Expected value problem... Probability...

Nappy

ok i dont have the mathlab that enables me to write the symbols correctly but the question is looking for the expected value of the binomial distribution. Ill do my best to make it readable..

P(Y=j) = (n choose j).p(to the power of j). ( 1 -p )(to the power of (n - j))

Would anyone be able to give me a walkthrough of this question. I know the answer is np but I dont understand the solution I have here in front of me.

Thanks alot...

Fremen

I guess this is the derivation you're using, right?
http://en.wikipedia.org/wiki/Binomial_distribution#Algebraic_derivations_of_mean_and_variance

The trick is to rewrite [latex]k {{n}\choose{k}}[/latex] in a clever way.
[latex]k {{n}\choose{k}} = k \frac{n!}{k!(n-k)!}[/latex] by definition.

But if we cancel the Ks and factor out an n,
[latex] k \frac{n!}{k!(n-k)!} = n \frac{(n-1)!}{(k-1)!(n-k)!}[/latex]

But this is just
[latex]n {{n-1}\choose{k-1}}[/latex]

(check this!)

Now if you factor out the n and a P from the sum, what you're left with sums to 1, because it's a sum of binomial probabilities (with parameters p and (n-1))

Post back if you didn't follow, I'll try to clarify.

Nappy

thats was exactly the problem Thanks a mill, All sorted now!