# Math Help - statistics question

1. ## statistics question

The general form for the pmf of X = number of children born upto and including the first boy is

p(x) = p * (1-p)^(x-1) where x = 1,2,3,/......

and p(x) = 0 otherwise

find the expected value E(X).

Okay,so this is an example from the book.But what I dont really get is why are they differentiating the p(x) inside the summation.

2. I'd be surprised if they are really differentiating. This is a discrete distribution. Would you please do the following?

E(x) = sum x*p(x) = sum x*p * (1-p)^(x-1)
= p sum x*(1-p)^(x-1)

To compute sum x*(1-p)^(x-1). Let A=sum x*(1-p)^(x-1)
write x*A/(1-p) and subtruct it from A.

I mean write: x*A/(1-p) - A

Do you see any terms canceling each other? There should be many.
If you can't see, write them in two different lines, and let the same terms (x, x^2, x^2,...) come one top of another and see how they cancel each other.

Good luck.

3. Originally Posted by NidhiS
The general form for the pmf of X = number of children born upto and including the first boy is

p(x) = p * (1-p)^(x-1) where x = 1,2,3,/......

and p(x) = 0 otherwise

find the expected value E(X).

Okay,so this is an example from the book.But what I dont really get is why are they differentiating the p(x) inside the summation.
For $|r| < 1$ you should know (infinite geometric series) that

$S = \sum_{x=1}^{+\infty} r^x = \frac{r}{1-r}$ .... (1)

Differentiate equation (1) with respect to r: $\frac{dS}{dr} = \sum_{x=1}^{+\infty} x r^{x-1} = \frac{1}{(1-r)^2}$ .... (2)

$E(X) = \sum_{x=1}^{+\infty} x p (1 - p)^{x-1} = p \sum_{x=1}^{+\infty} x (1 - p)^{x-1}$.

Now let $r = 1 - p$ and use equation (1) to evaluate the sum in this expected value.

4. Differentiate equation (1) with respect to x
must be:
Differentiate equation (1) with respect to r.

$
\frac{dS}{dr} = \sum_{x=1}^{+\infty} x r^{x-1} = \frac{1}{(1-r)^2}
$

-O

5. Originally Posted by oswaldo
Differentiate equation (1) with respect to x
must be:
Differentiate equation (1) with respect to r.

$
\frac{dS}{dr} = \sum_{x=1}^{+\infty} x r^{x-1} = \frac{1}{(1-r)^2}
$

-O
Aw rats! The deliberate mistake I always make to see if the OP thinks about and understands what I post has been exposed. Second time this week it's happened, dran it. Now I gotta go fix it and put in a new mistake ....

6. Why did you guys differentiate it ..that was my question initially

okay,I see what's happening...thanks!