# Solve a sum in prob. question

1. Oct 24, 2011

### ArcanaNoir

1. The problem statement, all variables and given/known data

Finding the expected value of x, with poisson distribution. I don't follow the sum. It goes like this:

$$E(x)= \sum_{x=0}^{\infty} \frac{xe^{-\lambda}\lambda^x}{x!}$$
$$= e^{-\lambda} \sum_{x=0}^{\infty} \frac{x\lambda^x}{x(x-1)!}$$
$$= \lambda e^{-\lambda} \sum_{x=1}^{\infty} \frac{\lambda^{x-1}}{(x-1)!}$$
$$= \lambda e^{-\lambda} \sum_{k=0}^{\infty} \frac{\lambda^{k}}{k!}$$
$$= \lambda e^{-\lambda}e^{\lambda} = \lambda$$

So basically the part I don't get is why they say
$$\sum_{k=0}^{\infty} \frac{\lambda^{k}}{k!} = e^{\lambda}$$

2. Oct 24, 2011

### micromass

That is just the Taylor series expansion of $e^x$. Remember that

$$f(x)=\sum_{k=0}^{+\infty}{\frac{f^{(k)}(0)}{k!}x^k}$$

So if $f(x)=e^x$, then

$$e^x=\sum_{k=0}^{+\infty}{\frac{x^k}{k!}}$$

3. Oct 24, 2011

### ArcanaNoir

Thanks micro. This chapter is going to be the death of me. All kinds of crazy sums that end up with specific values that I'm supposed to remember from two weeks in calc II. Doomed! I hate sums.