Please consider the problem and the answer below. The problem is from a
text book on statistics. I believe that my answer is correct but it seems
to easy so I am concerned that it is wrong. Is it wrong?
Thanks,
Bob
Problem:
If $Y$ has a binomial distribution with $n$ trials and the probability of
success $p$, show that the moment generating function for $Y$ is:
\begin{eqnarray}
m(t) &=& (pe^t+q)^n
\end{eqnarray}
where $q = 1 – p$.
Answer:
The moment generating function is defined to be $m(t) = E(e^{ty})$. This
gives us the following:
\begin{eqnarray*}
m(t) &=& E(e^{ty}) \\
m(t) &=& \sum_{i = 0}^{n} { {n \choose i} p^i q^{n-i} e^{ti} } \\
\end{eqnarray*}
Now, I apply the binomial theorem to equation 1 and get:
\begin{eqnarray*}
m(t) &=& \sum_{i = 0}^{n} { {n \choose i} p^i q^{n-i} e^{ti} } \\
\end{eqnarray*}
Q.E.D.
[Math] Binomial Distribution and the Moment Generating Function
probabilitystatistics
Best Answer
You certainly have the right idea. It isn't any harder than that. Just maybe work on presentation.
To put that line of thought together neatly I suggest something like:
$$\begin{align}M_X(t) &= \mathsf E(e^{tX}) &&\text{by definition of mgf}\\[1ex] & = \sum_{k} e^{tk}\mathsf P(X=k) && \text{by definition of expectation for discrete random variables} \\[1ex] & = \sum_{k=0}^n e^{tk}\binom{n}{k} p^k q^{n-k} & & \text{using the pmf for Binomial Distribution} \\[1ex] & = \sum_{k=0}^n \binom{n}{k} (pe^t)^k q^{n-k} & & \text{Algebra} \\[1ex] & = (pe^t+q)^n && \text{via the Binomial Theorem}\end{align}$$