Thread: Moment generating function of the Binomial distribution

1. [SOLVED] Moment generating function of the Binomial distribution

Suppose that $Y~Bin(n,p)$, with probability mass function given by

$p_Y(y)=P(Y=y)=\left( \begin{array}{c}n \\ y \end{array} \right) p^y(1-p)^{n-y}, y=0,1,...,n$

and $0$ otherwise. Show that the moment generating function of $Y$ is given by

$M_y(t)=\{pe^t+(1-p)\}^n$

I keep getting majorly lost with this. I've tried Googling to find a solution to this, since it's a fairly generic question, but to no avail. Can anyone show me how to show this?

2. Originally Posted by chella182
Suppose that $Y~Bin(n,p)$, with probability mass function given by

$p_Y(y)=P(Y=y)=\left( \begin{array}{c}n \\ y \end{array} \right) p^y(1-p)^{n-y}, y=0,1,...,n$

and $0$ otherwise. Show that the moment generating function of $Y$ is given by

$M_y(t)=\{pe^t+(1-p)\}^n$

I keep getting majorly lost with this. I've tried Googling to find a solution to this, since it's a fairly generic question, but to no avail. Can anyone show me how to show this?
One way to do this is to consider $X_i$ independent bernoulli trials, where the pdf is $p(x)=p^x(1-p)^{1-x}$ for $x=0,1$. Also, recall that binomial is the sum of the bernoulli trials, in this case $\displaystyle Y=\sum\limits_{i=1}^n X_i$.

The MGF of a bernoulli trial is $\displaystyle M_{X_i}(t)=\sum\limits_{x=0}^{1}e^{tx}p^x(1-p)^{1-x}=pe^t+(1-p)$.

Therefore, the MGF of the binomial is

\begin{aligned}M_Y(t)&= M_{\sum\limits_{i=1}^n X_i}(t)\\ &=\prod_{i=1}^{n}M_{X_i}(t)\quad (by\,a\,theorem)\\ &=\prod_{i=1}^n\left[pe^t+(1-p)\right]\\ &= \left[pe^t+(1-p)\right]^n\end{aligned}

which is the desired result.

Does this make sense?

3. Yeah that's great thanks, should be sufficient if the question comes up again today