Q. 8.13 - A First Course in Probability

Question

Show that if $E\left[X\right]<0$and $\theta \ne 0$ is such that$E\left[e^{\theta X}\right]=1$, then$\theta >0$.

Step-by-Step Solution

Answer

Consider the function$f\left(y\right)=-\ln \left(y\right),y>0$. Since this function is convex, if we let$Y=e^{\theta X}$, using Jensen's inequality$E\left[f\right(Y\left)\right]\ge f\left(E\right[Y\left]\right)$,

Step 1 Given Information.

 $E\left[X\right]<0$ and $\theta \ne 0$is such that$E\left[e^{\theta X}\right]=1$, then$\theta >0$.

Step 2 Explanation.

Assume that$E\left[X\right]<0$ and $\theta \ne 0$such that$E\left[e^{\theta X}\right]=1$, whereby $\theta$is a scalar. We want to show that$\theta >0$. To do this, we will use Jensen's inequality. It says that for a convex function $f$ and an arbitrary random variable $Y$is

$E\left[f\right(Y\left)\right]\ge f\left(E\right[Y\left]\right)$

Consider the function$f\left(y\right)=-\ln \left(y\right),y>0$. Since this function is convex, if we let$Y=e^{\theta X}$, using Jensen's inequality, we get:

$E\left[-\ln \left(e^{\theta X}\right)\right]\ge -\ln \left(E\left[e^{\theta X}\right]\right)$

Since$\ln \left(e^u\right)=u$, for each$u$, and using the information$E\left[e^{\theta X}\right]=1$, we get:

$E[-\theta X]\ge -\ln \left(1\right)=0$

if properties of expectation

$-\theta E\left[X\right]\ge 0$

$\Downarrow$ it is given$E\left[X\right]<0$

$-\theta <0$

$-\theta <0$

$\theta >0\text{. }$