# N: P(N=n)=(3/5)[(2/5)^n], find E(N)

• Nov 15th 2012, 04:32 PM
tcRom
N: P(N=n)=(3/5)[(2/5)^n], find E(N)
Suppose the number of children N of a random family satifies P(N=n)=(3/5)[(2/5)^n] where n=0,1,2,... Compute E(N).

I went through the math and I come up with 1, which is the probability space. So, I know I'm not going about this the right way.

$E(N=n)=\frac{3}{5}(\frac{2}{5})^0 + \frac{3}{5}(\frac{2}{5})^1 + \frac{3}{5}(\frac{2}{5})^2 + ...$

$=\sum_{n=0}^{\infty}\frac{3}{5}(\frac{2}{5})^n$

$=\frac{3}{5}\sum_{n=0}^{\infty}(\frac{2}{5})^n$

$=\frac{3}{5}\frac{5}{3}$

$=1$

Could somone point out the basic understanding of expected value that I seem to be missing in this problem?
• Nov 15th 2012, 04:50 PM
Re: N: P(N=n)=(3/5)[(2/5)^n], find E(N)
I'm going to assume you mean expected value. Here is the definition

$\operatorname{E}[N] = n_1p_1 + n_2p_2 + \dotsb + n_kp_k \;.$

What you have is the sum of all possible probabilities, which of course must be equal to 1!
However, each of your probabilities P[N=i] has a weight ni
• Nov 15th 2012, 04:59 PM
tcRom
Re: N: P(N=n)=(3/5)[(2/5)^n], find E(N)
$E(N=n)=0*\frac{3}{5}(\frac{2}{5})^0 + 1*\frac{3}{5}(\frac{2}{5})^1 + 2*\frac{3}{5}(\frac{2}{5})^2 + ...$

$=\sum_{n=0}^{\infty}n*\frac{3}{5}(\frac{2}{5})^n$

$=\frac{2}{3}$

At least I was right about missing a basic understanding. The unfortunate part is that I need to learn how to go from the summation to the answer using paper and pencil, I won't always have access to a calculator that can do the sum for me...
• Nov 15th 2012, 05:14 PM