# The expected value of a Geometric Series

1. Sep 29, 2004

### relinquished™

I'm supposed to prove that in a geometric distribution, the expected value,

$$\mu = \frac{1}{p}$$

without the use of moment generating functions (whatever that is)

I start off with the very definition of the expected value.

$$\mu_x = E(x) = \sum x \cdot p \cdot (1-p)^{x-1}$$

$$\mu_x = p \sum x \cdot (1-p)^{x-1}$$

$$\mu_x = p \sum x \cdot (1-p)^x \cdot (1-p)^{-1}$$

$$\mu_x = \frac{p}{1-p} \sum x \cdot (1-p)^x$$

Now I get stuck because I don't know how to evaluate the summation. Can anyone help me out?

btw, x starts from 1 to n

2. Sep 29, 2004

### matt grime

can you sum y^r as r goes from 1 to n. what if you differentiate both sides?

3. Sep 29, 2004

### relinquished™

I am assuming that $$y^r$$ is $$(1-p)^x$$. If I would convert the summation into its series and differentiate both sides, what would be the derivative of $$\mu_x$$?

4. Sep 29, 2004

### matt grime

erm, what? i indicated to you how to sum a certain kind of series, the series you wanted to sum. i'm not doing anything with differentiating mu_x.

5. Sep 29, 2004

### relinquished™

err... sorry, my bad. So, when you said differentiate both sides I thought both sides of the equation. What you really mean is that in order to evaluate the summation you need to differentiate, am I understanding it right?

6. Sep 30, 2004

### matt grime

you know a formula :

S(n) = sum 1 to n of y^r

that is anequation in y, diff wrt to y and you'll find a formula for a sum that looks a lot like the one you want to sum in your problem. you've pulled that factor of 1/(1-p) out when you shouldn't have: it'll make it more transparent when you put it back in.