Let R(t)= ln M(t), where M(t) is the moment-generating function of a random variable of the continuous type. How do you show the following?
a) $\displaystyle \mu=R'(0)$
b) $\displaystyle \sigma^{2}=R''(0)$
Hello,
We know that $\displaystyle M(0)=1~,~M'(0)=\mu~,~M''(0)=\mathbb{E}(X^2)$
And remember that $\displaystyle \sigma^2=\mathbb{E}(X^2)-\mu^2$
a)
$\displaystyle R'(t)=\frac{M'(t)}{M(t)} \Rightarrow R'(0)=\frac{\mu}{1}=\mu$
b)
$\displaystyle R''(t)=\frac{M''(t)M(t)-M'(t)M'(t)}{M(t)^2}$ (using quotient rule for $\displaystyle \frac{M'(t)}{M(t)}$)
Thus $\displaystyle R''(0)=\frac{\mathbb{E}(X^2) \cdot 1-\mu^2}{1^2}=\dots$