An elementary construction of the Wiener measure
R. P. Pakshirajan* and M. Sreehari**

00footnotetext: *R. P. Pakshirajan, 227, 18th Main, 6th Block, Koramangala,
Bengaluru -560095, India.
Corresponding author:** M. Sreehari, 6-B, Vrundavan Park, New Sama Road,
Vadodara, 390024, India.
E-mail addresses<EMAIL_ADDRESS>(R. P. Pakshirajan);
<EMAIL_ADDRESS>(M. Sreehari)

Abstract.
Our construction of the Wiener measure on $\mathfrak{C}$ consists in first
defining a set function $\varphi$ on the class of all compact sets based on
certain $n$-dimensional normal distributions, $n=1,\ 2,\ldots$ using the
structural relation at (1) below. This structural relation, discovered by the
first author, is recorded in his book [2] on page 130. We then define a
measure $\mu$ on the Borel $\sigma$-field of subsets of C which is the Wiener
measure.

AMS Subject Classification 60J65; 60G15.
Keywords: Construction of Wiener measure, Brownian Motion Process, A
structural relation.

##  1\. Introduction and preliminary results

Let $\textbf{C}=\textbf{C}[0,\ 1]$ denote the space of real valued continuous
functions defined on $[0,\ 1]$, all vanishing at $0$ and endowed with the norm
$\Arrowvert x\Arrowvert=\sup\limits_{0\leq t\leq 1}|x(t)|,\ x\in\textbf{C}$.
Write $\rho(x,\ y)=\Arrowvert x-y\Arrowvert,\ x,\ y\in\textbf{C}$. Let
$\mathfrak{C}$ denote the Borel $\sigma$-field (i.e., the $\sigma$-field
generated by the open subsets) of C. For $t_{j}\in(0,\ 1],\ 1\leq j\leq k$,
define projection operators $\pi_{t_{1},\ t_{2},\ldots,\ t_{k}}$ thus. For
$x\in\textbf{C},\ \pi_{t_{1},\ t_{2},\ldots,\ t_{k}}(x)=\big{(}x(t_{1}),\
x(t_{2}),\ldots,\ x(t_{k})\big{)}$. This maps C into $R^{k}$. Assume $R^{k}$
is endowed with the usual metric and denote the resulting Borel $\sigma$-field
by $\mathfrak{R}^{k}$. We note $\pi_{t_{1},\ t_{2},\dots.,\ t_{k}}$ is a
continuous map and $\mathfrak{C}$ measurable.

Let $\nu_{t_{1},\ t_{2},\ldots,\ t_{k}}$ denote the measure generated on
$\mathfrak{R}^{k}$ by a $k$-variate normal distribution whose components have
zero means and the covariance between the $\text{j}^{\text{th}}$ and the
$\text{r}^{\text{th}}$ components is $\min\\{t_{j},\ t_{r}\\}$. We note the
variance covariance matrix is positive definite and hence the distribution is
non-singular. Denote the corresponding frequency function by $f_{t_{1},\
t_{2},\ldots.,\ t_{k}}$. Let $\alpha_{t_{1},\ t_{2},\ldots,\ t_{k}}$ denote
the measure generated on the sub $\sigma$-field $\pi^{-1}_{t_{1},\
t_{2},\ldots,\ t_{k}}(\mathfrak{R}^{k})$ by the mapping $\pi_{t_{1},\
t_{2},\ldots,\ t_{k}}$. Note $\alpha_{t_{1},t_{2},\ldots,t_{k}}(A)$ decreases
as $k$ increases for
$A\in\pi_{t_{1},t_{2},\ldots,t_{k}}^{-1}(\mathcal{R}^{k})$.

For the usual method of construction of the Brownian motion process / Wiener
measure, refer to chapter 2 in Karatzas and Shreve [1].

Let $T=\\{t_{1},t_{2},\ldots\\}$ be a countable dense subset of $[0,\ 1]$. Let
$K$ be a compact subset of C. Hence $\pi_{t_{1},t_{2},\ldots,t_{n}}(K)$ is a
compact subset of $R^{n}$. Then the following structural relation holds: (ref.
pp 130-131 in Pakshirajan [2])

$K=\bigcap_{n=1}^{\infty}\pi^{-1}_{t_{1},t_{2},\ldots,t_{n}}\pi_{t_{1},t_{2},\ldots,t_{n}}(K).$
(1)

For completeness we include proof of (1).
Since
$K\subset\pi^{-1}_{t_{1},t_{2},\ldots,t_{n}}\pi_{t_{1},t_{2},\ldots,t_{n}}(K)$,
it follows that $K$ is a subset of the set on the right side of (1).
Let now $x$ be an arbitrary member of the set on the right side of (1). Hence
for every $n$,
$x\in\pi^{-1}_{t_{1},t_{2},\ldots,t_{n}}\pi_{t_{1},t_{2},\ldots,t_{n}}(K)$.
Then there exists $y_{n}\in K$ such that $\pi_{t_{1},\ t_{2},\ ....,\
t_{n}}x=\pi_{t_{1},\ t_{2},\ ....,\ t_{n}}y_{n}$. Since $K$ is a compact set
the sequence $(y_{n})$ admits a convergent subsequence, say, ($y_{m}$)
converging to, say, $y_{0}\in K$ in the metric $\rho$. This implies
$y_{m}(t)\rightarrow y_{0}(t)$ for all $t\in[0,1]$. Fix $r$. Then for all
$m>r$, $x(t_{r})=y_{m}(t_{r})$. Letting $m\rightarrow\infty$ we observe
$x(t_{r})=y_{0}(t_{r})$. Since this is true for every $t_{r}\in T$, since $T$
is a dense subset of $[0,1]$ and since the functions are continuous, we get
$x\equiv y_{0}$. Thus $x\in K$ and the result follows.

######  Theorem 1.1.

$\alpha_{t_{1},\ t_{2},\ldots.,\ t_{n}}(\pi^{-1}_{t_{1},\ t_{2},\ldots,\
t_{n}}\pi_{t_{1},\ t_{2},\ldots,\ t_{n}}(K)),\ n=1,\ 2,\ldots.$ is a monotonic
decreasing sequence of numbers.

Proof.

$\alpha_{t_{1},t_{2},\ldots,t_{n+1}}\big{(}\pi^{-1}_{t_{1},t_{2},\ldots,t_{n+1}}\pi_{t_{1},t_{2},\ldots,t_{n+1}}(K)\big{)}$
$=\int_{\pi_{t_{1},t_{2},\ldots,t_{n+1}}(K)}f_{t_{1},t_{2},\ldots,t_{n+1}}(u_{1},u_{2},\ldots,u_{n+1})du_{1}du_{2}\dots,du_{n+1}$
$\leq\int_{\pi_{t_{1},t_{2},\ldots,t_{n}(K)\times(-\infty,\infty)}}f_{t_{1},t_{2},\ldots,t_{n+1}}(u_{1},u_{2},\ldots,\
u_{n+1})du_{1}du_{2}\ldots,du_{n+1}$
$\leq\int_{\pi_{t_{1},t_{2},\ldots,t_{n}}(K)}f_{t_{1},t_{2},\ldots,t_{n}}(u_{1},u_{2},\ldots,u_{n})du_{1}du_{2}\ldots
du_{n}$
$\leq\alpha_{t_{1},t_{2},\ldots,t_{n}}\big{(}\pi^{-1}_{t_{1},t_{2},\ldots,t_{n}}\pi_{t_{1},t_{2},\ldots,t_{n}}(K))$

completing the proof.
Define set function $\varphi$ on the compact sets $K$ of C :

$\varphi(K)=\lim_{n\rightarrow\infty}\alpha_{t_{1},\ t_{2},\ ....,\
t_{n}}\big{(}\pi^{-1}_{t_{1},\ t_{2},\ ....,\ t_{n}}\pi_{t_{1},\ t_{2},\
....,\ t_{n}}(K)\big{)}.$ (2)

Since $\pi_{t_{1},\ t_{2},\ ....,\ t_{n}}(\textbf{C})=R^{n}$ and since (1) is
true with $K$ replaced by C, we set $\varphi(\textbf{C})=1$.

We shall now discuss some properties of the set function $\varphi$.

######  Lemma 1.2.

Let $K_{1},\ K_{2}$ be two disjoint compact subsets of C. Then
$\varphi(K_{1}\cup K_{2})=\varphi(K_{1})+\varphi(K_{2}).$

Proof.
Since $K_{1}\cup K_{2}$ is a compact set

$\varphi(K_{1}\cup
K_{2})=\lim_{n\rightarrow\infty}\alpha_{t_{1},t_{2},\ldots,t_{n}}\big{(}\pi^{-1}_{t_{1},t_{2},\ldots,t_{n}}\pi_{t_{1},t_{2},\ldots,t_{n}}(K_{1}\cup
K_{2})\big{)}$
$=\lim_{n\rightarrow\infty}\alpha_{t_{1},t_{2},\dots,t_{n}}\big{(}\pi^{-1}_{t_{1},t_{2},\ldots,t_{n}}\\{\pi_{t_{1},t_{2},\dots,t_{n}}(K_{1})\cup\\\
\pi_{t_{1},t_{2},\dots,t_{n}}(K_{2})\\}\big{)}$
$=\lim_{n\rightarrow\infty}\Big{[}\alpha_{t_{1},t_{2},\dots,t_{n}}(\pi^{-1}_{t_{1},t_{2},\ldots,t_{n}}\\{\pi_{t_{1},t_{2},\ldots,t_{n}}(K_{1})\\})+$
$+\alpha_{t_{1},t_{2},\ldots,t_{n}}(\pi^{-1}_{t_{1},t_{2},\ldots,t_{n}}\\{\pi_{t_{1},t_{2},\ldots,t_{n}}(K_{2})\\})\Big{]}$

since the events are mutually exclusive and since
$\alpha_{t_{1},t_{2},\ldots,t_{n}}$ is a measure. Then

$\varphi(K_{1}\cup
K_{2})=\lim_{n\rightarrow\infty}\alpha_{t_{1},t_{2},\ldots,t_{n}}(\pi^{-1}_{t_{1},t_{2},\ldots,t_{n}}\\{\pi_{t_{1},t_{2},\ldots,t_{n}}(K_{1})\\})+$
$+\lim_{n\rightarrow\infty}\alpha_{t_{1},t_{2},\ldots,t_{n}}(\pi^{-1}_{t_{1},t_{2},\ldots,t_{n}}\\{\pi_{t_{1},t_{2},\ldots,t_{n}}(K_{2})\\})$

since the individual limits exist. We thus have $\varphi(K_{1}\cup
K_{2})=\varphi(K_{1})+\varphi(K_{2}).$

Remark 1. We have the following observations from the earlier discussion:
a) $\varphi$ is finitely additive on the collection of compact sets.
b) $\varphi(K)\geq 0$ for all compact sets $K$.
c) If $K_{1},K_{2}$ are compact sets and $K_{1}\subset K_{2}$ then from (2),
$\varphi(K_{1})\leq\varphi(K_{2})$.
We now discuss some limiting properties of $\varphi(K_{n})$.

Definition. We call a set, in a topological space, a boundary set if it is a
closed set with a null interior. The boundary of a set $A$ (i.e.,
$\overline{A}\sim\textit{Int}\;A$ ) will be denoted by $\partial A$.
We note that the boundary of a set is a boundary set.

######  Lemma 1.3.

Let $f$ be a continuous map of a topological space $X$ into a topological
space $Y$. If $A\subset X$ is a boundary set and if $f(A)$ is a closed set,
then $f(A)$ is a boundary set in $Y$.

Proof.
If $f(A)$ contains an open set $G$, then it is clear that $f^{-1}(G)$ is an
open subset of $A$, which is not possible since it is a boundary set.

######  Theorem 1.4.

(i) If $E\subset\textbf{C}$ is a compact boundary set, then $\varphi(E)=0$.
(ii) Let $K_{n},\ n\geq 1$ be compact boundary sets and let $K_{n}\rightarrow
K$ where $K$ is a compact set. Then $\varphi(K)=0$.
(iii) Let $K_{n},n\geq 1$, be compact sets such that $K_{n}\downarrow K$. Then
$\varphi(K_{n})\downarrow\varphi(K).$

Proof.
(i) The compact set $\pi_{t_{1},t_{2},\ldots,t_{r}}E$ is, by Lemma 1.3, a
boundary set in $R^{r}$. Hence its Lebesgue measure is zero. Since the measure
generated by the multivariate normal distribution is absolutely continuous
with respect to the Lebesgue measure, it follows that
$\alpha_{t_{1},t_{2},\ldots,t_{r}}(\pi^{-1}_{t_{1},t_{2},\ldots,t_{r}}\pi_{t_{1},t_{2},\ldots,t_{r}}E)=0$.
Since this is true for every $r\geq 1$, we conclude $\varphi(E)=0$.
(ii) We note that, by part (i), $\varphi(K_{n})=0,\ n\geq 1$. Further for all
$r\geq 1,\ n\geq 1,\
\alpha_{t_{1},t_{2},\ldots,t_{r}}(\pi^{-1}_{t_{1},t_{2},\ldots,t_{r}}\pi_{t_{1},t_{2},\ldots,t_{r}}K_{n})=0$
by an appeal to the Lemma 1.3 and arguing as in Part (i). If possible let
$\varphi(K)>0$. Hence for all $r\geq 1$,
$\alpha_{t_{1},t_{2},\ldots,t_{r}}(\pi^{-1}_{t_{1},t_{2},\ldots,t_{r}}\pi_{t_{1},t_{2},\ldots,t_{r}}K)>0$.
Since the convergence
$\alpha_{t_{1},\ldots,t_{r}}(\pi^{-1}_{t_{1},t_{2},\ldots,t_{r}}\pi_{t_{1},t_{2},\ldots,t_{r}}K_{n})\rightarrow\alpha_{t_{1},t_{2},\ldots,t_{r}}(\pi^{-1}_{t_{1},t_{2},\ldots,t_{r}}\pi_{t_{1},t_{2},\ldots,t_{r}}K)$
holds and since this leads to the absurd result that a sequence of zeros
converges to a positive number, it follows that $\varphi(K)=0$.
(iii) Since the sequence $(\varphi(K_{n}))$ is monotonic decreasing, it is
enough to show that, given $\varepsilon>0$, there exists $K_{N}$ such that
$\varphi(K_{N})<\varphi(K)+\varepsilon$.
We note $K$ is compact. Hence given $\varepsilon>0$, we can find $r\geq 1$
such that

$\varphi(K)>\alpha_{t_{1},t_{2},\ldots,t_{\ell}}\big{(}\pi^{-1}_{t_{1},t_{2},\ldots,t_{\ell}}\pi_{t_{1},t_{2},\ldots,t_{\ell}}(K)\big{)}-\varepsilon\
$ (3)

for all $\ell\geq r.$ Since $K_{n}\downarrow K$, for all $\ell\geq 1$ we have

$\pi^{-1}_{t_{1},t_{2},\ldots,t_{\ell}}\pi_{t_{1},t_{2},\ldots,t_{\ell}}(K_{n})\downarrow\pi^{-1}_{t_{1},t_{2},\ldots,t_{\ell}}\pi_{t_{1},t_{2},\ldots,t_{\ell}}(K).$

For fixed $\ell$ we then have, as $n\rightarrow\infty$

$\alpha_{t_{1},t_{2},\ldots,t_{\ell}}\big{(}\pi^{-1}_{t_{1},t_{2},\ldots,t_{\ell}}\pi_{t_{1},t_{2},\ldots,t_{\ell}}(K_{n})\big{)}\downarrow\alpha_{t_{1},t_{2},\ldots,t_{\ell}}\big{(}\pi^{-1}_{t_{1},t_{2},\ldots,t_{\ell}}\pi_{t_{1},t_{2},\ldots,t_{\ell}}(K)\big{)}.$

Take $\ell=r$. We can find $N=N(r)$ large such that

$\alpha_{t_{1},t_{2},\ldots,t_{r}}\big{(}\pi^{-1}_{t_{1},t_{2},\ldots,t_{r}}\pi_{t_{1},t_{2},\ldots,t_{r}}(K_{N})\big{)}<\alpha_{t_{1},t_{2},\ldots,t_{r}}\big{(}\pi^{-1}_{t_{1},\
t_{2},\ldots,t_{r}}\pi_{t_{1},t_{2},\ldots,t_{r}}(K)\big{)}+\varepsilon.$

This, together with (3), yields

$\varphi(K)+\varepsilon>\alpha_{t_{1},t_{2},\ldots,t_{r}}\big{(}\pi^{-1}_{t_{1},t_{2},\ldots,t_{r}}\pi_{t_{1},t_{2},\ldots,t_{r}}(K_{N})\big{)}-\varepsilon>\varphi(K_{N})-\varepsilon.$

Since $\varepsilon>0$ is arbitrary, the claim follows.

##  2\. The Wiener measure

In this Section we introduce a new set function in terms of $\varphi$ on the
Borel $\sigma$-field $\mathfrak{C}$ and study its properties to show that it
is indeed the Wiener measure.
For arbitrary measurable sets $A\in\mathfrak{C}$ define

$\mu(A)=\sup\limits_{K\subset A,\ K\ \text{compact}}\varphi(K).$ (4)

At the outset we observe that for compact sets $K$, $\mu(K)=\varphi(K)$ and
hence all the properties noted in the previous Section for $\varphi$ also hold
for $\mu$. Further the definition implies (i) that if $A\subset
B,A,B\in\mathfrak{C}\ \text{then}\ \mu(A)\leq\mu(B)$ and (ii) that there
exists an increasing sequence ($K_{n}$) of compact sets, $K_{n}\subset A$ such
that $\mu(A)=\lim\limits_{n\rightarrow\infty}\mu(K_{n})$.
The sets $K_{n}$ can be chosen to be monotonic increasing.
Remark 2. This does not mean that $K_{n}\uparrow A$. i.e.,
$\bigcup\limits_{n=1}^{\infty}K_{n}$ can be a proper subset of $A$. To see
this, take $v\in\textbf{C},\|v\|=1.$ Let $K_{n}=\\{\lambda
v,\;0\leq\lambda\leq 1-\frac{1}{n}\\}$ and $A=\\{\lambda v,\;0\leq\lambda\leq
1\\}.$ However, if $K_{n}=\\{\lambda v,\;0\leq\lambda\leq
1-\frac{1}{n}\\}\cup\\{v\\}$ then both $K_{n}$ and $A$ are compact and
$K_{n}\uparrow A$.

We next discuss further properties of $\mu$ that enable us to claim that $\mu$
is indeed a probability measure.

######  Lemma 2.1.

Let $A,\ B\in\mathfrak{C},\ A\cap B=\emptyset$. Then $\mu(A\cup
B)=\mu(A)+\mu(B)$.

Proof.
Let $E\subset A,\ F\subset B$ be compact sets. We have, from Lemma 1.2
$\mu(E)+\mu(F)=\mu(E\cup F)\leq\mu(A\cup B)$.
Thus $\mu(A\cup B)\geq\mu(A)+\mu(B)$. It remains to be shown that $\mu(A\cup
B)\leq\mu(A)+\mu(B).$
Given $\varepsilon>0$, we can find a compact set $K,\;\ K\subset A\cup B$ such
that
$\mu(A\cup\ B)-\varepsilon<\mu(K)$.
Case 1. The distance $d(A,\ B)=q>0$.
Consider an arbitrary sequence $(x_{n})\ \text{in}\ K\cap A$. Since it is a
sequence in $K$, it contains a convergent subsequence, converging to, say,
$x_{0}$. This $x_{0}$ has to be in $K\cap A$ or in $K\cap B$. Since the
sequence lies in $K\cap A$ and since $d(K\cap A,\ K\cap B)\geq q>0$, we
conclude $x_{0}\in K\cap A$. Thus we see that every sequence in $K\cap A$
contains a convergent subsequence converging to a point in $K\cap A$. This
means $K\cap A$ is a compact set. Similarly, $K\cap B$ is a compact set.
Summarising, we conclude that every compact subset of $A\cup B$ is the union
of a compact subset $E$ of $A$ and a compact subset $F$ of $B$. We get
$\mu(A\cup\ B)-\varepsilon<\mu(K)=\mu(E\cup
F)=\mu(E)+\mu(F)\leq\mu(A)+\mu(B)$. That $\mu(A\cup B)\leq\mu(A)+\mu(B)$ is
now immediate.
Case 2. $d(A,\ B)=0$.
This case assumption implies that $Q=\bar{A}\cap\bar{B}\neq\emptyset$. Again
in this case one or both the sets $K\cap A,\ K\cap B$ can fail to be compact.
Since the other case admits to being similarly argued, let us assume that
neither of the two sets is compact. $K\subset A\cup B$ can not be compact if
any convergent sequence in it converges to a point outside $K$. i.e., if
convergent sequences in $K\cap A$ or in $K\cap B$ converge to points outside
these sets. Thus $K$ can be a compact subset only if $E=K\cap A$ and $F=K\cap
B$ are compact. And the arguments and the conclusion in case 1 hold.
With this the proof is complete.
Remark 3.
Immediate consequences of Lemma 2.1 are :
a) If $A_{k},\ 1\leq k\leq n$ is a collection of $n$ mutually exclusive sets,
then $\mu(\bigcup\limits_{k=1}^{n}A_{k})=\sum\limits_{k=1}^{n}\mu(A_{k})$ and
b) If $A,\ B\in\mathfrak{C}$ and if $A\subset B$, then $\mu(B\sim
A)=\mu(B)-\mu(A)$.

######  Theorem 2.2.

(i) Let $K_{n}\uparrow,\;K_{n}$ be a compact boundary set for $n\geq 1$. If
$A=\bigcup_{n=1}^{\infty}K_{n}$, then $\mu(A)=0.$
(ii) Suppose $K,K_{n},n\geq 1$ are compact subsets with $K_{n}\uparrow K$.
Then $\mu(K_{n})\uparrow\mu(K)$.
(iii) Let $K_{n}\uparrow,\ K_{n}$ be compact. Let
$A=\bigcup_{n=1}^{\infty}K_{n}$. Then
$\mu(A)=\lim_{n\rightarrow\infty}\mu(K_{n})$.

Proof.
(i) If $\mu(A)>0$ then given $\varepsilon>0$ we can find a compact set
$K\subset A$ such that $\mu(A)<\mu(K)+\varepsilon.$ We note $\mu(K\bigcap
K_{n})=0$ and $K\bigcap K_{n}\uparrow K$. Now appeal to Theorem 1.4 (ii) and
conclude $\mu(K)=0.$ This implies $\mu(A)<\varepsilon.$ Since $\varepsilon$ is
arbitrary it follows $\mu(A)=0.$
(ii) We are to prove $\mu(K)-\mu(K_{n})\rightarrow 0$. Write $A_{n}=K\sim
K_{n}\downarrow\emptyset$. Interior $B_{n}$ of $A_{n}$ is $(IntK)\cap(K\sim
K_{n})$, since $K_{n}^{\prime}$ is an open set. We note that
$\bar{A}_{n}=\bar{B}_{n}=A_{n}\cup C_{n}$ where $C_{n}=\partial K_{n}$.
We note that $A_{n}\cap C_{n}=\emptyset$, that $\bar{A}_{n}$ is a compact set,
that the ($C_{n}$) is a sequence of disjoint compact boundary sets and that
$(\bar{A}_{n})$ is a monotonic decreasing sequence, the last assertion being
true since $A_{n}\downarrow$. We will show $\mu(\bar{A}_{n})\rightarrow 0$ and
that will complete the proof of this part.
Now, if $A=\bigcap\limits_{n=1}^{\infty}\bar{A}_{n}$, then by Theorem
1.4(iii), $\lim_{n\rightarrow\infty}\mu(\bar{A}_{n})=\mu(A)$.
Let $D=\bigcup\limits_{n=1}^{\infty}C_{n}$. We note that
$\bar{A}_{n}=A_{n}\cup C_{n}\subset A_{n}\cup D$. Hence
$A\subset(\bigcap\limits_{n=1}^{\infty}A_{n})\cup D=D$, leading to
$\mu(A)\leq\mu(D)$.
We note that for each $n,\ \mu(C_{n})=0$ since $C_{n}$ is a compact boundary
set. (ref. Theorem 1.4, part (i) ) If $D_{n}=\bigcup_{j=1}^{n}C_{j}$ then, by
part (a) of Remark 1, $\mu(D_{n})=\sum\limits_{j=1}^{n}\mu(C_{j})=0$. $D_{n}$,
being the finite union of disjoint compact boundary sets, is a compact
boundary set. Since $D_{n}\uparrow D$, it follows by part(i) above that
$\mu(D)=0$. Thus $\mu(A)=0$. With this the proof of (ii) is complete.
(iii) The hypothesis implies

$\lim_{n\rightarrow\infty}\mu(K_{n})\leq\mu(A).$ (5)

If $A$ is a compact set, then the claim follows from part (ii). If $A$ is not
a compact set, then, given $\varepsilon$, a compact set $K\subset A$ can be
found such that $\mu(K)>\mu(A)-\varepsilon$. Now, by part (ii),
$\lim\limits_{n\rightarrow\infty}\mu(K\cap K_{n})=\mu(K)>\mu(A)-\varepsilon$.
Hence
$\lim_{n\rightarrow\infty}\mu(K_{n})\geq\lim_{n\rightarrow\infty}\mu(K\cap
K_{n})\geq\mu(A)-\varepsilon$
This, together with (5) and use of the fact that $\varepsilon>0$ is arbitrary
completes the proof of (iii).
Remark 4
The result (ii) above is for compact sets but we could not get it in Section 1
because its proof needs the result in (i).
We now extend some of the above results (proved for compact sets) to
measurable sets.

######  Theorem 2.3.

(i) Let $A_{n}\in\mathfrak{C},\ n\geq 1;\ K$ be compact. Let $A_{n}\uparrow
K$. Then $\mu(A_{n})\uparrow\mu(K)$.
(ii) Let $A_{n},\ A\in\mathfrak{C}$, with $A_{n}\uparrow A$. Then
$\mu(A_{n})\uparrow\mu(A)$.

Proof.

(i) The hypothesis $(A_{n}\uparrow K)\ \Longleftrightarrow(E_{n}=K\sim
A_{n}\downarrow\emptyset)$. Note that $\mu(A_{n})\uparrow.$ We note
$\bar{E}_{n}\subset K$ is a compact set for each $n$ and
$\bar{E}_{n}\downarrow$. Now,

$\bigcap_{n=1}^{\infty}\bar{E}_{n}=\bigcap_{n=1}^{\infty}\big{(}E_{n}\cup\partial
E_{n}\big{)}\subset\bigcap_{n=1}^{\infty}(E_{n}\cup D)=D$

where $D=\bigcup_{n=1}^{\infty}\partial E_{n}$ since
$\bigcap\limits_{n=1}^{\infty}E_{n}=\emptyset$. Theorem 1.4 (iii) applies and
we get
$\lim_{n\rightarrow\infty}\mu(\bar{E}_{n})=\mu\big{(}\bigcap_{n=1}^{\infty}\bar{E}_{n}\big{)}\leq\mu(D)$.
Since $D$ is the union of a countably many disjoint compact boundary sets, it
follows from Theorem 1.4(ii) that $\mu(D)=0$. Thus
$\lim_{n\rightarrow\infty}\mu(\bar{E}_{n})=0$. Since $\partial E_{n}$ is a
boundary set, its $\mu$-measure is zero. Hence $\mu(E_{n})\rightarrow 0$.
i.e., $\mu(K)-\mu(A_{n})\rightarrow 0$, as was to be shown.
(ii) Given $\varepsilon>0$ find compact set $K\subset A$ such that
$\mu(A)-\varepsilon<\mu(K)$. Since $K\cap A_{n}\uparrow K$, part (i) applies
and we get

$\mu(K)=\lim_{n\rightarrow\infty}\mu(K\cap
A_{n})\leq\lim_{n\rightarrow\infty}\mu(A_{n}).$

Thus $\lim_{n\rightarrow\infty}\mu(A_{n})\geq\mu(K)\geq\mu(A)-\varepsilon$,
leading to $\lim_{n\rightarrow\infty}\mu(A_{n})\geq\mu(A)$. This together with
the obvious inequality that $\lim_{n\rightarrow\infty}\mu(A_{n})\leq\mu(A)$
completes the proof.

######  Theorem 2.4.

$\mu$ defined as above at (4) on $\mathfrak{C}$ is a probability measure.

Proof.
Let $A_{n}\in\mathfrak{C},\ n\geq 1$ be a sequence of mutually exclusive
events. Let $A=\bigcup_{n=1}^{\infty}A_{n}=\bigcup_{n=1}^{\infty}B_{n}$ where
$B_{n}=\bigcup_{k=1}^{n}A_{k}$. Since $B_{n}\uparrow A$, Theorem 2.3 (i)
applies and then we have by the observations following (4)

$\mu(A)=\lim_{n\rightarrow\infty}\mu(B_{n})=\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\mu(A_{k})=\sum\limits_{k=1}^{\infty}\mu(A_{k}).$

i.e., $\mu$ on $\mathfrak{C}$ is countably additive. Since $\mu(A)\geq 0$ for
$A\in\mathfrak{C}$ and $\mu(\textbf{C})=1$ it follows that $\mu$ is a
probability measure.
Remark 5.
(i) Let $\\{u_{1},u_{2},\ldots,u_{p}\\}\subset[0,1]$ be arbitrary. We can find
$\\{t_{1,n},t_{2,n},\ldots,t_{p,n}\\}\subset T$ such that $t_{j,n}\rightarrow
u_{j},\;1\leq j\leq p$. Then we will have
$\pi_{t_{1,n},t_{2,n},\ldots,t_{p,n}}\;x\rightarrow\pi_{u_{1},u_{2},\ldots,u_{p}}\;x$
for every $x\in\bf{C}$. Hence the random vector
$\pi_{t_{1,n},t_{2,n},\ldots,t_{p,n}}\;\stackrel{{\scriptstyle
D}}{{\rightarrow}}\pi_{u_{1},u_{2},\ldots,u_{p}}\;$. Using characteristic
functions, we note $\pi_{u_{1},u_{2},\ldots,u_{p}}$ has, under $\mu$, a multi-
variate normal distribution with
$cov(\pi_{u_{j}},\pi_{u_{r}})=\min\\{u_{j},u_{r}\\}$. It follows now that
$\mu$ is the Wiener probability measure.
(ii) The co-ordinate process $\\{\pi_{t},\ t\geq 0\\}$ is known as the
Brownian motion process.

References

1\. Karatzas, I. Shreve, S. E.: Brownian motion and stochastic calculus,
Springer (1988).
2\. Pakshirajan, R. P.: Probability Theory (A foundational course), Hindustan
Book Agency, New Delhi (2013).


#  A Stress-Strength Reliability Model using Exponential-Gamma$(3,\lambda)$
Distribution

Beenu Thomas  V. M. Chacko
St. Thomas’ College (Autonomous)Thrissur, Kerala, India

###### Abstract

One of the important problem in reliability analysis is computation of stress-
strength reliability. But it is impractical to compute it in certain
situations. So the estimation stay as an alternative solution to get an
approximate value of the reliability. There are research papers which deals
with stress-strength reliability analysis using statistical distributions. In
this paper, a stress-strength reliability model for exponential-
gamma$(3,\lambda)$ distribution is introduced. The maximum likelihood
estimator (MLE) for the model parameters is derived. Asymptotic distribution
and confidence interval for the maximum likelihood estimates of stress-
strength reliability, $R=P(X>Y)$, are given. The numerical illustration is
performed using Monte Carlo simulations. The results are analyzed with real
data analysis.

###### keywords:

Stress-strength reliability, Exponential-Gamma distribution , Statistical
inference , Maximum likelihood estimators

##  1 Introduction

The strength of a unit can be treated as a random variable. Due to
uncertainty, the unit’s stress should also be treated as a random variable in
its operating context. Let X represent a unit’s strength and Y represent the
random stress that the operational environment imposes on the unit. $R=P(X>Y)$
is the formula for defining a unit’s stress-strength reliability (R). A
substantial amount of literature exists regarding the problems associated with
the estimation of reliability for single-component stress-strength models. The
Exponential-Gamma$(3,\lambda)$ distribution has been proposed by Thomas and
Chacko (2020) in conjunction with a bathtub-shaped failure rate function. In
this paper, we denote the Exponential-Gamma Distribution$(3,\lambda)$ by
$EGD(3,\lambda)$. Specifically, $EGD(3,\lambda)$ has the PDF

$f(x)=\frac{\lambda^{2}}{1+\lambda}(1+\frac{\lambda}{2}x^{2})e^{-\lambda
x},x>0,\lambda>0.$ (1)

It should be noted that $EGD(3,\lambda)$ is a mixture of exponential
distribution with a scale parameter of $\lambda$ and gamma distribution with a
shape parameter of 3 and a scale parameter of $\lambda$ with mixing proportion
$\frac{\lambda}{1+\lambda}$. Consider two independent random variables X and Y
from the $EGD(3,\lambda)$ with different parameters $\lambda_{1}$ and
$\lambda_{2}$. This paper focuses on the estimation of the parameter
$R=P(X<Y)$. Typically, the problem of estimating R arises when dealing with
the reliability of a component of strength X subject to a load or stress Y.
The component will fail if $X<Y$. As a result, R can be viewed as a measure of
component reliability. There has been a long history of stress-strength (SS)
reliability, beginning with the pioneering work of Birnbaum (1956) and
Birnbaum and McCarty (1958). Church and Harris (1970) are credited with
introducing the term stress-strength. Kotz et al. (2003) provide an excellent
overview of the various stress-strength models up to 2001. Several
publications on the stress-strength model have been published recently,
including Gupta and Brown (2001), Raqab and Kundu (2005), Kundu and Gupta
(2005, 2006), Krishnamoorthy et al. (2007), Kundu and Raqab (2009), Sharma et
al.(2015) and their references. Based on a bivariate Pareto model, Hanagal
(1997) calculated the maximum likelihood estimate (MLE) of the SS parameter R.
The estimation of R for three-parameter generalized exponential distributions
was investigated by Raqaab et al. (2008). Balakrishnan and Lai (2009)
estimated R in models with correlated stress and strength. A study by Al-
Mutairi et al. (2013) examined R-estimates for Lindley distributions. The
estimation of reliability $R=P(Y<X)$ where X and Y are independent random
variables that follow the Kumaraswamy distribution with varying parameters was
discussed by Nadar et al. (2014). Ghitany et al. (2015) discussed the
reliability of SS systems based on power Lindley distributions. For a
transmuted Rayleigh distribution, Dey et al. (2017) calculated the SS
reliability parameter R. Deepthi and Chacko (2020) discussed the single-
component SS reliability procedure and multi-component SS reliability
estimation for the three-parameter generalized Lindley distribution. Varghese
and Chacko (2022) examined the reliability of the SS model for the Akash
distribution. Our goal in this article is to estimate the reliability of SS
reliability when both stress and strength follow EGD with different parameters
$\lambda_{1}$ and $\lambda_{2}$. Section 2 considers the SS reliability of
$EGD(3,\lambda)$. In section 3, the MLE of R, asymptotic distribution, and CI
for the MLE of R are obtained. Section 4 illustrates the extensive simulation
study. In section 5, the results are presented with real data analysis. In the
final section, conclusions are given.

##  2 Stress-Strength Reliability of $EGD(3,\lambda)$ Distribution

SS reliability is estimated using the EGD distribution in this section. The SS
reliability for the independent random variables X and Y is given by

$R=\int_{-\infty}^{\infty}f_{X}(x)\>F_{Y}(x)\>dx,$

where $f_{X}(x)$ and $F_{Y}(x)$ are the marginal PDF of X and marginal CDF of
Y, respectively. We next consider the EGD distribution. Consider X and Y as
independent random variables complying with the EGD distribution with
parameters $\lambda_{1}$ and $\lambda_{2}$, respectively. Thus, we denote
$X\sim EGD(3,\lambda_{1})$ and $Y\sim EGD(3,\lambda_{2})$. Then, SS
reliability is

$R=\int_{0}^{\infty}\frac{\lambda_{1}^{2}}{(1+\lambda_{1})}\bigg{(}1+\frac{\lambda_{1}}{2}x^{2}\bigg{)}e^{-\lambda_{1}x}\bigg{[}1-\frac{(\lambda_{2}(x(\lambda_{2}x+2)+2)+2)e^{-\lambda_{2}x}}{2(1+\lambda_{2})}\bigg{]}dx$
$=\frac{\lambda_{1}^{2}}{2(1+\lambda_{1})(1+\lambda_{2})}\int_{0}^{\infty}\bigg{(}1+\frac{\lambda_{1}x^{2}}{2}\bigg{)}e^{-\lambda_{1}x}[2(1+\lambda_{2})-(\lambda_{2}^{2}x^{2}+2x\lambda_{2}+2\lambda_{2}+2)e^{-\lambda_{2}x}]dx$

$\displaystyle=\frac{\lambda_{1}^{2}}{(1+\lambda_{1})}\int_{0}^{\infty}\bigg{(}1+\frac{\lambda_{1}}{2}x^{2}\bigg{)}e^{-\lambda_{1}x}dx$
$\displaystyle-\frac{\lambda_{1}^{2}}{2(1+\lambda_{1})(1+\lambda_{2})}\int_{0}^{\infty}\bigg{(}1+\frac{\lambda_{1}}{2}x^{2}\bigg{)}$
$\displaystyle
e^{-(\lambda_{1}+\lambda_{2})x}(\lambda_{2}^{2}x^{2}+2x\lambda_{2}+2\lambda_{2}+2)dx$
$\displaystyle=1-$
$\displaystyle\biggr{[}\frac{\lambda_{1}^{2}(\lambda_{2}^{2}+\lambda_{1}\lambda_{2}+\lambda_{1})}{(1+\lambda_{1})(1+\lambda_{2})(\lambda_{1}+\lambda_{2})^{3}}+\frac{\lambda_{1}^{2}\lambda_{2}}{(1+\lambda_{1})(1+\lambda_{2})(\lambda_{1}+\lambda_{2})^{2}}+\frac{\lambda_{1}^{2}}{(1+\lambda_{1})(\lambda_{1}+\lambda_{2})}$
$\displaystyle+\frac{6\lambda_{1}^{3}\lambda_{2}^{2}}{(1+\lambda_{1})(1+\lambda_{2})(\lambda_{1}+\lambda_{2})^{5}}+\frac{3\lambda_{1}^{3}\lambda_{2}}{(1+\lambda_{1})(1+\lambda_{2})(\lambda_{1}+\lambda_{2})^{4}}\biggr{]}$
$\displaystyle=$
$\displaystyle\frac{\lambda_{2}(10\lambda_{1}^{2}\lambda_{2}^{2}+5\lambda_{1}\lambda_{2}^{3}+\lambda_{2}^{4}+12\lambda_{1}^{2}\lambda_{2}^{3}+6\lambda_{1}\lambda_{2}^{4})}{(1+\lambda_{1})(1+\lambda_{2})(\lambda_{1}+\lambda_{2})^{5}}+$
(2)
$\displaystyle\frac{\lambda_{2}(+3\lambda_{1}^{4}\lambda_{2}+10\lambda_{1}^{3}\lambda_{2}^{2}+\lambda_{2}^{5}+\lambda_{1}^{5}\lambda_{2}+4\lambda_{1}^{4}\lambda_{2}^{2}+6\lambda_{1}^{3}\lambda_{2}^{3}+4\lambda_{1}^{2}\lambda_{2}^{4}+\lambda_{1}\lambda_{2}^{5})}{(1+\lambda_{1})(1+\lambda_{2})(\lambda_{1}+\lambda_{2})^{5}}$

##  3 Maximum Likelihood Estimate of R

Let us suppose that $x_{1},x_{2},\dots,x_{n}$ is a random sample of size n
from $EGD(3,\lambda_{1})$ and $y_{1},y_{2},\dots,y_{m}$ is a random sample of
size m from $EGD(3,\lambda_{2})$. The likelihood function is

$L=\frac{\lambda_{1}^{2n}}{(1+\lambda_{1})^{n}}e^{-\lambda_{1}\sum_{i=1}^{n}x_{i}}\frac{\lambda_{2}^{2m}}{(1+\lambda_{2})^{m}}e^{-\lambda_{2}\sum_{j=1}^{n}y_{j}}\prod_{i=1}^{n}\big{(}1+\frac{\lambda_{1}}{2}x_{i}^{2}\big{)}\prod_{j=1}^{m}\big{(}1+\frac{\lambda_{2}}{2}y_{j}^{2}\big{)}$
(3)

The log-likelihood associated with the above equation is given by

$\displaystyle\log
L=2n\log\lambda_{1}-n\log(1+\lambda_{1})+2m\log\lambda_{2}-m\log(1+\lambda_{2})-\lambda_{1}\sum_{i=1}^{n}x_{i}-\lambda_{2}\sum_{j=1}^{m}y_{j}$
(4)
$\displaystyle+\sum_{i=1}^{n}\log(1+\frac{\lambda_{1}}{2}x_{i}^{2})+\sum_{j=1}^{m}\log(1+\frac{\lambda_{2}}{2}y_{j}^{2})$

The first derivative of Eq.4 with respect to the unknown parameters
$\lambda_{1}$ and $\lambda_{2}$ are respectively given by

$\frac{\partial\log
L}{\partial\lambda_{1}}=\frac{2n}{\lambda_{1}}-\frac{n}{1+\lambda_{1}}-\sum_{i=1}^{n}x_{i}+\sum_{i=1}^{n}\frac{x_{i}^{2}}{2(1+\frac{\lambda_{1}}{2}x_{i}^{2})}$

$\frac{\partial\log
L}{\partial\lambda_{2}}=\frac{2m}{\lambda_{2}}-\frac{m}{1+\lambda_{2}}-\sum_{j=1}^{m}+\sum_{j=1}^{m}\frac{y_{j}^{2}}{2(1+\frac{\lambda_{2}}{2}y_{j}^{2})}.$

The second derivative of Eq.4 with respect to the unknown parameters
$\lambda_{1}$ and $\lambda_{2}$ are respectively given by

$\frac{\partial^{2}\log
L}{\partial\lambda_{1}^{2}}=\frac{n}{(1+\lambda_{1})^{2}}-\frac{2n}{\lambda_{1}^{2}}-\sum_{i=1}^{n}\frac{x_{i}^{4}}{4(1+\frac{\lambda_{1}}{2}x_{i}^{2})^{2}}$
$\frac{\partial^{2}\log
L}{\partial\lambda_{2}^{2}}=\frac{m}{(1+\lambda_{2})^{2}}-\frac{2m}{\lambda_{2}^{2}}-\sum_{j=1}^{m}\frac{y_{j}^{4}}{4(1+\frac{\lambda_{2}}{2}y_{j}^{2})^{2}}$

According to Eq.2, MLE of SS reliability, $\hat{R}_{ML}$, can be calculated as
follows:

${\hat{R}_{ML}=\frac{\lambda_{2}(10\lambda_{1}^{2}\lambda_{2}^{2}+5\lambda_{1}\lambda_{2}^{3}+\lambda_{2}^{4}+12\lambda_{1}^{2}\lambda_{2}^{3}+6\lambda_{1}\lambda_{2}^{4}+3\lambda_{1}^{4}\lambda_{2}+10\lambda_{1}^{3}\lambda_{2}^{2}+\lambda_{2}^{5}+\lambda_{1}^{5}\lambda_{2}+4\lambda_{1}^{4}\lambda_{2}^{2}+6\lambda_{1}^{3}\lambda_{2}^{3}+4\lambda_{1}^{2}\lambda_{2}^{4}+\lambda_{1}\lambda_{2}^{5})}{(1+\lambda_{1})(1+\lambda_{2})(\lambda_{1}+\lambda_{2})^{5}}}$
(5)

### Asymptotic Distribution and Confidence Intervals

The asymptotic distribution and confidence interval (CI) for the MLE of R are
presented in this section. Let us represent the Fisher information matrix of
$\lambda=(\lambda_{1},\lambda_{2})$ as $I(\lambda)$ in order to obtain the
asymptotic variance of the MLE of R, $\hat{R}_{ML}$, in Eq.5.

$I(\lambda)=E\left[{\begin{array}[]{cc}-\frac{\partial^{2}\log
L}{\partial\lambda_{1}^{2}}&-\frac{\partial^{2}\log
L}{\partial\lambda_{1}\partial\lambda_{2}}\\\\[8.0pt] -\frac{\partial^{2}\log
L}{\partial\lambda_{2}\partial\lambda_{1}}&-\frac{\partial^{2}\log
L}{\partial\lambda_{2}^{2}}\\\ \end{array}}\right].$

We further define R to establish its asymptotic normality as

$d(\lambda)=\bigg{(}\frac{\partial R}{\partial\lambda_{1}},\frac{\partial
R}{\partial\lambda_{2}}\bigg{)}^{\prime}=(d_{1},d_{2})^{\prime}$

where

${\frac{\partial
R}{\partial\lambda_{1}}=-\frac{\lambda_{1}\lambda_{2}^{2}(\lambda_{1}^{5}+(4\lambda_{2}+6)\lambda_{1}^{4}+(+\lambda_{2}^{2}+20\lambda_{2}+3)\lambda_{1}^{3}+(4\lambda_{2}^{3}+24\lambda_{2}^{2}+48\lambda_{2})\lambda_{1}^{2}+(\lambda_{2}^{4}+12\lambda_{2}^{3}+21\lambda_{2}^{2}+30\lambda_{2})\lambda_{1}+2\lambda_{2}^{4}+6\lambda_{2}^{3})}{(1+\lambda_{1})^{2}(1+\lambda_{2})(\lambda_{1}+\lambda_{2})^{6}}}$

${\frac{\partial
R}{\partial\lambda_{2}}=\frac{\lambda_{1}^{2}\lambda_{2}(\lambda_{2}^{5}+2(2\lambda_{1}+3)\lambda_{2}^{4}+(6\lambda_{1}^{2}+20\lambda_{1})\lambda_{2}^{3}+4\lambda_{1}(\lambda_{1}^{2}+6\lambda_{1}+12)\lambda_{2}^{2}+\lambda_{1}(\lambda_{1}^{3}+12\lambda_{1}^{2}+21\lambda_{1}+30)\lambda_{2}+2\lambda_{1}^{4}+6\lambda_{1}^{3}}{(1+\lambda_{1})^{2}(1+\lambda_{2})(\lambda_{1}+\lambda_{2})^{6}}}$

As a result, we can find the asymptotic distribution of $\hat{R}_{ML}$ as

$\sqrt{n+m}(\hat{R}_{ML}-R)\rightarrow^{d}N(0,d^{\prime}(\lambda)\>I^{-1}(\lambda)\>d(\lambda))$

We obtain the asymptotic variance of $\hat{R}_{ML}$ as follows:

$\displaystyle AV(\hat{R}_{ML})$
$\displaystyle=\frac{1}{n+m}d^{\prime}(\lambda)\>I^{-1}(\lambda)\>d(\lambda)$
$\displaystyle=V(\hat{\lambda_{1}})d_{1}^{2}+V(\hat{\lambda_{1}})d_{2}^{2}+2d_{1}d_{2}(\hat{\lambda_{1}}\hat{\lambda_{1}}).$

Asymptotic $100(1-\omega)\%$ CI for R can be obtained as

$\hat{R}_{ML}\pm Z_{\omega/2}\sqrt{AV(\hat{R}_{ML})}$

where $Z_{\omega/2}$ is the upper $\omega/2$ quantile of the standard normal
distribution.

##  4 Simulation Study

This section presents some results related to the performance of estimators in
R using the Newton-Raphson method. Using independent $EGD(3,\lambda_{1})$ and
$EGD(3,\lambda_{2})$ distributions, we have generated 1000 samples for this
purpose. The parameter values, $(\lambda_{1},\lambda_{2})$, examined in this
study were: (0.5, 1.5), (1, 1.5), and (1, 0.5), and various sample sizes (n,
m): (10,10), (15,15), (25,25), (30,30), (50,50), and (75,75). According to
these parameter values, R values are 0.8391, 0.6405, and 0.2551, respectively.

Tables 1\- 3 provide estimates of R based on the MLE method along with average
biases, mean square errors (MSEs), and 95% CIs. Simulated results indicate
that biases and MSEs decrease with increasing sample size (n,m).

Table 1: MLE, average(Avg) bias, and MSEs of different estimators of R when $\lambda_{1}=0.5$ and $\lambda_{2}=1.5$. (n,m) | Estimates | Avg Bias | MSEs | 95% CI
---|---|---|---|---
(10,10) |  $\hat{\lambda_{1}}$=0.52881 | 0.02881 | 0.01272 | (0.00714, 1.67700)
$\hat{\lambda_{2}}$=1.62901 | 0.12901 | 0.19516 |
(15,15) |  $\hat{\lambda_{1}}$=0.51921 | 0.019208 | 0.00808 | (-0.16250, 1.84527)
$\hat{\lambda_{2}}$=1.58803 | 0.08803 | 0.11591 |
(25,25) |  $\hat{\lambda_{1}}$=0.51427 | 0.01427 | 0.00453 | (0.29010, 1.38693)
$\hat{\lambda_{2}}$=1.54604 | 0.04604 | 0.05462 |
(30,30) |  $\hat{\lambda_{1}}$=0.50998 | 0.00998 | 0.00389 | (0.08218, 1.59616)
$\hat{\lambda_{2}}$=1.53623 | 0.03623 | 0.04227 |
(50,50) |  $\hat{\lambda_{1}}$=0.50531 | 0.00531 | 0.00221 | (0.41299, 1.26788)
$\hat{\lambda_{2}}$=1.53003 | 0.03003 | 0.02474 |
(75,75) |  $\hat{\lambda_{1}}$=0.50529 | 0.00529 | 0.00150 | (0.42865, 1.24868)
$\hat{\lambda_{2}}$=1.51549 | 0.01549 | 0.01516 |
Table 2: MLE, average(Avg) bias, and MSEs of different estimators of R when $\lambda_{1}=1$ and $\lambda_{2}=1.5$. (n,m) | Estimates | Avg Bias | MSEs | 95% CI
---|---|---|---|---
(10,10) |  $\hat{\lambda_{1}}$=1.07005 | 0.07005 | 0.07308 | (-0.95120, 2.23538)
$\hat{\lambda_{2}}$=1.62033 | 0.12033 | 0.17647 |
(15,15) |  $\hat{\lambda_{1}}$=1.03367 | 0.03367 | 0.03485 | (-0.30554, 1.59834)
$\hat{\lambda_{2}}$=1.58321 | 0.08321 | 0.11195 |
(25,25) |  $\hat{\lambda_{1}}$=1.02501 | 0.02501 | 0.02182 | (-0.36708, 1.65794)
$\hat{\lambda_{2}}$=1.56419 | 0.06418 | 0.06247 |
(30,30) |  $\hat{\lambda_{1}}$=1.01862 | 0.01862 | 0.01575 | (-0.05805, 1.34368)
$\hat{\lambda_{2}}$=1.54095 | 0.04095 | 0.04532 |
(50,50) |  $\hat{\lambda_{1}}$=1.01643 | 0.01643 | 0.01089 | (0.18678, 1.09807)
$\hat{\lambda_{2}}$=1.53552 | 0.03552 | 0.02541 |
(75,75) |  $\hat{\lambda_{1}}$=1.01383 | 0.01383 | 0.00614 | (0.17796, 1.10334)
$\hat{\lambda_{2}}$=1.52276 | 0.02276 | 0.01679 |
Table 3: MLE, average(Avg) bias, and MSEs of different estimators of R when $\lambda_{1}=1$ and $\lambda_{2}=0.5$. (n,m) | Estimates | Avg Bias | MSEs | 95% CI
---|---|---|---|---
(10,10) |  $\hat{\lambda_{1}}$=1.07666 | 0.07666 | 0.07914 | (-0.10393, 0.60624)
$\hat{\lambda_{2}}$=0.52771 | 0.02771 | 0.01282 |
(15,15) |  $\hat{\lambda_{1}}$=1.03821 | 0.03821 | 0.03830 | (-0.12139, 0.63169)
$\hat{\lambda_{2}}$=0.51733 | 0.01753 | 0.00846 |
(25,25) |  $\hat{\lambda_{1}}$=1.02654 | 0.02654 | 0.02262 | (-0.11027, 0.62711)
$\hat{\lambda_{2}}$=0.51750 | 0.01750 | 0.00496 |
(30,30) |  $\hat{\lambda_{1}}$=1.02066 | 0.02066 | 0.01738 | (-0.02649, 0.53561)
$\hat{\lambda_{2}}$=0.50841 | 0.00841 | 0.00363 |
(50,50) |  $\hat{\lambda_{1}}$=1.01033 | 0.01033 | 0.00939 | (0.07393, 0.43790)
$\hat{\lambda_{2}}$=0.50595 | 0.00595 | 0.00217 |
(75,75) |  $\hat{\lambda_{1}}$=1.01027 | 0.01027 | 0.00697 | (0.09505, 0.41531)
$\hat{\lambda_{2}}$=0.50472 | 0.00472 | 0.00132 |

##  5 Applications

In the following section, we examine two real data sets that show the breaking
strength of jute fiber at different gauge lengths of 10mm and 20 mm (see Xia
et al. (2009)). We present two datasets as follows:

### Dataset 1 : Jute fiber breaking strength of 10 mm

693.73, 704.66, 323.83, 778.17, 123.06, 637.66, 383.43, 151.48, 108.94, 50.16,
671.49, 183.16, 257.44, 727.23, 291.27, 101.15, 376.42, 163.40, 141.38,
700.74,
262.90, 353.24, 422.11, 43.93, 590.48, 212.13, 303.90, 506.60, 530.55, 177.25.

### Dataset 2 : Jute fiber breaking strength of 20 mm

71.46, 419.02, 284.64, 585.57, 456.60, 113.85, 187.85, 688.16, 662.66, 45.58,
578.62, 756.70, 594.29, 166.49, 99.72, 707.36, 765.14, 187.13, 145.96, 350.70,
547.44, 116.99, 375.81, 581.60, 119.86, 48.01, 200.16, 36.75, 244.53, 83.55.

Based on the assumption that the two independent samples have been drawn from
$EGD(3,\lambda_{1})$ and $EGD(3,\lambda_{2})$, respectively, we provide the
MLE estimates for parameters $\lambda_{1}$ and $\lambda_{2}$ and the goodness-
of-fit tests Kolmogorov-Smirnov (K-S) and Cramer-Von Mises (CVM) in Table 4.
As a result, the MLE of R is $\hat{R}=0.5319$, and the 95% CI for R is
(0.3936,0.6702).

Table 4: MLE, CVM, and KS goodness of fit tests Data | Estimates | CVM (p-value) | KS (p-value)
---|---|---|---
Length 10 mm | 0.008149069 | 0.13151 (0.4533) | 0.1393 (0.5584)
Length 20 mm | 0.008725855 | 0.41935 (0.06361) | 0.20661 (0.1336)

##  6 Conclusion

There have been several well-developed estimation techniques for SS models
with single components that follow well-known lifetime distributions. However,
there has been no consideration of the EGD model. In this paper, we consider
the problem of estimating SS reliability with an EGD distribution in a single-
component SS model for independent stress and strength random variables. We
obtain the MLE of SS reliability, $\hat{R}_{ML}$. Extensive simulation causes
MSE and average biases caused by the estimates to approach zero when sample
sizes are increased. Analyses are conducted on real-life datasets.

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