Question

Let$X$ have moment generating function $M\left(t\right)$, and define$\Psi \left(t\right)=\log M\left(t\right)$. Show that${\left.{\Psi }^{\text{'}\text{'}}\left(t\right)\right|}_{t=0}=Var\left(X\right)$.

Short answer

The second derivative value of $\Psi \left(t\right)$and plug in$t=0$.

Step-by-step solution

Given Information

Let's $X$have moment generating function$M\left(t\right)$show that ${\left.{\Psi }^{\text{'}\text{'}}\left(t\right)\right|}_{t=0}=Var\left(X\right)$̣

Explanation 

The expression for the second derivative of $\Psi \left(t\right)=\log M\left(t\right)$,

${\Psi }^{\text{'}}\left(t\right)=\frac{M^{\text{'}}\left(t\right)}{M\left(t\right)}$

Which implies

${\Psi }^{\text{'}\text{'}}\left(t\right)=\frac{M^{\text{'}\text{'}}\left(t\right)M\left(t\right)-M^{\text{'}}\left(t\right)M^{\text{'}}\left(t\right)}{M^2\left(t\right)}$.

Explanation

Now, plug $t=0$into the previous expression to obtain that,

${\left.{\Psi }^{\text{'}\text{'}}\left(t\right)\right|}_{t=0}=\frac{M^{\text{'}\text{'}}\left(0\right)M\left(0\right)-M^{\text{'}}\left(0\right)M^{\text{'}}\left(0\right)}{M^2\left(0\right)}$

$=E\left(X^2\right)-E(X{)}^2=Var(X)$

so we have proved the claimed.

Final answer

The second derivative value of $\Psi \left(t\right)$and plug in$t=0$.