How to prove three properties of the moment generating function? [duplicate]

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.

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

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.

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