1
$\begingroup$

The moment generating function of a random variable $X$ is defined to be the function $$M_{X}(t)=E(e^{tX})=\sum_{n=0}^{\infty}\frac{E(X^n)}{n!}t^n.$$ Let $I=\{t\in\mathbb R:M_{X}(t)<\infty\}.$

I wish to show that

  1. $I$ is possibly a degenerate interval and $0\in I$. (Degenerate means the interval includes only one real number.)

  2. $M_{X}(t)$ is a convex function on $I$.

  3. If $0$ is an interior point of $I$, then $E(X^k)\lt\infty$ for all $k\in \mathbb N$; i.e., $X$ has finite moments of all orders.

$\endgroup$
  • $\begingroup$ All of this, and more, is collected in this post. $\endgroup$ – cardinal Nov 14 '13 at 14:51
3
$\begingroup$

Let me answer the questions in order.

  • Claim 1. We can show that if the distribution of X is Cauchy distribution, then $I = \{0\}$.

    Obviously, $M_X(0)=1$. When $t>0$, $M_X(t) \geq C \int_1^\infty \exp(tx)/x^2 dx = \infty.$ The same observation applies for $t<0$ case.

  • Claim 2. Convexity of $M_X$

    Let $t_1, t_2$ be two points in $I$. for $0 \leq \lambda \leq 1$, $\exp(\lambda t_1 + (1-\lambda) t_2) \leq \lambda \exp(t_1) + (1-\lambda) \exp(t_2)$ holds because of the convexity of the exponential function. Taking the expectation of this inequality, we obtain the convexity of $M_X$.

  • Claim 3. If I contains open interval, then all moments exists.

    We can take small $\epsilon > 0$ so that $\pm \epsilon \in I$. Take any natural number $k$. We have $\exp(\epsilon X) + \exp(- \epsilon X) \geq C |X|^k$. Taking the expectation of this inequality, we see that $E[|X|^k] < \infty.$ Hence the claim holds.

$\endgroup$
  • 5
    $\begingroup$ Hi, and welcome to CrossValidated! When we're presented with requests for homework assistance (which are often, but not always, tagged as "self-study") we generally try to help people get to the solutions on their own without taking them all the way there. In the future, please resist the urge to actually do someone's homework for them, and instead consider asking questions or illustrating examples that will nudge them toward the correct solution or highlight flaws in their current thought process. $\endgroup$ – David Marx Nov 14 '13 at 14:16

Not the answer you're looking for? Browse other questions tagged or ask your own question.