Variance

StupidIdiot · #1 $Old$ November 22nd, 2008, 10:22 AM

There are x amount of balls in a lovely green bag. Each ball has an number on it. 1 ball is drawn at random, what is the expected value and variance?

Here's what I have so far:

Possible values: 1,2,3,4,5.....x
Probability of each: 1/x

Expected Value:
(1/x + 2/x + 3/x + 4/x + ... + x/x)

Sum of series = (n(t1-tn))/2
= (x(x+1))/2
= (1+x)/2

Variance will be (1/x)((1-(1+x)/2)^2) + (1/x)((2-(1+x)/2)^2) + ... + (1/x)((x -(1+x)/2)^2) but I can't sum the series (if that's what you're even supposed to do. How do I do that?

Can anyone help? I would massively appreciate it (also sorry for the lack of proper [math] thingys and all that, haven't been on in ages and I can't remember how to do it at all).

Thanks

Plato · #2 $Old$ November 22nd, 2008, 10:53 AM

You have done the expected value correctly although there is a simpler way.
$E(X) = \sum\limits_{k = 1}^x {\frac{k}{x}} = \frac{1}{x}\sum\limits_{k = 1}^x k = \frac{1}{x}\left( {\frac{{x\left( {x + 1} \right)}}{2}} \right) = \frac{{\left( {x + 1} \right)}}{2}$ .

Now to do variance: $V(X) = E\left( {X^2 } \right) - E^2 (X)$ .
$E\left( {X^2 } \right) = \sum\limits_{k = 1}^x {\frac{{k^2 }}{x}}$ .