Q.7.7 - A First Course in Probability

Question

We say that $X$is stochastically larger than $Y$, written $X{\ge }_{\text{st }}Y$, if, for all $t$,

$P\{X>t\}\ge P\{Y>t\}$

Show that if $X{\ge }_{\mathrm{st}}Y$then $E\left[X\right]\ge E\left[Y\right]$ when

(a) $X$ and $Y$ are nonnegative random variables;

(b) $X$ and $Y$are arbitrary random variables. Hint:

Write $X$ as

$X=X^{+}-X^{-}$

where

$X^{+}=\left\{\begin{array}{l}X\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ if }X\ge 0\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ if }X<0\end{array},X^{-}=\left\{\begin{array}{r}0\text{ if }X\ge 0\\ -X\text{ if }X<0\end{array}\right.\right.$

Similarly, represent $Y$ as $Y^{+}-Y^{-}$. Then make use of part (a).

Step-by-Step Solution

Answer

a) The values when$X$and$Y$are non negative random variables are $E\left(X\right)\ge E\left(Y\right)$ 

b) The values when$X$and$Y$are arbitrary random variables are $E\left(X\right)\ge E\left(Y\right)$

Step 1: Given Information (Part a)

$X$ is stochastically larger than $Y$

$X^{+}=\left\{\begin{array}{l}X\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ if }X\ge 0\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ if }X<0\end{array},X^{-}=\left\{\begin{array}{r}0\text{ if }X\ge 0\\ -X\text{ if }X<0\end{array}\right.\right.$

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

Step 2: Explanation (Part a)

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

$\Rightarrow P\{X>t\}\ge P\{Y>t\}$

$\Rightarrow 1-P\{X>t\}\le 1-P\{Y>t\}$

$\Rightarrow P\{X\le t\}\le P\{Y\le t\}\dots \dots .\left(1\right)$

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

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

$I(t<x)=\left\{\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ if }t<x\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ if }t\ge x\end{array}\right.$

For any $x>0,$

$x={\int }_0^{\infty }I(t<x)dt .....\left(2\right)$

Step 3: Explanation (Part a)

Now,

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

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

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

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

$={\int }_0^{\infty }\left({\int }_0^t\left[0\times f_X\left(x\right)\right]dx+{\int }_t^{\infty }\left[1\times f_X\left(x\right)\right]dx\right)dt$

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

$={\int }_0^{\infty }P(X>t)dt$

Step 4: Explanation (Part a)

Similarly,

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

$I(t<y)=\left\{\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ if }t<y\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ if }t\ge y\end{array}\right.$

For any $y>0$,

$y={\int }_0^{\infty }I(t<y)dt....\left(3\right)$

Now,

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

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

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

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

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

$={\int }_0^{\infty }\left({\int }_0^t\left[0\times f_Y\left(y\right)\right]dy+{\int }_1^{\infty }\left[1\times f_Y\left(y\right)\right]dy\right)dt$

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

$={\int }_0^{\infty }P(Y>t)dt$

Step 5: Explanation (Part a)

From equation (1), we can get

$\Rightarrow 1-P(X\le t)\ge 1-P(Y\le t)$

$\Rightarrow P(X>t)\ge P(Y>t)$

Apply integration on both sides with respective $t,$

$\Rightarrow {\int }_0^{\infty }P(X>t)dt\ge {\int }_0^{\infty }P(Y>t)dt$

$\Rightarrow E\left(X\right)\ge E\left(Y\right)$

Step 6: Final Answer (Part a)

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

Step 1: Given Information (Part b)

$X$ is stochastically larger than $Y$

$X^{+}=\left\{\begin{array}{l}X\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ if }X\ge 0\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ if }X<0\end{array},X^{-}=\left\{\begin{array}{r}0\text{ if }X\ge 0\\ -X\text{ if }X<0\end{array}\right.\right.$

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

Step 2: Explanation (Part b)

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

Here,

$X^{+}=\left\{\begin{array}{ll}X& \text{ if }X\ge 0\\ 0& X<0\end{array}\right.$

$X^{-}=\left\{\begin{array}{ll}0& \text{ if }X\ge 0\\ -X& X<0\end{array}\right.$

And, 

$\begin{array}{r}{Y}^{+}=\left\{\begin{array}{ll}Y& \text{ if }Y\ge 0\\ 0& Y<0\end{array}\right.\\ Y^{-}=\left\{\begin{array}{ll}0& \text{ if }Y\ge 0\\ -Y& Y<0\end{array}\right.\end{array}$

Now, 

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

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

$\begin{array}{r}=\{0\times P(X<0)+X\times P(X\ge 0\left)\right\}-\{-X\times P(X<0)+0\times P(X\ge 0\left)\right\}\end{array}$

$=XP(X\ge 0)-XP(X<0)$

$=X\left[P\right(X\ge 0)-P(X<0\left)\right]$

Step 3: Explanation (Part b)

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

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

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

$=YP(Y\ge 0)-YP(Y<0)$

$=Y\left[P\right(Y\ge 0)-P(Y<0\left)\right]$

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

So, $X{\ge }_{st}Y$

$X{\ge }_{st}Y\Rightarrow P[X>t]\ge P[Y>t]$

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

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

$P(X\ge 0)-P(X<0)\ge P(Y\ge 0)-P(Y<0)\dots \left(4\right)$

$X{\ge }_{st}Y\Rightarrow X\ge Y\dots .\dots \left(5\right)$

From equation (4) and (5)

$X\left\{P\right(X\ge 0)-P(X<0\left)\right\}\ge Y\left\{P\right(Y\ge 0)-P(Y<0\left)\right\}$

$E\left(X\right)\ge E\left(Y\right)$

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

If $a>b$and $c>d$ then $a c>b d$

Step 3: Final Answer (Part b)

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