If you flip a fair coin 1000 times, what is the expected value of the product of the number of
heads and the number of tails?

Question

If you flip a fair coin 1000 times, what is the expected value of the product of the number of
heads and the number of tails?

anonymous · Answer

need help?

anonymous · Answer

okay buddy

anonymous · Answer

60%

hartnn · Answer

2 equally likely possibilities
so expected number will also be same
500 and 500
25 0000

yttrium · Answer

no sir you got it wrong

hartnn · Answer

fair coin ...
P(head) = P(tail) = 0.5

Expected Heads E(Heads) = P(Head)*1000  = 500

right till here ?

hartnn · Answer

ohh!
you're asking 
E(Head*Tail)
but I answered
E(Head)*E(Tail)
isn't it ?

hartnn · Answer

can't imagine P(head*tail) in one coin toss :P

hartnn · Answer

so if i am wrong I mis-interpreted the question! :O

hartnn · Answer

whats the correct approach ?

anonymous · Answer

Let $X$ denote the number of heads, and $Y$ the number of tails. Hence $Y=1-X$. We want to find $E[X(1-X)]$ @hartnn

This is the same as $E(X)-E(X^2)=E(X)-\bigg(V(X)+[E(X)]^2\bigg)$. Both $X$ and $Y$ are binomially distributed with $p=0.5$ and $n=1000$, and the expected value for a binomial distribution such as this is $np$, or $500$. The variance $V(X)$ is given by $np(1-p)$, or $250$.
$$E[X(1-X)]=500-250-500^2=-249,750$$
Is this right @Yttrium ?

hartnn · Answer

what does negative Expectation Indicate ?

anonymous · Answer

Not really sure, I never had a good notion of what the product of random variables might represent unless they explicitly refer to things like length and width.