Question

Show that if$E\left[X\right]<0$androle="math" localid="1649871241073" $\theta \ne 0$is such that$E\left[e^{\theta X}\right]=1$, then$\theta >0$.

Short 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-by-step solution

Given Information.

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

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$E\left[X\right]<0$

$-\theta <0$

$-\theta <0$

$\theta >0\text{.}$