1

Suppose $X \sim \text{Bin}(n,p)$, and look at the distribution of $Y = (X - n p)^2$.

  • Does $Y$ follow any known classical distribution / how much do we know about it ?
  • Can I bound its moment generating function? I am interested in upper bounding the upper tail $\mathbf{P}(\sum Y_i > t)$ with $Y_i = (X_i - n p_i)^2$ and $X_i \sim \text{Bin}(n,p_i)$ where $\sum X_i = n$. Markov's inequality gives me $$\mathbf{P}\Big(\sum Y_i > t\Big) \leq \frac{\sum \mathbf{Var}[X_i]}{t} = \frac{\sum np_i(1-p_i)}{t},$$ but I would like to get something that is exponentially decaying.
Teddy38
  • 3,309
John
  • 11

1 Answers1

0

You can calculate the mean and variance of the distribution of Y for part (I). Whether it follows any classical distribution or not I do not know.

Mean of Y

$ \mathbb{E}[(X-np)^2] =\mathbb{E}[X^2-2npX+n^2p^2] \\ =\mathbb{E}[X^2]+(-2np)\mathbb{E}[X]+n^2p^2 \\ = \mathbb{E}[X^2] - 2np(np)+ n^2p^2\\ $

$\mathbb{E}[(X)^2] -(\mathbb{E}[(X)])^2 = Var(X)$

$\mathbb{E}[(X)^2]= Var(X)+(\mathbb{E}[(X)])^2$

$\mathbb{E}[(X)^2] = np(1-p) + n^2p^2$

($Var(X) = np(1-p)$)

$ \mathbb{E}[(X-np)^2]=np(1-p) + n^2p^2 - 2n^2p^2 +n^2p^2 \\= np(1-p) = \sigma_X^2$

The Variance of Y, I would like to allude to the following post and get the result of this.

$Var(Y) = \mathbb{E}[Y^2] - (\mathbb{E}[Y])^2$

$\mathbb{E}[Y^2] = n\left[p(1-p)^4 + p^4(1-p)\right] + 3n(n-1)p^2(1-p)^2$

How to calculate the $4$th central moment of binomial distribution?

Now $Var(Y) = n\left[p(1-p)^4 + p^4(1-p)\right] + 3n(n-1)p^2(1-p)^2 - n^2p^2(1-p)^2$

If you simplify this you get

$Var(Y) = np(1-p)\left[(1+2(n-3)pq)\right] = \sigma_X^2\left[(1+2(n-3)pq)\right]$

Satish Ramanathan
  • 11,984
  • For the mean of $Y$, you could just note that $E[Y]=E[(X-np)^2]=E[(X-E(X))^2]=var[X]$. – r.e.s. Nov 15 '17 at 16:06
  • You are right, could be one liner!!, – Satish Ramanathan Nov 15 '17 at 16:08
  • Simplifying your next-to-last line, I get $Var[Y]=npq[1+2(n-3)pq]$, where $q=1-p$. This checks when $p=q=\frac{1}{2}$, giving $\frac{n(n-1)}{8}$, whereas your result does not. – r.e.s. Nov 15 '17 at 17:36
  • Also, note that if $p=q=\frac{1}{2}$ and $n=1$, then $Var[Y]=Var[(X-\frac{1}{2})^2]=0$, because in this case $X$ takes values only in ${0,1}$, so $(X-\frac{1}{2})^2$ always equals $\frac{1}{4}$. (The variance of a constant is of course $0$.) Your result does not agree with this, as it gives a non-zero variance – r.e.s. Nov 15 '17 at 18:06
  • I edited it. Thanks for correcting @r.e.s. – Satish Ramanathan Nov 15 '17 at 22:17