A question about moment generating function

Question

I am reading the lecture notes where proposition 6.3.1. says:

Suppose that the moment-generating function $m_Y(t)$ of a random variable $Y$ admits an expansion into a power series. Then the coefficients are related to the moments of $Y$ in the following way: $$ m_Y(t)=\sum_{k=0}^\infty\frac{\mu_k}{k!}t^k, $$ where $\mu_k=\mathbb E[Y^k]$ is the $k$-th moment of $Y$.

And it is said: "A fully rigorous argument of this proposition is beyond the scope of these notes".

But it is followed by the following line:

$$ m_Y(t)=\mathbb E[e^{tY}]=\mathbb E[\sum_{k=0}^\infty\frac{(tY)^k}{k!}]=\sum_{k=0}^\infty t^k\frac 1{k!}\mathbb E[Y^k]=\sum_{k=0}^\infty \frac {\mu_k}{k!}t^k. $$

I think this is already rigorous if say $Y$ is positive by the Fubini-Tonelli theorem. But why does this also hold even when $Y$ is an arbitrary random variable?

Answer 1

You are right about the positive case. In general $Y$ need not have finite moments and even when the moments exits you need conditions on the moments for this to hold.

Suppose $Ee^{c|Y|} <\infty$ for some $c >0$ (equivalently, $\sum \frac {|\nu_k|} {k!} c^{k} <\infty$ where $\nu_k$ is the $k-$th absolute moment of $Y$). Then DCT can be used to justify the equation $m_Y(t)=\sum \frac {\mu_k t^{k}} {k!}$ for $|t| <c$.

Answer 2

Let $X$ be any random variable with $\mathbb E\left[e^{tX}\right]<\infty$ for all $|t|<a$, for some $a>0$.

Therefore,

$$\mathbb E\left[e^{|tX|}\right]\le\mathbb E\left[e^{tX}\right]+\mathbb E\left[e^{-tX}\right]<\infty \quad,\,\forall \,|t|<a \tag{1}$$

Now $e^{|tX|}\ge \frac{|tX|^k}{k!}$ for every $k\in \mathbb N$. So by choosing a $t$ in $(0,a)$, we have $$\mathbb E|X|^k <\infty \quad,\forall \,k\ge 1$$

Also for every $n\in\mathbb N$, $$\left|\sum_{k=0}^n \frac{(tX)^k}{k!}\right|\le \sum_{k=0}^n \frac{|tX|^k}{k!}\le e^{|tX|} \tag{2}$$

$(1)$ and $(2)$ ensure that we can write the 2nd equality here by applying DCT:

\begin{align} \mathbb E\left[e^{tX}\right]&=\mathbb E\left[\lim_{n\to\infty}\sum_{k=0}^n \frac{(tX)^k}{k!}\right] \\&=\lim_{n\to\infty}\mathbb E\left[\sum_{k=0}^n \frac{(tX)^k}{k!}\right] \\&=\lim_{n\to\infty} \sum_{k=0}^n\frac{t^k}{k!} \mathbb E\left[X^k\right] \\&=\sum_{k=0}^\infty \frac{t^k}{k!}\mathbb E\left[X^k\right] \end{align}