Question

Show that $X$is stochastically larger than $Y$if and only if$E\left[f\right(X\left)\right]\ge E\left[f\right(Y\left)\right]$

for all increasing functions $f.$.

Hint: Show that $X{\ge }_{\text{st}}Y$, then $E\left[f\right(X\left)\right]\ge E\left[f\right(Y\left)\right]$by showing that $f\left(X\right){\ge }_{st}f\left(Y\right)$and then using Theoretical Exercise 7.7. To show that if $E\left[f\right(X\left)\right]\ge E\left[f\right(Y\left)\right]$for all increasing functions $f$, then $P\{X>t\}\ge P\{Y>t\}$, define an appropriate increasing function $f$.

Short answer

It has been show that $X$is stochastically larger than $Y$if and only if$E\left[f\right(X\left)\right]\ge E\left[f\right(Y\left)\right]$for all increasing functions$f.$

Step-by-step solution

Given Information

$X$is stochastically larger than $Y$if and only if$E\left[f\right(X\left)\right]\ge E\left[f\right(Y\left)\right]$.

Explanation

Case 1: If$X{\ge }_{st}Y$

Then$f\left(x\right){\ge }_{st}f\left(y\right)$($\therefore \mathrm{f}$is an increasing function)

$E\left[f\right(\mathrm{X}\left)\right]\ge .E\left[f\right(\mathrm{X}\left)\right]$(using the result of positive exercise)

Explanation

Case 2: If $\mathrm{E}\left[f\right(\mathrm{X}\left)\right]\ge \mathrm{E}\left[f\right(\mathrm{X}\left)\right]$

$\mathrm{E}\left[f\right(\mathrm{X}\left)\right]={\int }_{-\infty }^{\infty }\mathrm{P}\left[f\right(x)>t]dt$

$\mathrm{E}\left[f\right(\mathrm{Y}\left)\right]={\int }_{-\infty }^{\infty }\mathrm{P}\left[f\right(\mathrm{Y})>t]dt$

As $\mathrm{E}\left[f\right(\mathrm{X}\left)\right]\ge E\left[f\right(\mathrm{Y}\left)\right]$

$\Rightarrow \mathrm{E}\left[f\right(\mathrm{X})>t]\ge P\left[f\right(\mathrm{Y})>t]\forall t$

As$f$is an increasing function

$\mathrm{P}[\mathrm{X}>t]\ge \mathrm{P}[\mathrm{Y}>t]$

$\Rightarrow \mathrm{X}{\ge }_{st}\mathrm{Y}$

Final Answer

Hence, it has been shown that $X$is stochastically larger than $Y$if and only if$E\left[f\right(X\left)\right]\ge E\left[f\right(Y\left)\right].$