. We state and prove versions of Monotone Convergence Theorem,
Dominated Convergence Theorem and others and we will make a frequent use of
Fubini Theorem.

In section 4 we introduce the notion of generating function in Riesz spaces
equipped with a conditional expectation operator and study its basic
properties. We prove, in particular, that a natural element $x$ belongs to
$L^{p}\left(T\right)$ if and only if its generating function $g_{x}$ is an
order deravative of rank $k$ at $1.$

The last section contains some applications. In particular we give a new short
proof of a result established by Kuo, Vardy and Watson in [15] concernin
Poisson approximation and we state and prove a Riesz space version of Chernoff
Inequality, generalizing a result obtained recently by by Ben Amor and Omrani
in [6].

For more information about generating function classical probability theory we
refer the reader to Feller [9], [17]. All unexplained terminology and notation
concerning Riesz spaces an be found in standard references [1] and [19].

##  2 Preliminaries

###### Definition 1

We say that $\left(E,e,T\right)$ is a conditional Riesz triple if $E$ is a
Dedekind complete Riesz space, $e$ is a weak order unit and $T$ is a
conditional expectation operator, that is, a positive order continuous
projection on $E$ satisfying $Te=e,$ with range, $R\left(T\right),$ a Dedekind
complete Riesz subspace. We will require in this paper that $T$ is strictly
positive, which means that if $f>0$ then $Tf>0$ (see [5]).

Independence. The concept of independence with respect to a conditional
expectation has been introduced in [14, Definition 4.1 ]. See also [12,
Definition 4.1]. We recall here some facts.

A family $(P_{\alpha})_{\alpha\in A}$ of band projections is said to be
$T$-independent if for each finite subset $\\{\alpha_{1},...,\alpha_{n}\\}$ of
$A$ we have
$TP_{\alpha_{1}}P_{\alpha_{2}}....P_{\alpha_{n}}T=TP_{\alpha_{1}}TP_{\alpha_{2}}...TP_{\alpha_{2}}T,$
or equivalently,

$TP_{\alpha_{1}}P_{\alpha_{2}}....P_{\alpha_{n}}e=TP_{\alpha_{1}}TP_{\alpha_{2}}...TP_{\alpha_{2}}e.$

A family of closed Riesz subspaces $\left(E_{\alpha}\right)$ is said to be
$T$-independent if every family of band projections such that $P_{\alpha}e\in
E_{\alpha}$ for every $\alpha$ is $T$-independent.

A family of elements $(f_{i})$ in $E$ is independent with respect to $T$ and
$e$ if the family $\left\\{\left\langle
f_{i},R\left(T\right)\right\rangle:i\in I\right\\}$ of Dedekind complete Riesz
subspaces is independent, where $\left\langle f,R\left(T\right)\right\rangle$
denotes the closed Riesz subspace generated by $f$ and $R\left(T\right).$ It
is worth noting that if $x,y\in L^{1}\left(T\right)$ are independent then
$xy\in L^{1}\left(T\right)$ and

$T\left(xy\right)=TxTy.$ (1)

Observe that the second member exists as $R\left(T\right)$ is an
$f$-subalgebra of $E^{u}$ (see [5, Theorem 3.1]). The formula is true by
definition if $x$ and $y$ are components, and hence it extends by linearity
for step elements, so for positive elements by Freudenthal Theorem and order
continuity of $T.$ Finally we use the linearity to extend it for all elements
in $L^{1}\left(T\right).$ Using the fact that $T$ can be extended to the whole
space of $E^{s}$ (see [4, Corollary 6]) it easy to see that the formula (1)
holds for all elements in $E_{+}^{u}..$ Here we use the fact that if $x$ and
$y$ are $T$-independent then $x\wedge ne$ and $y\wedge ne$ are $T$-independent
as well. For more information about $T$-independence the reader is referred to
[12].

Bands and projections. As $E$ is Dedekind complete with weak order unit, every
band in $E$ is a principal projection band. If $P$ is the band projection on
the band $B$ then $P^{d}=I-P$ is the band projection on $B^{d}.$ The
projection band generated by $x$ is denoted by $B_{x}$ and the corresponding
band projection is denoted by $P_{x}.$ We recall that for $x,y\in E_{+}$ we
have $B_{x\vee y}=B_{x+y}.$ It is easily seen that
$P_{x}^{d}P_{y}^{d}=P_{x\vee y}^{d}.$ If $x,y\in E$ we put
$P_{x=y}=P_{\left(x-y\right)^{+}}^{d}P_{\left(y-x\right)^{+}}^{d}=P_{\left|x-y\right|}^{d}.$
We shall adopt some notations used by Grobler (see for example [11]):

$P_{x\geq y}=P_{\left(y-x\right)^{+}}^{d},$
$P_{x>y}=P_{\left(x-y\right)^{+}},$

$P_{x\geq y,a\geq b}=P_{\left(y-x\right)^{+}}^{d}P_{\left(b-a\right)^{+}}^{d}$
etc.

Daniell integral. For an element $x\in E$ we denote by $I_{x}$ the Daniell
integral with respect to the right continuous spectral system
$(e_{t}^{x})_{t\in\mathbb{R}}:=(P_{x\leq te}e)_{t\in\mathbb{R}}$ of $x$,
initially defined on the Riesz space of $\mathcal{F}(\mathbb{R})$-step
functions on $\mathbb{R}$ denoted by $L=L(\mathcal{F}(\mathbb{R}))$ and
extended to $L_{2}(I_{x}),$ where $\mathcal{F}(\mathbb{R})$ is the algebra
consisting of all finite unions of disjoint left open right closed intervals
$(a,b],(a,\infty),(-\infty,b]$ with $a,b\in\mathbb{R}.$ We just recall that
$I_{x}\left(f\right)$ is denoted by $f\left(x\right)$ and that
$f\left(x\right)=e_{b}^{x}-e_{a}^{x}$ for $f=\mathbf{1}_{(a,b]}.$ We refer the
reader to [11] and [5] for more details of the subject. We consider here the
following spaces

$L^{\uparrow}=L^{\uparrow}(I_{x}):=\left\\{f\in\overline{\mathbb{R}}^{\mathbb{R}}:\exists\left(f_{n}\right)\subset
L,\text{ }f_{n}\uparrow f\right\\},$

$L_{1}^{\uparrow}=L_{1}^{\uparrow}(I_{x}):=\left\\{f\in\mathbb{R}_{+}^{\mathbb{R}}:f\in
L^{\uparrow}(I_{x})\text{ \ and \ }I_{x}(f)\in E\right\\},$

$L_{1}=L_{1}(I_{x}):=L_{1}^{\uparrow}(I_{x})-L_{1}^{\uparrow}(I_{x}),$

$L_{2}=L_{2}(I_{x}):=L^{\uparrow}(I_{x})-L_{1}(I_{x})=L^{\uparrow}(I_{x})_{+}-L_{1}(I_{x}).$

###### Lemma 2

Let $\left(E,e,T\right)$ be a conditional Riesz triple and $x,y,w\in E.$ Then
the following hold.

  1. (i)

$P_{\left|x-y\right|\vee(x-w)^{+}}=P_{\left|x-y\right|\vee(y-w)^{+}}$;

  2. (ii)

$P_{x=y,x=w}=P_{x=y,y=w};$

  3. (iii)

$P_{x\geq y}P_{x=w}=P_{w\geq y}P_{x=w}$ and $P_{y\geq x}P_{x=w}=P_{y\geq
w}P_{x=w}.$

Proof. (i) It follows from the inequality

$(x-w)^{+}\leq(x-y)^{+}+(y-w)^{+}\leq\left|x-y\right|+(y-w)^{+}.$

that $B_{\left|x-y\right|\vee(x-w)^{+}}\subset
B_{\left|x-y\right|+(y-w)^{+}}=B_{\left|x-y\right|\vee(y-w)^{+}}.$ We
interchange $x$ and $y,$ obtaining the converse inclusion and so the desired
equality follows.

(ii) As $P_{x=y,x=w}=P_{\left|x-w\right|\vee\left|x-y\right|}^{d}$ it is
sufficient to show the inequality

$P_{\left|x-w\right|\vee\left|x-y\right|}\leq
P_{\left|x-y\right|\vee\left|y-w\right|},$

as the converse inequality is similar. But this follows easily from the
inequality $\left|x-w\right|\leq\left|x-y\right|+\left|y-w\right|.$ The proof
of (iii) is similar.

We need in our work some results on calculus functional from the paper of
Grobler [11] and Azouzi and Trabelsi [5]. The next result will be useful and
can not be found in those papers and is more general than [5, Lemma 2.7] and
[11, Theorem 3.9].

###### Theorem 3

Let $E$ be a Dedekind complete Riesz space with weak order unit $e.$

  1. (i)

If $f,g\in L_{2}\left(I_{x}\right)^{+}$ then
$\left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right).$

  2. (ii)

Assume $g\in L_{1}\left(I_{x}\right),$ $f\in L_{2}\left(I_{x}\right),$
$g\left(t\right)\in dom\left(f\right)$ for every $t.$ Then $\left(f\circ
g\right)\left(x\right)=f\left(g\left(x\right)\right)\in E^{s}.$

Proof. (i) Assume first that $f=\mathbf{1}_{\left(-\infty,a\right)}$ and
$g=\mathbf{1}_{\left(-\infty,b\right)}.$ Then
$fg=\mathbf{1}_{\left(-\infty,a\wedge b\right)}.$ Moreover
$f\left(x\right)=p_{a},$ $g\left(x\right)=p_{b}$ and
$\left(fg\right)\left(x\right)=p_{a\wedge b}=p_{a}p_{b}.$ The formula is then
true for all positive linear combination of indicator functions.

If $f\in L^{\uparrow}$ there exists a sequence of step functions
$\left(f_{n}\right)$ such that $f_{n}\uparrow f$ and $g_{n}\uparrow g.$ Now
using the previous step we deduce that

$\left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right).$

This can be extended for every $f,g\in L_{2}\left(I_{x}\right)^{+}.$

(ii) This is true if $f$ is continuous by [11, Theorem 3.9]. To see that this
is also true for every $f\in L^{\uparrow}$ it is only sufficient to observe
that for every $f\in L^{\uparrow}$ there is a sequence of positive continuous
functions $f_{n}$ such that $f_{n}\uparrow f.$ Now by taking the supremum over
$n$ in the equality $\left(f_{n}\circ
g\right)\left(x\right)=f_{n}\left(g\left(x\right)\right)$ we obtain
$\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right).$ The
result for general $f\in L_{2}\left(I_{x}\right)$ follows easily.

###### Lemma 4

Let $E$ be a Dedekind complete Riesz space with weak order unit $e$ and
$x,y\in E.$ Then the following statements hold.

  1. (i)

$P_{x=y}f\left(x\right)=P_{x=y}f\left(y\right),$ for all $f\in
L_{2}\left(I_{x}\right).$ In particular $P_{x=y}x=P_{x=y}y.$

  2. (ii)

$P_{f(x)^{+}}g(x)=\left(g\mathbf{1}_{(0,\infty)}\circ f\right)(x)$ and
$P_{f(x)^{-}}g(x)=\left(g\mathbf{1}_{(-\infty,0)}\circ f\right)(x),$ for all
$f,g\in L_{1}\left(I_{x}\right)$ for all $f,g\in L_{1}(I_{x}).$

Proof. (i) We prove first the equality for $f=\mathbf{1}_{(-\infty,a]},$
$a\in\mathbb{R}.$ But in this case $f(x)=P_{\left(x-ae\right)^{+}}^{d}$ and
the result follows from Lemma 2(i). So it can be extended by linearity and
then using the continuity of order projection and monotone convergence of
Daniell integral to every function in $L_{2}(I_{x}).$

(ii) We have

$P_{f(x)^{+}}g(x)=g(x)P_{f(x)^{+}}e=g(x).\mathbf{1}_{\left(0,\infty\right)}\left(f(x)\right)=\left(g.\mathbf{1}_{(0,\infty)}\circ
f\right)(x),$

The second part is similar.

###### Remark 5

It should be noted that (ii) is not true for $f$ in $L_{2}\left(I_{x}\right).$
Take for example, $x=e,$ $g=1,$ $f=\infty.$ Then $f\in
L_{2}\left(I_{x}\right)$ and
$e=P_{f(x)^{+}}g(x)\neq\left(g\mathbf{1}_{(0,\infty)}\circ
f\right)\left(x\right)=0$.

###### Definition 6

Let $\left(E,e,T\right)$ be a conditional Riesz triple and $x\in E.$

  1. 1.

The function

$\pi_{x}:t\in\mathbb{R}\longmapsto TP_{x=te}e=F_{x}(t)-F_{x}(t^{-}).$

is called the $(T,e)$-mass function of $x,$ where $F_{x}:t\longmapsto
TP_{x\leq te}e$ is the $T$-distribution function of $x$ and
$F_{x}(t^{-})=\lim\limits_{s\uparrow
t}F_{x}\left(s\right)=TP_{\left(x-te\right)^{-}}e.$

  2. 2.

An element $x$ of $E$ is said to be natural if
${\textstyle\sum\limits_{n=0}^{\infty}}\pi_{x}(n)=e.$

  3. 3.

Note that if $x$ is a natural element in $E,$ then its $T$-distribution
function $F_{x}$ is piecewise-constant. Bands $\left\\{B_{x=ne}:n\geq
0\right\\}$ are disjoint and it follows from the equality
$e={\textstyle\sum\limits_{n=1}^{\infty}}\pi_{x}\left(n\right)$ that
$E=\left[\left\\{B_{x=ne}:n\geq 0\right\\}\right].$ We derive that for every
$a\in\left(n,n+1\right),$ $B_{ne<x\leq ae}=0,$ which means that
$F_{x}\left(a\right)=F_{x}\left(n\right)$ for every $a\in\left(n,n+1\right).$

  4. 4.

In the definition above an element $x$ is said to be natural if
${\textstyle\sum\limits_{n\in\mathbb{N}}}\pi_{x}(n)=e.$ As $T$ is supposed to
be strictly positive this condition is equivalent to the fact that
${\textstyle\sum\limits_{n\in\mathbb{N}}}P_{x=ne}e=e.$ But as it is not common
to add this assumption we will keep the first definition.

Let $\left(E,e,T\right)$ be a conditional Riesz triple. We say that two
natural vectors $x$ and $y$ in $E$ have the same conditional distribution with
respect to $T$ and we write $x\overset{d}{=}_{T}y$ if they have the same
distribution function. Bernoulli processes and Poisson processes have been
introduced in [15]. Each one of them is defined as a sequence of elements in
$E$ having the same distribution despite the notion of distribution was not
explicitly used. Next we define in a more general manner these notions and
then explain the link between our definitions and those introduced in [15].

###### Definition 7

We say that an element $x\in E$ has Bernoulli distribution with parameter
$0<p<e$ with respect to $T,$ and write $x\sim\mathcal{B}(p).$ whenever
$\pi_{x}(1)=p$ and $\pi_{x}(0)=e-p.$ Binomial distribution with parameters
$n\in\mathbb{N}$ and $p\in\left(0,e\right)$ and Poisson distribution with
parameter $g\in E,$ $\ g>0$ are defined respectively by their mass function

$\displaystyle\pi_{x}\left(k\right)$
$\displaystyle=\dbinom{n}{k}p^{k}\left(e-p\right)^{n-k},\qquad k=0,1,2,...n,$
$\displaystyle\pi_{x}\left(k\right)$
$\displaystyle=\dfrac{g^{k}}{k!}e^{-g},\qquad k=0,1,2,....$

We write $x\sim\mathcal{B}(n,p)$ and $x\sim\mathcal{P}(p).$

Next we show that our definition of Bernoulli processes in vector lattices
coincides with that considered in [15].

###### Proposition 8

Let $\left(E,e,T\right)$ be a conditional Riesz triple and $x\in E.$ Then
$x\sim_{T}B(p)$ if and only if $x=Qe$ for some band projection $Q$ with
$TQe=p.$

Proof. If $x=Qe$, then
$\pi_{x}(1)=TP_{x=e}e=TP_{Qe=e}e=TP_{e-Qe}^{d}e=TQe,$and
$\pi_{x}(0)=TP_{x=0}e=TP_{Qe=0}e=TP_{Qe}^{d}e=TQ^{d}e=e-TQe,$ Hence
$x\sim_{T}B(p)$ with $p=TQe.$ Conversely assume that $\ x\sim_{T}B(p).$ Then
$e=TP_{x=e}e+TP_{x=0}e=T\left(P_{x=e}e+P_{x=0}\right)e.$ We will show now that
$P_{x=e}e\wedge P_{x=0}e=0.$ But this follows from Lemma 2.(ii) as
$P_{x=0}P_{x=e}=P_{0=e}P_{x=e}=0.$ Hence $e\geq P_{x=e}e+P_{x=0}e;$ which
gives

$0=Te-TP_{x=e}e-TP_{x=0}e=T\left(e-P_{x=e}e-P_{x=0}e\right).$

As $T$ is strictly positive and $e-P_{x=e}e-P_{x=0}e\geq 0$ we get
$e=P_{x=e}e+P_{x=0}e.$ Thus

$x=P_{x=e}x+P_{x=0}x=P_{x=e}e,$

and, as required, the band projection $Q=P_{x=e}$ satisfies $TQe=TP_{x=e}e=p$.

We end this preliminary section with a brief discussion concerning order
differentiability in Riesz spaces. We need this essentially to get in terms of
differentiability of the generating function in $1$ of the existence of a
moments of a given order for a natural element in $E.$

###### Definition 9

Let $J$ be an interval of $\mathbb{R}$ not reduced to a point, $s_{0}$ a fixed
point in $J$ and $E$ a Riesz space. A function $f:J\longrightarrow E$ is said
to be (order) differentiable at $s_{0}$ if the order limit
$\lim\limits_{s\longrightarrow s_{0}}\dfrac{f(s)-f(s_{0})}{s-s_{0}}$ exists in
$E.$ This limit is called the (order) derivative of $f$ at $s_{0}$ and is
denoted by $f^{{}^{\prime}}(s_{0})$.

Notions of right and left differentiability can be defined in an obvious way.
We can define by induction the nth derivative of $f$ by putting
$f^{(n+1)}:=(f^{(n)})^{\prime}$ for $n=0,1,2,...$

For general information about Riesz spaces we refer the reader to [19] and
[1]. We will make use of the notion of sup-completion, a comprehensive
presentation of the subject can be found in [2] and [4].

##  3 Series in Riesz spaces

We will prove in this section some results concerning series in Riesz spaces.
We consider only convergence in order, so
${\textstyle\sum\limits_{n=1}^{\infty}}x_{n}$ means the order limit of the
sequence $\left(S_{n}\right),$ where
$S_{n}={\textstyle\sum\limits_{k=1}^{n}}x_{k}.$ We say that the series $\sum
x_{n}$ is absolutely convergent if the series $\sum\left|x_{n}\right|$ is
order convergent. In $\sigma$-order complete Riesz spaces every absolutely
convergent series is order convergent. In particular if
$\left|x_{n}\right|\leq y_{n}$ for each $n$ and if the series $\sum y_{n}$ is
order convergent, then $\sum x_{n}$ is order convergent as well. When
$\left(f_{n}\right)$ is a sequence of functions from an interval $I$ to a
Riesz space $E,$ we say that $\sum f_{n}$ is order convergent uniformly on $t$
if the series $\sum f_{n}\left(t\right)$ converges in order for every $t\in I$
and there is a net $\left(y_{\alpha}\right)$ such that $y_{\alpha}\downarrow
0$ and for every $\alpha$ there exists $n_{0}$ such that
$\left|\sum\limits_{k=n}^{\infty}f_{n}\left(t\right)\right|\leq y_{\alpha}$
for all $n\geq n_{0}.$ We are interested in this work only in Dedekind
complete Riesz spaces; in such spaces the former property is equivalent to the
fact that $\sup\limits_{n\geq k}\sup\limits_{t\in
I}\left|\sum\limits_{j=n}^{\infty}f_{j}\left(t\right)\right|\downarrow 0$ as
$k\longrightarrow\infty.$ The following is a version of Monotone Convergence
Theorem for series in Riesz spaces.

###### Theorem 10

Let $E$ be a Dedekind complete Riesz space and
$\left(x_{\alpha}\left(n\right)\right)_{\alpha\in\Gamma}$ be a sequence of
increasing net in $E_{+}^{s}.$ Then

$\lim\limits_{\alpha}{\textstyle\sum\limits_{n=1}^{\infty}}x_{\alpha}\left(n\right)={\textstyle\sum\limits_{n=1}^{\infty}}\lim\limits_{\alpha}x_{\alpha}\left(n\right)={\textstyle\sum\limits_{n=1}^{\infty}}x\left(n\right).$

Proof. Recall that suprema commute in $E_{+}^{s}$ and that the limit of an
increasing net is its supremum. As $x_{\alpha}\left(n\right)$ is positive for
all $n,\alpha$ we have

$\displaystyle{\textstyle\sum\limits_{n=1}^{\infty}}x\left(n\right)$
$\displaystyle=\sup\limits_{N}{\textstyle\sum\limits_{n=1}^{N}}x\left(n\right)=\sup\limits_{N}\sup\limits_{\alpha}{\textstyle\sum\limits_{n=1}^{N}}x_{\alpha}\left(n\right)$
$\displaystyle=\sup\limits_{\alpha}\sup\limits_{N}{\textstyle\sum\limits_{n=1}^{N}}x_{\alpha}\left(n\right)=\sup\limits_{\alpha}{\textstyle\sum\limits_{n=1}^{\infty}}x_{\alpha}\left(n\right)$
$\displaystyle=\lim\limits_{\alpha}{\textstyle\sum\limits_{n=1}^{\infty}}x_{\alpha}\left(n\right),$

and we are done.

Next we state a Dominated Convergence Theorem for series in Riesz spaces.

###### Theorem 11

Let $\left(x_{\alpha}\left(n\right)\right)_{\alpha\in\Gamma}$ be a sequence of
nets in $E$, where $E$ is a Dedekind complete Riesz space. We assume that the
following conditions are satisfied.

  1. (i)

$x_{\alpha}\left(n\right)\overset{o}{\longrightarrow}x\left(n\right)$ for each
$n.$

  2. (ii)

$\left|x_{\alpha}\left(n\right)\right|\leq y\left(n\right)$ for every
$\alpha,$ with ${\textstyle\sum\limits_{n=1}^{\infty}}y\left(n\right)\in
E^{u}.$
Then
$\lim\limits_{\alpha}{\textstyle\sum\limits_{n=1}^{\infty}}x_{\alpha}\left(n\right)={\textstyle\sum\limits_{n=1}^{\infty}}\lim\limits_{\alpha}x_{\alpha}\left(n\right).$

Proof. It is sufficient to show that

$\limsup\limits_{\alpha}\left|{\textstyle\sum\limits_{n=1}^{\infty}}x_{\alpha}\left(n\right)-{\textstyle\sum\limits_{n=1}^{\infty}}x\left(n\right)\right|=0.$

Observe first that $\left|x\left(n\right)\right|\leq y\left(n\right),$ so the
series $\sum x\left(n\right)$ is order convergent. Fix an integer $N.$ Then it
is clear that

$\lim\limits_{\alpha}{\textstyle\sum\limits_{n=1}^{N}}\left|x_{\alpha}\left(n\right)-x\left(n\right)\right|=0.$

It is enough to prove that

$A:=\limsup\limits_{\alpha}\left|{\textstyle\sum\limits_{n=1}^{\infty}}x_{\alpha}\left(n\right)-{\textstyle\sum\limits_{n=1}^{\infty}}x\left(n\right)\right|=0.$

For this end observe that

$\displaystyle A$
$\displaystyle\leq\limsup\limits_{\alpha}{\textstyle\sum\limits_{n=1}^{\infty}}\left|x_{\alpha}\left(n\right)-x\left(n\right)\right|$
$\displaystyle\leq\limsup\limits_{\alpha}{\textstyle\sum\limits_{n=N+1}^{\infty}}\left|x_{\alpha}\left(n\right)-x\left(n\right)\right|+\limsup\limits_{\alpha}{\textstyle\sum\limits_{n=1}^{N}}\left|x_{\alpha}\left(n\right)-x\left(n\right)\right|$
$\displaystyle\leq
2{\textstyle\sum\limits_{n=N+1}^{\infty}}2\left|y\left(n\right)\right|.$

As this happens for every $N,$ we get $A=0$ as required.

We end this section by a version of Fubini Theorem in Riesz spaces. For a
family $\left(x_{i}\right)_{i\in J}$ of positive elements in $E$ we write:

${\textstyle\sum\limits_{i\in I}}x_{i}=\sup\limits_{F\text{finite }\subset
I}{\textstyle\sum\limits_{k\in F}}x_{k}\in E^{s}.$

The following properties are very similar to the real case and their proofs
will be omitted.

###### Theorem 12

Let $E$ be a Dedekind complete Riesz space and $\left(x_{i}\right)_{i\in I}$ a
family in $E_{+}^{s}.$ Then the following hold.

  1. 1.

If $I=\mathbb{N}$ then $\sum_{i\in
I}x_{i}=\lim\limits_{n\longrightarrow\infty}{\textstyle\sum\limits_{k=1}^{n}}x_{k}.$

  2. 2.

If $\left(J_{\alpha}\right)_{\alpha}is$ a partition of $I$ then
${\textstyle\sum\limits_{i\in
I}}x_{i}={\textstyle\sum\limits_{\alpha}}{\textstyle\sum\limits_{i\in
J_{\alpha}}}x_{i}.$ In particular we have

${\textstyle\sum\limits_{\left(i,j\right)\in\mathbb{N}^{2}}}x_{i,j}={\textstyle\sum\limits_{i\in\mathbb{N}}}\left({\textstyle\sum\limits_{j\in\mathbb{N}}}x_{i,j}\right)={\textstyle\sum\limits_{j\in\mathbb{N}}}{\textstyle\sum\limits_{i\in\mathbb{N}}}x_{i,j}.$
(2)

  3. 3.

In the last statement if the family is in $E$ then if
${\textstyle\sum\limits_{\left(i,j\right)\in\mathbb{N}^{2}}}\left|x_{i,j}\right|\in
E^{u}$ then the formula (2) holds.

##  4 Generating function

We assume throughout that $\left(E,e,T\right)$ is a conditional Riesz triple.

###### Definition 13

Let $x$ be a $T$-natural vector in $E.$

  1. (i)

The $T$-generating function of $x$, is the function
$g_{x}:[0,\infty)\longrightarrow E^{s}$ defined by

$g_{x}\left(s\right)={\textstyle\sum\limits_{n\geq
0}}s^{n}TP_{x=ne}e={\textstyle\sum\limits_{n\geq 0}}s^{n}\pi_{x}(n).$

  2. (ii)

The generalized $T$-generating function of $x$ is the function
$\widetilde{g}_{x}:E_{+}\longrightarrow E^{s}$ defined by

$\widetilde{g}_{x}\left(u\right)={\textstyle\sum\limits_{n\geq
0}}TP_{x=ne}u^{n}={\textstyle\sum\limits_{n\geq 0}}\pi_{x}(n)u^{n}.$

###### Example 14

We refer to Example 7 for the notations below.

  1. (i)

If $x\sim B(p)$ then $g_{x}(s)=sp+e-p.$

  2. (ii)

If $x\sim B(n,p)$ then $g_{x}(s)=\left(sp+e-p\right)^{n}.$

  3. (iii)

If $x\sim\mathcal{P}\left(p\right)$ then
$g_{x}\left(s\right)=\exp(\left(s-1\right)g).$

The following result collects basic properties of generating functions.

###### Proposition 15

Let $\left(E,e,T\right)$ be a conditional Riesz triple and let $x$ be a
natural element in $E.$ Then the following hold.

  1. (i)

$g_{x}(0)=\pi_{x}(0),$ $g_{x}(1)=e$ and $g_{x}(s)=\widetilde{g}_{x}(se);$

  2. (ii)

$0\leq g_{x}(s)\leq e,$ $0\leq s\leq 1.$ and $0\leq\widetilde{g}_{x}(u)\leq
e,$ $0\leq u\leq e.$

  3. (iii)

$g_{x}(s)=Ts^{x},$ for all $s\geq 0.$

  4. (iv)

$g_{x}$ and $\widetilde{g}_{x}$ are non-decreasing. Moreover, if $x>0$ then
$g_{x}$ and $\widetilde{g}_{x}$ are strictly increasing.

  5. (v)

$g_{x}$ and $\widetilde{g}_{x}$ are convex. Moreover, if $TP_{x\leq e}e<e$
then $g_{x}$ and $\widetilde{g}_{x}$ are strictly convex.

Proof. (i), (ii) and (iv) are obvious.

(iii) As $s^{ne}=s^{n}e,$ it follows from Lemma 4 that
$P_{x=ne}s^{n}e=P_{x=ne}s^{x}.$ Hence

$\displaystyle g_{x}\left(s\right)$
$\displaystyle={\textstyle\sum\limits_{n=1}^{\infty}}TP_{x=ne}s^{n}e={\textstyle\sum\limits_{n=1}^{\infty}}TP_{x=ne}s^{x}$
$\displaystyle=T\left[\left({\textstyle\sum\limits_{n=1}^{\infty}}P_{x=ne}\right)s^{x}\right]=Ts^{x}.$

(v) This follows from the fact that the map $s\longmapsto s^{n}$ is convex for
$n\geq 0$ and strictly convex for $n\geq 2$. Observe that the condition
$TP_{x\leq e}e<e$ means that $\pi_{x}\left(n\right)>0$ for some $n\geq 2.$

Chernoff Inequality is stated in general in terms of exponential generating
functions (see for instance [16]). We state a version of Chernoff Inequality
in Riesz spaces.

###### Theorem 16

(Chernoff Bounds) Let $\left(E,e,T\right)$ be a Riesz conditional triple, $x$
a natural element in $E$ and $u\in R\left(T\right).$ Then the following hold.

$TP_{x\geq u}e\leq s^{u}g_{x}(s),\ \ \forall s>1,\ $ $TP_{x\leq u}e\leq
s^{-u}g_{x}(s),\ \ \forall 0<s<1.$

Proof. Let $s>1.$ According to Lemma 2.(iii) we have

$P_{x\geq u}P_{x=ne}s^{u}=P_{ne\geq u}P_{x=ne}s^{u}e\leq P_{ne\geq
u}P_{x=ne}s^{n}e.$

It follows that

$\displaystyle P_{x\geq u}s^{u}e$
$\displaystyle=\sum\limits_{n=1}^{\infty}P_{x\geq
u}P_{x=ne}s^{u}e\leq\sum\limits_{n=0}^{\infty}P_{ne\geq u}P_{x=ne}s^{n}e$
$\displaystyle=P_{ne\geq u}\sum\limits_{n=0}^{\infty}P_{x=ne}s^{n}e=P_{ne\geq
u}x^{s}.$

Applying $T$ to both sides, we get the first inequality. The second one can be
proved in a similar way.

###### Corollary 17

Let $x$ be a natural vector in $E$ and $\alpha\in\mathbb{R}_{+}.$ Then the
following hold.

$TP_{x\geq\alpha e}e\leq\dfrac{g_{x}(s)}{s^{\alpha}},\ \ \forall s>1\ $
$TP_{x\leq\alpha e}e\leq\dfrac{g_{x}(s)}{s^{\alpha}},\ \ \forall 0<s<1.$

We will show now that the main result in [6] can be deduced from the previous
corollary.

###### Theorem 18

[6, Theorem 5.3]Let $\left(E,e,T\right)$ be Riesz conditional triple. Let
$\left(P_{j}\right)$ be a Benoulli process with $TP_{j}=f$ for all
$j\in\mathbb{N}$ for some $f\in E.$ Let
$S_{n}={\textstyle\sum\limits_{k=1}^{n}}P_{k}.$ Then

$TP_{\left(S_{n}-te\right)^{+}}e\leq\left(\dfrac{n\left\|f\right\|_{u}\exp
1}{t}\right)^{t}\exp\left(-nf\right).$

holds for all $t>n\left\|f\right\|_{e},$ where

$\left\|f\right\|_{e}:=\inf\left\\{\beta\in\mathbb{R}:\left|f\right|\leq\beta
e\right\\}.$

It should be noted that in [6] the authors assume that $e$ is a strong unit
for $E.$ This is the case if we work in the ideal generated by $e.$

Proof. As $S_{n}\sim_{T}B(n,f)$ we have

$g_{S_{n}}(s)=\left(sf+e-f\right)^{n}=\left(e+\left(s-1\right)f\right)^{n}.$

In the sequel we will use the inequality $\log\left(e+x\right)\leq x$ for all
$x\in E_{+}$ and the formula (see [11, Theorem 3.9])

$\exp\left[n\log x\right]=x^{n}$

true for any $x\geq\alpha e,$ where $\alpha\in\left(0,\infty\right).$ Note
also that $e$ is a multiplicative unit in the $f$-algebra $E_{e}.$ The real
$s=\dfrac{t}{n\left\|f\right\|_{e}}$ satisfies $s>1$ and we have then

$\displaystyle g_{S_{n}}(s)$
$\displaystyle=\left(e+\left(s-1\right)f\right)^{n}=\exp\left[n\log\left(e+\left(s-1\right)f\right)\right]$
$\displaystyle\leq\exp\left[n\left(s-1\right)f\right]=\exp\left[nsf\right]\exp\left[-nf\right]$
$\displaystyle\leq\exp\left(ns\left\|f\right\|_{e}e\right)\exp\left(-nf\right)=\exp\left(t\right)\exp\left(-nf\right).$

Now using Corollary 17 to derive that

$\displaystyle TP_{\left(S_{n}-te\right)^{+}}e$ $\displaystyle\leq
TP_{S_{n}\geq te}e\leq\dfrac{g_{S_{n}}(s)}{s^{t}}\leq\dfrac{\exp
t.\exp\left(-nf\right)}{s^{t}}$ $\displaystyle=\exp
t.\left(\frac{n\left\|f\right\|_{e}}{t}\right)^{t}\exp\left(-nf\right),$

which completes the proof.

###### Lemma 19

Let $\left(E,e,T\right)$ be a conditional Riesz triple and $x$ a natural
element in $E.$ Then the following hold.

  1. (i)

$g_{x}$ is continuous on $\left[0,1\right]$ and infinitely differentiable on
$[0,1).$ Moreover we have the formula

$g_{x}^{(k)}(s)={\textstyle\sum\limits_{n=k}^{\infty}}n(n-1)...(n-k+1)s^{n-k}\pi_{x}(n)$
(3)

for every integer $k\geq 0$ and for all $s\in[0,1).$

  2. (ii)

For each integer $k$, we have

$g_{x}^{(k)}(s)=T\left[x(x-e)...(x-(k-1)e)s^{x-ke}\right].$

In particular, $g_{x}^{(k)}(0)=k!\pi_{x}(k).$

It follows then from the above lemma that

$g_{x}(s)={\textstyle\sum\limits_{n=0}^{\infty}}s^{n}\dfrac{g_{x}^{(k)}(0)}{k!},\qquad
s\in[0,1).$

Proof. (i) As
$\left|s^{k}\pi_{x}\left(k\right)\right|\leq\pi_{x}\left(k\right)$ and
$\sum\pi_{x}\left(k\right)$ order converges, it follows from Theorem 11 that

$\lim\limits_{s\longrightarrow
s_{0}}\sum\limits_{n=0}^{\infty}s^{n}\pi_{x}\left(k\right)=\sum\limits_{n=0}^{\infty}\lim\limits_{s\longrightarrow
s_{0}}s^{n}\pi_{x}\left(k\right)=\sum\limits_{n=0}^{\infty}s_{0}^{n}\pi_{x}\left(k\right),$

which proves that $g_{x}$ is continuous on $\left[0,1\right].$ Let us now fix
a real $s_{0}\in[0,1)$ and a compact neighborhood $K$ of $s_{0}$ in $[0,1).$
We have for $k\in\mathbb{N}$ and $s\in K,$

$\left|\dfrac{s^{k}\pi_{x}\left(k\right)-s_{0}^{k}\pi_{x}\left(k\right)}{s-s_{0}}\right|\leq
kM^{k-1}\pi_{x}\left(k\right)\leq kM^{k-1}e,$

where $M=\sup\limits_{t\in K}\left|t\right|<1.$ Using Theorem 11 again we
derive that $g_{x}$ is differentiable on $[0,1)$ and that formula 3 holds for
$k=1$. We can then prove by induction on $k$ that $g_{x}$ is
$k$-differentiable and the formula 3 is valid.

(ii) Let $0\leq s<1.$ We have

$\displaystyle g_{x}^{(k)}(s)$
$\displaystyle={\textstyle\sum\limits_{n=k}^{\infty}}n(n-1)...(n-k+1)s^{n-k}\pi_{x}(n)$
$\displaystyle=Th_{k}(x)=T\left[x(x-e)...(x-(k-1)e)s^{x-ke}\right],$

where $h_{k}(u)=u(u-1)...(u-k+1)\mathbf{1}_{\left[k,\infty\right)}(u).$

The next lemma generalizes a well-known fact in Probability theory and will be
needed in the proof of Proposition 21 below.

###### Lemma 20

If $x$ is a natural element in $E$ then

$Tx={\textstyle\sum\limits_{n=1}^{\infty}}TP(x\geq ne)e.$

Proof. We have

$Tx={\textstyle\sum\limits_{n=1}^{\infty}}n\pi_{x}(n)={\textstyle\sum\limits_{n=1}^{\infty}}{\textstyle\sum\limits_{k=1}^{n}}\pi_{x}(n)={\textstyle\sum\limits_{k=1}^{\infty}}{\textstyle\sum\limits_{n=k}^{\infty}}\pi_{x}(n)={\textstyle\sum\limits_{k=1}^{\infty}}TP(x\geq
ke)e,$

which proves the lemma.

###### Proposition 21

Let $\left(E,e,T\right)$ be a conditional Riesz triple and let $x$ be a
natural element in $E.$ For each $n\in\mathbb{N}$ the following are
equivalent:

  1. (i)

$Tx^{n}\in E^{u}$ (i.e. $x\in L^{n}(T)$);

  2. (ii)

$g_{x}^{(n)}(1)$ exists in $E^{u}$ (in the sense that the left limit
$\lim_{s\uparrow 1}g_{x}^{(n)}(s)$ exists in $E^{u}$).
In either case, we have

$\displaystyle g_{x}^{(n)}(1)$
$\displaystyle=T\left[x(x-e)...(x-(n-1)e)\right]$
$\displaystyle={\textstyle\sum\limits_{k=n}^{\infty}}k(k-1)...(k-n+1)\pi_{x}(k).$

The quantities $Tx^{n},$ $n\geq 1,$ are called $T$-moments of $x,$ and the
quantities $Tx,Tx(x-e),Tx(x-e)(x-2e),...$are called $T$-factorials moments of
$\ x.$

Proof. Observe that

$\displaystyle\dfrac{g_{x}(s)-g_{x}(1)}{s-1}$
$\displaystyle=\dfrac{e-g_{x}(s)}{1-s}={\textstyle\sum\limits_{n=1}^{\infty}}\frac{1-s^{n}}{1-s}\pi_{x}(n)$
$\displaystyle={\textstyle\sum\limits_{n=1}^{\infty}}\left(1+s+...+s^{n-1}\right)\pi_{x}(n).$

So according to Monotone Convergence Theorem (Theorem 10) we have

$\displaystyle\lim\limits_{s\uparrow 1}\dfrac{g_{x}(s)-g_{x}(1)}{s-1}$
$\displaystyle={\textstyle\sum\limits_{n=1}^{\infty}}\lim\limits_{s\uparrow
1}\left(1+s+...+s^{n-1}\right)\pi_{x}(n)$
$\displaystyle={\textstyle\sum\limits_{n=1}^{\infty}}n\pi_{x}(n)=Tx.$

This proves the equivalence for $n=1.$

Assume now that the result is true for rank $n.$ Let us prove it for rank
$n+1.$ Assume first that $g^{\left(n+1\right)}\left(1\right)$ exists. Then

$\displaystyle g^{\left(n+1\right)}\left(1\right)$
$\displaystyle=\lim\limits_{s\uparrow
1}\dfrac{g_{x}^{\left(n\right)}(s)-g_{x}^{\left(n\right)}(1)}{s-1}=\lim\limits_{s\uparrow
1}{\textstyle\sum\limits_{k=n}^{\infty}}k\left(k-1\right)...\left(k-n+1\right)\dfrac{\left(1-s^{n-k}\right)}{1-s}\pi_{x}(k)$
$\displaystyle=\left(\ast\right){\textstyle\sum\limits_{k=n}^{\infty}}\lim\limits_{s\uparrow
1}k\left(k-1\right)...\left(k-n+1\right)\left(k-n\right)\pi_{x}(k)=T\left[x\left(x-e\right)...\left(x-ne\right)\right],$

where the equality $\left(\ast\right)$ follows from Theorem 10. Now it follows
from Lyapunov inequality that if $x\in L^{q}\left(T\right)$ then $x\in
L^{p}\left(T\right)$ for all $p\leq q.$ So as $x\in L^{n}\left(T\right)$ we
have $Tx^{n+1}\in E^{u}$ if and only if
$T\left[x\left(x-e\right)...\left(x-ne\right)\right]\in E^{u}.$This proves the
result.

Assume now that $Tx^{n+1}\in E^{u}$ then by the above remark, $Tx^{n}\in
E^{u}$ and then by the induction hypothesis,
$g^{\left(n\right)}\left(1\right)$ exists. Also
$T\left[x\left(x-e\right)...\left(x-ne\right)\right]\in E^{u},$ and then the
use of Theorem 10 gives again that

$g^{\left(n+1\right)}\left(1\right)=T\left[x\left(x-e\right)...\left(x-ne\right)\right]\in
E^{u}.$

The proof is now complete.

###### Corollary 22

Let $\left(E,e,T\right)$ be a conditional Riesz triple and $x$ a natural
element in $E.$ Then $Tx=g_{x}^{\prime}(1)$ and
$Tx^{2}=g_{x}^{\prime}(1)+g_{x}^{\prime\prime}(1)$ if $Tx^{2}\in E^{u}.$

##  5 Generating function and T-independence

###### Theorem 23

Let $\left(x_{k}\right)_{1\leq k\leq n}$ be a sequence of $T$-independent
natural elements in $E,$ and put
$S_{n}:={\textstyle\sum\limits_{i=1}^{n}}x_{i},$ then the $T$-generating
function of $S_{n}$ is given by

$g_{S_{n}}(s)=Ts^{S_{n}}=g_{x_{1}}(s)...g_{x_{n}}(s).$

Proof. It is enough to prove the result for the case $n=2.$ Let $x$ and $y$ be
two natural elements in $E$ that are $T$-independent. We have

$\displaystyle g_{x+y}(s)$ $\displaystyle={\textstyle\sum\limits_{n\geq
0}}s^{n}TP_{x+y=ne}e={\textstyle\sum\limits_{n\geq
0}}s^{n}{\textstyle\sum_{k=0}^{n}}TP_{y=ke}P_{x+y=ne}e$
$\displaystyle={\textstyle\sum\limits_{n\geq
0}}s^{n}{\textstyle\sum\limits_{k=0}^{n}}TP_{y=ke}P_{x=\left(n-k\right)e}e={\textstyle\sum\limits_{n\geq
0}}s^{n}{\textstyle\sum\limits_{i+j=n}}TP_{y=je}e.TP_{x=ie}e$
$\displaystyle=\left({\textstyle\sum\limits_{i\geq
0}}s^{i}TP_{x=ie}e\right).\left({\textstyle\sum\limits_{j\geq
0}}s^{j}TP_{y=je}e\right)=g_{x}(s).g_{y}(s).$

This proves that
$g_{x+y}\left(s\right)=g_{x}\left(s\right)g_{y}\left(s\right)$ as required.

Let $N,x_{0},x_{1},x_{2},...$ be natural elements in $E$ with $x_{0}=0.$ Put
$x_{N}:={\textstyle\sum\limits_{n\geq 1}}P_{N=ne}x_{n},$
$S_{n}:={\textstyle\sum\limits_{k=0}^{n}}x_{k}$ and
$S_{N}:={\textstyle\sum\limits_{n\geq 1}}P_{N=ne}S_{n}.$ It follows from these
definitions that $P_{N=ne}x_{N}=P_{N=ne}x_{n}$ and
$P_{N=ne}S_{N}=P_{N=ne}S_{n}.$

###### Lemma 24

For any $f\in L_{2}(x)$ the following statements hold.

  1. (i)

We have for every $n,k\in\mathbb{N}$

$P_{N=ne}f(x_{N})=P_{N=ne}f(x_{n}).$

In particular we have

$P_{N=ne}P_{x_{N}=ke}e=P_{N=ne}P_{x_{n}=ke}e.$

  2. (ii)

We have for every $k,n$ in $\mathbb{N},$

$P_{N=ne}f(S_{N})=P_{N=ne}f(S_{n})\ \text{and}\
P_{N=ne}P_{S_{N}=ke}e=P_{N=ne}P_{S_{n}=ke}e.$

Proof. (i) It suffices to show these equalities for
$f=\mathbf{1}_{\left(-\infty,a\right]},$ $a\in\mathbb{R}.$ By definition of
$x_{N}$ it follows that $P_{N=ne}x_{N}=P_{N=ne}x_{n}.$ Using this and noting
that $f(x)=P_{x\leq ae}e,$ we get

$\displaystyle P_{N=ne}f(x_{N})$
$\displaystyle=P_{N=ne}P_{\left(x_{N}-ae\right)^{+}}^{d}e=P_{N=ne}\left(I-P_{\left(x_{N}-ae\right)^{+}}\right)e$
$\displaystyle=P_{N=ne}-P_{P_{N=ne}\left(x_{N}-ae\right)^{+}}e.$

Now as

$\displaystyle P_{N=ne}\left(x_{N}-ae\right)^{+}$
$\displaystyle=\left(P_{N=ne}x_{N}-P_{N=ne}ae\right)^{+}$
$\displaystyle=\left(P_{N=ne}x_{n}-P_{N=ne}ae\right)^{+}=P_{N=ne}\left(x_{n}-ae\right)^{+}$

we deduce that

$P_{N=ne}f(x_{N})=P_{N=ne}-P_{P_{N=ne}\left(x_{n}-ae\right)^{+}}e=P_{N=ne}f(x_{n}).$

as required. To get the second equality we just apply the first one for
$f=\mathbf{1}_{\left\\{k\right\\}}.$

(ii) The proof is similar to (i).

###### Lemma 25

With the above notations we have:

  1. (i)

$x_{N}$ and $S_{N}$ are natural elements in $E.$

  2. (ii)

$S_{N}={\textstyle\sum\limits_{k\geq 1}}P_{N\geq ke}x_{k}.$

Proof. (i) We will show the requested property only for $x_{N};$ the case of
$S_{N}$ is similar. By Lemma 24, we get

$\displaystyle{\textstyle\sum\limits_{k\geq 0}}P_{x_{N}=ke}e$
$\displaystyle={\textstyle\sum\limits_{k\geq 0}}{\textstyle\sum\limits_{n\geq
0}}P_{N=ne}P_{x_{N}=ke}e={\textstyle\sum\limits_{k\geq
0}}{\textstyle\sum\limits_{n\geq 0}}P_{N=ne}P_{x_{n}=ke}e$
$\displaystyle={\textstyle\sum\limits_{n\geq 0}}P_{N=ne}{\textstyle\sum_{k\geq
0}}P_{x_{n}=ke}e={\textstyle\sum\limits_{n\geq 0}}P_{N=ne}e=e.$

This show that $x_{N}$ is natural.

(ii) By definition of $S_{N}$ we have

$\displaystyle S_{N}$ $\displaystyle={\textstyle\sum\limits_{n\geq
0}}P_{N=ne}S_{n}={\textstyle\sum\limits_{n\geq
0}}P_{N=ne}{\textstyle\sum\limits_{k=1}^{n}}x_{k}$
$\displaystyle={\textstyle\sum\limits_{k\geq 1}}\left({\textstyle\sum_{n\geq
k}}P_{N=ne}\right)x_{k}={\textstyle\sum\limits_{k\geq 1}}P_{N\geq ke}x_{k}.$

This gives the desired formula.

###### Theorem 26

Assume that $x=x_{1},x_{2},...$ are $T$-equidistributed and $T$-conditional
independent of $N.$ The generating function of $x_{N}$ is given by:

$g_{x_{N}}=T(I-P_{N=0})g_{x}+TP_{N=0}e.$

Moreover, if $TP_{N>0}e=e$ then $g_{x_{N}}=g_{x}.$

Proof. We have

$\displaystyle g_{x_{N}}(s)$ $\displaystyle={\textstyle\sum\limits_{n\geq
0}}s^{n}TP_{x_{N}=ne}e={\textstyle\sum\limits_{n\geq
0}}s^{n}{\textstyle\sum\limits_{k\geq 0}}TP_{N=ke}P_{x_{k}=ne}e$
$\displaystyle={\textstyle\sum\limits_{n\geq
0}}s^{n}TP_{N=0}P_{x_{0}=ne}e+{\textstyle\sum\limits_{n\geq
0}}s^{n}{\textstyle\sum\limits_{k\geq 1}}TP_{N=ke}.TP_{x_{k}=ne}e$
$\displaystyle=TP_{N=0}e+{\textstyle\sum\limits_{k\geq
1}}TP_{N=ke}.{\textstyle\sum\limits_{n\geq 0}}s^{n}TP_{x=ne}e.$

Now if we assume that $TP_{N>0}e=e,$ we get $TP_{N=0}e=0,$ and then
$g_{x_{N}}(s)=Tg_{x}(s)=g_{x}(s).$

###### Theorem 27

Let $\left(E,e,T\right)$ be a conditional Riesz triple. Assume that
$x_{1},x_{2},...$ are $T$-equidistributed and $T$-conditional independent of
$N.$ Then

  1. (i)

the generating function of $S_{N}$ is given by the following formula:

$g_{S_{N}}=\widetilde{g}_{N}\circ g_{x}.$

  2. (ii)

If $N$ and $x$ belong to $L^{1}\left(T\right),$ then
$TS_{N}=\left(TN\right).\left(Tx\right).$

  3. (iii)

If $N$ and $x$ belong to $L^{2}\left(T\right),$ then

$\operatorname*{Var}\nolimits_{T}\left(S_{N}\right)=\left(TN\right).\operatorname*{Var}\nolimits_{T}\left(x\right)+\operatorname*{Var}\nolimits_{T}\left(N\right).\left(Tx\right)^{2},$

where,

$\operatorname*{Var}\nolimits_{T}\left(y\right):=T\left(y-Ty\right)^{2}=Ty^{2}-\left(Ty\right)^{2}.$

Proof. (i) Using Lemma 24.(ii), we get

$\displaystyle g_{S_{N}}(s)$ $\displaystyle={\textstyle\sum\limits_{n\geq
0}}TP_{N=ne}s^{S_{n}}={\textstyle\sum\limits_{n\geq
0}}T\left(s^{S_{n}}.P_{N=ne}e\right)$
$\displaystyle={\textstyle\sum\limits_{n\geq
0}}Ts^{S_{n}}.TP_{N=ne}e={\textstyle\sum\limits_{n\geq
0}}\left(Ts^{x}\right)^{n}.TP_{N=ne}e=\widetilde{g}_{N}\left(g_{x}(s)\right).$

(ii) As $S_{n}$ and $N$ are $T$-independent we have for each $n,$
$TP_{N=ne}S_{n}=TS_{n}.TP_{N=ne}e.$ Hence

$\displaystyle TS_{N}$ $\displaystyle={\textstyle\sum\limits_{n\geq
1}}TP_{N=ne}S_{n}={\textstyle\sum\limits_{n\geq
1}}TS_{n}.TP_{N=ne}e={\textstyle\sum\limits_{n\geq 1}}nTx.TP_{N=ne}e$
$\displaystyle=Tx.{\textstyle\sum\limits_{n\geq 1}}nTP_{N=ne}e=Tx.TN.$

(iii) We have on one hand,

$\left(TS_{N}\right)^{2}=\left(Tx\right)^{2}.\left(TN\right)^{2}.$

and on the other hand we have

$\displaystyle T\left(S_{N}^{2}\right)$
$\displaystyle={\textstyle\sum\limits_{n\geq 1}}Tx_{n}^{2}.TP_{N\geq
ne}e+{\textstyle\sum\limits_{n\neq m\geq 1}}\left(Tx_{n}x_{m}\right).TP_{N\geq
ne}P_{N\geq me}e$ $\displaystyle=Tx^{2}.{\textstyle\sum\limits_{n\geq
1}}TP_{N\geq ne}e+\left(Tx\right)^{2}.{\textstyle\sum\limits_{n\neq m\geq
1}}TP_{N\geq ne}P_{N\geq me}e$
$\displaystyle=\left(Tx^{2}\right).\left(TN+Z\right).$

It is enough then to show that $Z:={\textstyle\sum\limits_{n\neq m\geq
1}}TP_{N\geq ne}P_{N\geq me}e=TN^{2}-TN.$ To this end observe that

$\displaystyle Z$ $\displaystyle={\textstyle\sum\limits_{k=n\vee m,n\neq
m}}TP_{N\geq ke}e={\textstyle\sum\limits_{m=2}^{\infty}}2(m-1)TP_{N\geq ke}e$
$\displaystyle=2{\textstyle\sum\limits_{k>1}}\left({\textstyle\sum\limits_{m=1}^{k-1}}m\right)P_{N=ke}e={\textstyle\sum\limits_{k=1}}\left(k^{2}-k\right)P_{N=ke}e=TN^{2}-TN.$

This completes the proof.

##  6 Application to convergence

It is easily seen that if $x$ and $y$ are natural elements in $E$ having the
same generating function then $x$ and $y$ have the same $T$-distribution,
namely $TP_{x=ke}e=TP_{y=ke}e$ for $k=0,1,...$ Several notions have been
generalized from the classical theory of probability and stochastic processes
to the setting of Riesz spaces. One of the challenges is to give a reasonable
definition of convergence in distribution. Many equivalent definitions are
known in the literature. In [18], David Williams defines the convergence in
distribution via distribution function: A sequence $\left(X_{n}\right)$ of
real random variables with distribution functions $\left(F_{X_{n}}\right)$ is
said to converge in distribution to $X$ if
$F_{X_{n}}\left(t\right)\longrightarrow F\left(t\right)$ for every
$t\in\operatorname*{Cont}\left(F\right),$ where $F_{X}$ is the distribution
function of $X$ and $\operatorname*{Cont}\left(F\right)$ is the set of points
on which $F$ is continuous. According to [3, Example 2.4], there is a Riesz
space $E$ and an element $x$ in $E$ such that $F_{x}$ is discontinuous
everywhere. So the definition above can not be translated to the theory of
Riesz spaces. We suggest, however, the following definition in the case of
natural elements.

###### Definition 28

Let $\left(E,e,T\right)$ be a conditional Riesz triple and
$\left(x_{\alpha}\right)$ a net of neutral elements in $E.$ We saythat the net
$\left(x_{\alpha}\right)_{\alpha\in A}$ converges in $T$-distribution to a
natural element $x$ in $E$ if
$TP_{x_{\alpha}=ke}e\overset{o}{\longrightarrow}TP_{x=ke}e$ for all $k\geq 0.$
We write $x_{\alpha}\overset{T-d}{\longrightarrow}x.$

###### Proposition 29

Let $\left(E,e,T\right)$ be a conditional Riesz triple,
$\left(x_{\alpha}\right)$ a net of natural elements in $E$ and $x\in E.$ Then
the following statements are equivalent.

  1. (i)

The net $\left(x_{\alpha}\right)$ converges to $x$ in $T$-distribution.

  2. (ii)

The net $\left(g_{x_{\alpha}}\right)$ converges to $g_{x}$ pointwise on
$\left[0,1\right]$.

Proof. Let us denote $TP_{x_{\alpha}=ke}e$ briefly by $p_{k,\alpha}$, and
$TP_{x=ke}e$ by $p_{k}.$

(i) $\Longrightarrow$ (ii) The conclusion is trivial for $s=1.$ Let $s$ be
fixed in $[0,1)$. We have for each $k,$ $\lim\limits_{\alpha}$
$s^{k}p_{k,\alpha}=s^{k}p_{k}$ and $\left|s^{k}p_{k,\alpha}\right|\leq
s^{k}e.$ Now as the series $\sum\limits_{k\geq 0}s^{k}e$ is order convergent
in $E_{e}$ we can apply Theorem 11 to get
$g_{x_{\alpha}}(s)\overset{o}{\longrightarrow}g_{x}(s).$

(ii) $\Longrightarrow$ (i) The proof will be done by induction on $k$. Observe
first that
$p_{0}^{(\alpha)}=g_{x_{\alpha}}\left(0\right)\overset{o}{\longrightarrow}g_{x}\left(0\right)=p_{0}$.
Assume by induction that $p_{i}^{(\alpha)}\overset{o}{\longrightarrow}p_{i}$
for all $i=0,1,...k$ and let $s\in\left(0,1\right).$ Define

$A_{n}^{\alpha}(s)=\left|p_{k+1}^{(\alpha)}-p_{k+1}+\sum\limits_{i\geq
k+2}\left(p_{i}^{(\alpha)}-p_{i}\right)s^{i-(k+1)}\right|.$

Then

$\displaystyle A_{n}^{\alpha}\left(s\right)$
$\displaystyle=\left|\frac{g_{x_{\alpha}}(s)-\sum_{i=0}^{k}p_{i}^{(\alpha)}s^{i}}{s^{k+1}}-\frac{g_{x}(s)-\sum_{i=0}^{k}p_{i}s^{i}}{s^{k+1}}\right|$
$\displaystyle\leq\left|\frac{g_{x_{\alpha}}(s)-g_{x}(s)}{s^{k+1}}\right|+\left|\frac{\sum_{i=0}^{k}p_{i}^{(\alpha)}s^{i}-\sum_{i=0}^{k}p_{i}s^{i}}{s^{k+1}}\right|\overset{o}{\longrightarrow}0.$

It follows that

$\limsup_{\alpha}\left|p_{k+1}^{(\alpha)}-p_{k+1}\right|\leq\left(\frac{s}{1-s}\right)e+\limsup_{\alpha}A_{n}^{\alpha}(s),$

for all $s\in\left(0,1\right).$ Hence
$p_{k+1}^{(\alpha)}\overset{o}{\longrightarrow}p_{k+1}$ as required and the
proof is complete.

###### Theorem 30

Suppose that $N\backsim_{T}\mathcal{P}(g)$, and that $x_{1},x_{2},...$ are
$T$-independent, identically distributed with Bernoulli distribution
$\mathcal{B}\left(p\right)$ and independent of $N$. Let
$S_{n}=\sum\limits_{i=1}^{n}x_{i}$. Then $S_{N}$ has the Poisson distribution
with parameter $pg$.

Proof. By Theorem 27,

$\displaystyle g_{S_{N}}(s)$
$\displaystyle=\widetilde{g}_{N}\left(g_{x_{1}}(s)\right)=\exp(\left(g_{x_{1}}(s)-e\right)g)$
$\displaystyle=\exp(\left(e-p+sp-e\right)g)=\exp(\left(-p+sp\right)g)$
$\displaystyle=\exp(-pg+spg)=\exp(-\left(1-s\right)pg).$

This shows that $S_{N}$ has the Poisson distribution with parameter $pg$ as
desired.

As the function $\exp$ is convex on $\mathbb{R}$ it follows from [11,
Corollary 4.3] that $\exp\left(x\right)$ is well defined in $E^{s}$ for all
$x\in E.$ Observe that if $x$ is positive then
$\exp\left(x\right)=\sup\limits_{n}v_{n},$ where
$v_{n}=\sum\limits_{k=0}^{n}\dfrac{x^{k}}{k!}.$ We will show that $\exp x$ is
in fact in $E^{u},$ the universal completion of $E.$ To this end it is
sufficient to prove that $\left(v_{n}\right)$ is unbounded order Cauchy (see
[10, Theorem 3.10] or [2, Theorem 17]). Now observe that for $p\leq q,$ and
$u\in E^{+},$

$\displaystyle\left|v_{p}-v_{q}\right|\wedge u$
$\displaystyle=P_{\left(x-te\right)^{+}}\left[\left(v_{q}-v_{p}\right)\wedge
u\right]+P_{\left(x-te\right)^{+}}^{d}\left[\left(v_{q}-v_{p}\right)\wedge
u\right]$ $\displaystyle\leq
P_{\left(x-te\right)^{+}}u+{\textstyle\sum\limits_{k=p+1}^{q}}\left(t^{k}/k!\right)e.$

Hence

$\limsup\limits_{p,q}\left[\left|v_{p}-v_{q}\right|\wedge u\right]\leq
P_{\left(x-te\right)^{+}}u.$

As this holds for every $t\geq 0$ we deduce that $\left(v_{n}\right)$ is
unbounded order Cauchy as required. It should be noted that if $x\in E_{e},$
the ideal generated by $e,$ then one can work in that ideal which is lattice
isomorphic to some $C\left(K\right)$-space (see for instance [13, Theorems 6.2
and 6.8]) and then use different tool which may seem easier. In both cases the
formula $\exp\left(x\right)\exp\left(-x\right)=e$ is valid. In the sequel we
extend a useful result verified by the exponentiation to the setting of Riesz
spaces.

###### Lemma 31

Let $E$ be a Dedekind complete Riesz space with weak order unit $e$ and
$\left(x_{n}\right)$ a sequence in $E$ such that $x_{n}^{n}\in E$ for each
$n.$ Then the following hold.

  1. (i)

If $x_{n}\overset{o}{\longrightarrow}0$ then
$x_{n}^{n}\overset{uo}{\longrightarrow}0$.

  2. (ii)

If $x_{n}\overset{o}{\longrightarrow}x$ in $E_{e}$ then
$\left(1+\dfrac{x_{n}}{n}\right)^{n\text{ \
}}\overset{o}{\longrightarrow}\exp(x)$ in $E^{u}.$

Proof. (i) As $e$ is a weak order unit it is sufficient to prove that
$x_{n}^{n}\wedge e\overset{o}{\longrightarrow}0$ [10, Corollary 3.5]. Let us
fix a real $t$ in $\left(0,1\right).$ As $\left(x_{n}\right)$ is order null,
$P_{\left(x_{n}-te\right)^{+}}e$ is order null as well. We derive that

$\displaystyle x_{n}^{n}\wedge e$ $\displaystyle=\left(x_{n}\wedge
e\right)^{n}=\left(P_{\left(x_{n}-e\right)^{+}}\left(x_{n}\wedge
e\right)+P_{\left(x_{n}-te\right)^{+}}^{d}\left(x_{n}\wedge
e\right)\right)^{n}$
$\displaystyle=P_{\left(x_{n}-te\right)^{+}}\left(x_{n}\wedge
e\right)^{n}+P_{\left(x_{n}-te\right)^{+}}^{d}\left(x_{n}\wedge e\right)^{n}$
$\displaystyle\leq
P_{\left(x_{n}-te\right)^{+}}e+P_{\left(x_{n}-te\right)^{+}}^{d}x_{n}^{n}\leq
P_{\left(x_{n}-te\right)^{+}}e+t^{n}e.$

This shows that $x_{n}^{n}\wedge e\overset{o}{\longrightarrow}0$ and then (i)
is proved.

(ii) Write
$\left(1+\dfrac{x_{n}}{n}\right)^{n}=\sum\limits_{k=0}^{\infty}f_{n}\left(k\right)$
where

$f_{n}\left(k\right)=\left\\{\begin{array}[c]{cc}\dbinom{n}{k}\dfrac{x_{n}^{k}}{n^{k}}&\text{if
}0\leq k\leq n\\\ 0&\text{otherwise}\end{array}\right..$

Let $M$ be an upper bound of $\left(\left|x_{n}\right|\right).$ Then it is
quite easy to see that $\left|f_{n}\left(k\right)\right|\leq\dfrac{M^{k}}{k!}$
and $f_{n}\left(k\right)\overset{o}{\longrightarrow}\dfrac{x^{k}}{k!}$ as
$n\longrightarrow\infty.$ It follows from Theorem 11 that
$\left(1+\dfrac{x_{n}}{n}\right)^{n}\overset{o}{\longrightarrow}e^{x}$ in
$E^{u}$ as required.

###### Remark 32

In (i) the order convergence can not be expected; it is only true in $E^{u}$
as order convergence and unbounded order convergence agree in $E^{u}$ for
sequences by a result of Grobler (see [10, Corollary 3.12]). The order
convergence of $\left(x_{n}\right)$ can not guarantee the order boundedness of
the sequence $\left(x_{n}^{n}\right).$ Take for instance $E=\ell^{\infty}$ and
$x_{n}=\left(1+1/\sqrt{n}\right)e_{n},$ where $\left(e_{n}\right)$ is the
standard basis. Then $x_{n}\overset{o}{\longrightarrow}0,$ however
$x_{n}^{n}\geq\sqrt{n}e_{n},$ hence $\left(x_{n}^{n}\right)$ is not order
bounded in $\ell^{\infty}.$

The following theorem generalizes a well-known result in probability theory
(see for instance, [17, Example 9.5.2]). A Riesz space version of this result
has been proved by Kuo, Vardy, and Watson [15, Theorem 5.6], which is a
particular case of our result.

###### Theorem 33

Let $\left(E,e,T\right)$ be a conditional Riesz triple and
$\left(y_{n}\right)$ a sequence of natural elements in $E.$ Assume that
$y_{n}$ has a Binomial $T$-distribution $\mathcal{B}(n,p_{n})$ such that
$np_{n}\overset{o}{\longrightarrow}g$ for some $g$ with $0<g\in E_{e}^{+}.$
Then $\left(y_{n}\right)$ converges in $T$-distribution to
$\mathcal{P}\left(g\right)$, that is,

$TP_{y_{n}=ke}e\overset{o}{\longrightarrow}\dfrac{1}{k!}g^{n}\exp\left(-g\right)$

for all integer $k.$

Proof. We have for all $s\in\left[0,1\right],$

$g_{y_{n}}(s)=\left(e-\left(1-s\right)p_{n}\right)^{n}=\left(e-\dfrac{\left(1-s\right)np_{n}}{n}\right)^{n},$

and by applying Lemma 31 to $x_{n}=\left(1-s\right)np_{n}$ and
$x=\left(1-s\right)g$ we get

$g_{y_{n}}(s)\overset{o}{\longrightarrow}\exp(\left(s-1\right)g),\ \text{for
all }0\leq s\leq 1,$

and the desired result follows.

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