I am working through an example in Wackerly's Mathematical Statistics and it asks to find the moment-generating function $m(t)$ for a Poisson distributed random variable with mean $\lambda$.
$$m(t)=E(e^{ty})=\sum_{y=0}^{\infty}e^{ty}p(y)=\sum_{y=0}^{\infty}e^{ty}\frac{\lambda^y e^{-\lambda}}{y!}$$ $$=\sum_{y=0}^{\infty}\frac{(\lambda e^t)^y e^{-\lambda}}{y!}= e^{-\lambda}\sum_{y=0}^{\infty}\frac{(\lambda e^t)^y }{y!}$$
Everything makes sense up to this point, but then it says "to complete the summation, find the Taylor series expansion".
There are two things I am confused about:
My idea of a Taylor series expansion was finding the Taylor series at a particular point to as many terms as desired to approximate the function locally, but in this case there is no point, and it seems that we are trying to find the end result of the sum as an exact expression. What exactly does this have to do with Taylor series? What is the "expansion"?
Is this a special case where you can use Taylor series, or is it common to use them when finding moment-generating functions?
Any help is appreciated.
Think of the Taylor series expansion of $f(x) = e^x$ around $x=a$.
$f(x) = f(a) + f'(a)(x-a) + {f''(a) \over 2!}(x-a)^2 + \ldots $
Using the fact that all derivatives of $f(x) = e^x$ are also $e^x$, we get:
$e^x = f(x) = e^a + e^a(x-a) + {e^a \over 2!}(x-a)^2 + \ldots$
Setting $a=0$, $e^x = 1 + x + {x^2 \over 2!} + \ldots$ (Maclaurin series)
Now, let's consider the MGF you computed:
$m(t) = e^{-\lambda} \sum_y \frac{(\lambda e^t)^y}{y!} $
Set $x = \lambda e^t$ to get:
$m(t) = e^{-\lambda} \sum_y {x^y \over y!} = e^{-\lambda} \left( 1 + x + {x^2 \over 2!} + \ldots \right) = e^{-\lambda} e^x = e^{-\lambda} e^{\lambda e^t}$ (using our Maclaurin series)