Variance of square of binomial distribution

Question

Let $q_{i}$ be a random variable having the Binomial distribution with number of trials $2^{n}$ and success probability $\frac{1}{2}$. I'm trying to find the variance of the random variable

\begin{equation} p_{i} = \bigg(\frac{2}{2^{n}}~ q_{i} - 1 \bigg)^{2}. \end{equation}

Answer 1

If you consider for $n=0,1,2,3$ so $2^n=1,2,4,8$, you get

n    2^n        E[p]               Var(p)
                                   
0     1           1                   0
1     2     0.5   = 1/2    0.25       = 1/2^2 
2     4     0.25  = 1/4    0.0935     = 3/2^5
3     8     0.125 = 1/8    0.02734375 = 7/2^8

which suggest a mean of $\dfrac{1}{2^n}$ and variance of $\dfrac{2^n-1}{2^{3n-1}}$

and if we let $m=2^n$ then they suggest a mean of $\dfrac{1}{m}$ and variance of $\dfrac{2}{m^2}-\dfrac{2}{m^3}$.

As for a proof, consider $q$ as a binomial random variable with parameters $m$ and $\frac12$:

$q$ has mean $\frac{m}{2}$ and variance $\frac{m}{4}$ and excess kurtosis $-\frac{2}{m}$,

so $\mathbb E\left[\frac2m q-1\right]=0$ and thus, with $p=\left(\frac2m q-1\right)^2$, you can say

• $\mathbb E[p]=\mathbb E\left[\left(\frac2m q-1\right)^2\right]=\text{Var}\left(\frac2m q-1\right)=\text{Var}\left(\frac2m q\right)=\left(\frac2m\right)^2\frac{m}{4}=\frac{1}{m},$

• $\mathbb E[p^2]=\mathbb E\left[\left(\frac2m q-1\right)^4\right]=\left(\frac2m\right)^4\left(3-\frac{2}{m}\right)\left(\frac{m}{4}\right)^2=\frac{3}{m^2}-\frac2{m^3},$

• $\text{Var}(p)= \left(\frac{3}{m^2}-\frac2{m^3}\right) -\left(\frac1m\right)^2 = \frac{2}{m^2}-\frac2{m^3}.\qquad \square$