Proposition Section
A moment-generating function uniquely determines the probability distribution of a random variable.
Proof
If the support \(S\) is \(\{b_1, b_2, b_3, \ldots\}\), then the moment-generating function:
\(M(t)=E(e^{tX})=\sum\limits_{x\in S} e^{tx} f(x)\)
is given by:
\(M(t)=e^{tb_1}f(b_1)+e^{tb_2}f(b_2)+e^{tb_3}f(b_3)+\cdots\)
Therefore, the coefficient of:
\(e^{tb_i}\)
is the probability:
\(f(b_i)=P(X=b_i)\)
This implies necessarily that if two random variables have the same moment-generating function, then they must have the same probability distribution.
Example 9-3 Section
If a random variable \(X\) has the following moment-generating function:
\(M(t)=\left(\dfrac{3}{4}+\dfrac{1}{4}e^t\right)^{20}\)
for all \(t\), then what is the p.m.f. of \(X\)?
Solution
We previously determined that the moment generating function of a binomial random variable is:
\(M(t)=[(1-p)+p e^t]^n\)
for \(-\infty<t<\infty\). Comparing the given moment generating function with that of a binomial random variable, we see that \(X\) must be a binomial random variable with \(n = 20\) and \(p=\frac{1}{4}\). Therefore, the p.m.f. of \(X\) is:
\(f(x)=\dbinom{20}{x} \left(\dfrac{1}{4}\right)^x \left(\dfrac{3}{4}\right)^ {20-x}\)
for \(x=0, 1, \ldots, 20\).
Example 9-4 Section
If a random variable \(X\) has the following moment-generating function:
\(M(t)=\dfrac{1}{10}e^t+\dfrac{2}{10}e^{2t} + \dfrac{3}{10}e^{3t}+ \dfrac{4}{10}e^{4t}\)
for all \(t\), then what is the p.m.f. of \(X\)?