moment generating function uniquely determines distribution

Question

Suppose that X and Y are random variables both taking possible values in {$0, 1, 2 ...$}. Further, suppose that X and Y have the same mgf for all t around $0$: Then it holds: $$\sum_{j=0}^{\infty}e^{tj}f_X(j)-\sum_{j=0}^{\infty}e^{tj}f_Y(j)=0$$

$$\Rightarrow\sum_{j=0}^{\infty}e^{tj}[f_X(j)-f_Y(j)]=0$$ $$\Rightarrow \sum_{j=0}^{\infty}e^{tj}c_j=0 $$ with $c_j:=f_X(j)-f_Y(j)$.

Is there a way to justify that $c_j=0$ for all $j \in \{0,1,...\}$

Answer 1

The series $g(z) = \mathbb E[z^X] = \sum_{j=0}^\infty z^j f_X(j)$ converges absolutely to an analytic function for $|z| < 1$, and the coefficients $f_X(j)$ can be obtainethbbd from the values of $g$ and its derivatives at $z=0$:

$$ f_X(j) = \frac{g^{(j)}(0)}{j!} $$

$g(z)$ for $z \ne 0$ can be obtained from the moment generating function: $$ g(e^{-t}) = \mathbb E[e^{-tX}] = \sum_{j=0}^\infty e^{-tj} f_X(j)$$

So the moment generating function determines $g$, and that determines the distribution.

Answer 2

First, notice that from the definition if $c_j$ we have $-1 \le c_j \le 1$ for all $j$. Now suppose $c_0 \ne 0$, and WLOG $c_0 > 0$. Then we have

\begin{align*} 0 &= \sum_{j=0}^\infty e^{tj}c_j = c_0 + \sum_{j=1}^\infty e^{tj}c_j \end{align*}

so $c_0 = -\sum_{j=1}^\infty e^{tj}c_j$ for all $t \in \mathbb{R}$. Since each $c_j$ is at least $-1$ this implies $$ 0 < c_0 \le \sum_{j=1}^\infty e^{tj} = \sum_{j=1}^\infty (e^t)^j,$$

but this is a geometric series and can be made arbitrarily small by choosing $t \ll 0$. This implies $c_0 = 0$, and we can use the same reasoning to show that $c_j = 0$ for all $j$ with induction.