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If we let $A$ be equal to the number of tails we get in $n$ flips of a coin that is fair, what is the expected value of $A^2$.

I am inclined to think that $E[A^2]$ is equal to $E[A]$ which is $ \frac{n}{2}$. Would this be the case?

Peter Peteson
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    The variance $V(A) = E(A^2) - (E(A))^2$. Since $A$ has a Binomial(n,1/2) distribution, we know $E(A) = n/2$ and $V(A)=n/4$, therefore ... – r.e.s. May 12 '21 at 03:34

2 Answers2

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$A\sim\text{Binomial}\left(n, \frac 12\right)$. We know for binomial distributions, $E(A)=np=\frac n2$, and $Var=np(1-p)=\frac n4$. We know the formula for variance $Var(A)=E(A^2)-[E(A)]^2$, so $E(A^2)=\frac n4+\frac{n^2}{4}=\frac{n(n+1)}{4}$.

Vons
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no that should be $n^2/4$ since the number of tails is $n/2$ and it squared will $n^2/4$

Wasif
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