91Ó°ÊÓ

$X$is stochastically larger than $Y$

$X^{+}=\left\{\begin{array}{l}Xâ€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots \text{if}Xâ‰\yen 0\\ 0â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots \text{if}X<0\end{array},X^{−}=\left\{\begin{array}{r}0\text{if}Xâ‰\yen 0\\ −X\text{if}X<0\end{array}\right.\right.$

When $X$and $Y$are negative random variables show that $E\left[X\right]â‰\yen E\left[Y\right]$.

It is given that $X$is stochastically larger than $Y$. So, $X{â‰\yen }_{st}Y$

$⇒P\{X>t\}â‰\yen P\{Y>t\}$

$⇒1-P\{X>t\}≤1-P\{Y>t\}$

$⇒P\{X≤t\}≤P\{Y≤t\}…….\left(1\right)$

From the known information if $X$and $Y$are non-negative random variables.

For any two numbers $x$and $t$, define

$I(t<x)=\left\{\begin{array}{l}1â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots \text{if}t<x\\ 0â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots \text{if}tâ‰\yen x\end{array}\right.$

For any $x>0,$

role="math" localid="1647341912052" $x={∫}_0^{∞}I(t<x)dt.....\left(2\right)$

Now,

$E\left(X\right)={∫}_{-∞}^{∞}x{f}_X\left(x\right)dx$$(X$is non-negative, $f_X\left(x\right)=0$for$\left.x≤0\right)$

$={∫}_0^{∞}\left({∫}_0^{∞}I(t<x)dt\right)f_X\left(x\right)dx$From equation (2)

$={∫}_0^{∞}{∫}_0^{∞}I(t<x)f_X\left(x\right)dtdx$

$={∫}_0^{∞}{∫}_0^{∞}I(t<x)f_X\left(x\right)dxdt$

$={∫}_0^{∞}\left({∫}_0^t\left[0×f_X\left(x\right)\right]dx+{∫}_t^{∞}\left[1×f_X\left(x\right)\right]dx\right)dt$

$={∫}_0^{∞}\left({∫}_t^{∞}{f}_X\left(x\right)dx\right)dt$

$={∫}_0^{∞}P(X>t)dt$

Similarly,

For any two numbers $y$and $t$, define

$I(t<y)=\left\{\begin{array}{l}1â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots \text{if}t<y\\ 0â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots \text{if}tâ‰\yen y\end{array}\right.$

For any $y>0$,

localid="1647342408010" $y={∫}_0^{∞}I(t<y)dt....\left(3\right)$

Now,

$E\left(Y\right)={∫}_{-∞}^{∞}y{f}_Y\left(y\right)dy$

$={∫}_0^{∞}y{f}_Y\left(y\right)dy$$(Y$is non-negative, $f_Y\left(y\right)=0$for $\left.y≤0\right)$

$={∫}_0^{∞}\left({∫}_0^{∞}I(t<y)dt\right)f_Y\left(y\right)dy$From equation (2)

$={∫}_0^{∞}{∫}_0^{∞}I(t<y)f_Y\left(y\right)dtdy$

$={∫}_0^{∞}{∫}_0^{∞}I(t<y)f_Y\left(y\right)dydt$

$={∫}_0^{∞}\left({∫}_0^t\left[0×f_Y\left(y\right)\right]dy+{∫}_1^{∞}\left[1×f_Y\left(y\right)\right]dy\right)dt$

$={∫}_0^{∞}\left({∫}_t^{∞}{f}_Y\left(y\right)dy\right)dt$

$={∫}_0^{∞}P(Y>t)dt$

From equation (1), we can get

$⇒1-P(X≤t)â‰\yen 1-P(Y≤t)$

$⇒P(X>t)â‰\yen P(Y>t)$

Apply integration on both sides with respective $t,$

$⇒{∫}_0^{∞}P(X>t)dtâ‰\yen {∫}_0^{∞}P(Y>t)dt$

$⇒E\left(X\right)â‰\yen E\left(Y\right)$

Hence, it has been shown that the values when$X$and $Y$are non negative random variables are $E\left(X\right)â‰\yen E\left(Y\right).$

$X$is stochastically larger than $Y$

$X^{+}=\left\{\begin{array}{l}Xâ€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots \text{if}Xâ‰\yen 0\\ 0â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots \text{if}X<0\end{array},X^{−}=\left\{\begin{array}{r}0\text{if}Xâ‰\yen 0\\ −X\text{if}X<0\end{array}\right.\right.$

When $X$ and $Y$ are arbitrary random variables, show that$E\left[X\right]â‰\yen E\left[Y\right]$

Let $X=X^{+}-X^{-}$and $Y=Y^{+}-Y^{-}$

Here,

role="math" localid="1647523042329" $X^{+}=\left\{\begin{array}{ll}X& \text{if}Xâ‰\yen 0\\ 0& X<0\end{array}\right.$

$X^{−}=\left\{\begin{array}{ll}0& \text{if}Xâ‰\yen 0\\ −X& X<0\end{array}\right.$

And,

$\begin{array}{r}{Y}^{+}=\left\{\begin{array}{ll}Y& \text{if}Yâ‰\yen 0\\ 0& Y<0\end{array}\right.\\ Y^{−}=\left\{\begin{array}{ll}0& \text{if}Yâ‰\yen 0\\ −Y& Y<0\end{array}\right.\end{array}$

Now,

$\begin{array}{r}E\left(X\right)=E\left(X^{+}\right)−E\left(X^{−}\right)\end{array}$

$\begin{array}{r}=∑x{P}_{x^{+}}\left(x\right)−∑x{P}_{X^{−}}\left(x\right)\end{array}$

$\begin{array}{r}=\{0×P(X<0)+X×P(Xâ‰\yen 0\left)\right\}−\{−X×P(X<0)+0×P(Xâ‰\yen 0\left)\right\}\end{array}$

role="math" localid="1647523002285" $=XP(Xâ‰\yen 0)−XP(X<0)$

$=X\left[P\right(Xâ‰\yen 0)−P(X<0\left)\right]$

Calculate the value of $E\left[Y\right],$

$\begin{array}{r}E\left(Y\right)=E\left(Y^{+}\right)−E\left(Y^{−}\right)y\end{array}$

$\begin{array}{r}=∑y{P}_{Y+}\left(y\right)−∑y{P}_{Y−}\left(y\right)\end{array}$

localid="1647523329814" $=YP(Yâ‰\yen 0)−YP(Y<0)$

$=Y\left[P\right(Yâ‰\yen 0)−P(Y<0\left)\right]$

From the known information$X$is stochastically larger than$Y.$

So, $X{â‰\yen }_{st}Y$

$X{â‰\yen }_{st}Y⇒P[X>t]â‰\yen P[Y>t]$

And $X$and $Y$are Arbitrary Random Variables.

$\begin{array}{r}P(X>0)â‰\yen P(Y>0)\text{and}P(X<0)<P(Y<0)\end{array}$

$P(Xâ‰\yen 0)−P(X<0)â‰\yen P(Yâ‰\yen 0)−P(Y<0)…\left(4\right)$

$X{â‰\yen }_{st}Y⇒Xâ‰\yen Y….…\left(5\right)$

From equation (4) and (5)

$X\left\{P\right(Xâ‰\yen 0)-P(X<0\left)\right\}â‰\yen Y\left\{P\right(Yâ‰\yen 0)-P(Y<0\left)\right\}$

$E\left(X\right)â‰\yen E\left(Y\right)$

Hint: Let, $a,b,c,d$are any non-negative integers.

If $a>b$and$c>d$then $ac>bd$

Hence, it has been shown that$E\left(X\right)â‰\yen E\left(Y\right)$When$X$and$Y$ are arbitrary random variables.