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I have a question that is: Given n iid Bernoulli(p) distributions: $X_1, X_2, \ldots, X_n$ and $S_n=\sum X_i$. Find $E[(S_n-np)^3]$. Hint: $S_n-np= \sum (X_i-p)$.

So far, I have gotten that $S_n$ is a Binomial$(n,p)$ distribution. I have also tried the Binomial expansion of $(S_n-np)^3$, and calculating the overall expectations as the sum of the expectations of each term, but that didn't appear to be too helpful. Given the hint that $S_n-np= \sum (X_i-p)$, I am assuming I need to use that.

So, using the formula for expectation, I found $E \left[(\sum(X_i-p))^3 \right]=\sum_{k=1}^n(k-p)^3\frac{n!}{k!(n-k)!}p^k(1-p)^{n-k}$

I can't see what I should do next though (if this is even on the right track)

Any help would be much appreciated Thanks!

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    $\begingroup$ Please add the self-study tag and read its tag wiki. $\endgroup$ – Glen_b Apr 1 '16 at 8:20
  • $\begingroup$ For the binomial it's generally easier to find the factorial moments. For $Y\sim \text{binomial}(n,p)$ find $E[Y(Y-1)(Y-2)]$ -- the expectation simplifies nicely; from there the third central moment is readily calculated $\endgroup$ – Glen_b Apr 1 '16 at 8:48
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A possible way to go is to use the moment-generating function.

By expanding $(S_n-np)^3$ you see that $E\bigl[(S_n-np)^3\bigr]$ can be obtained from the first three moments $E[S_n]$, $E[S_n^2]$ and $E[S_n^3]$.

The moments $E[S_n^k]$ can be derived with the help of the moment-generating function $M_{S_n}(t)=E(e^{tS_n})$ by the relation $\boxed{E[S_n^k] = M^{(k)}_{S_n}(0)}$ (the $k$-th derivative at $0$).

Since $S_n=\sum_{i=1}^n X_i$, one has $M_{S_n}(t)=E(e^{tX_1}\ldots e^{tX_n})$ and since the $X_i$ are independent this gives $M_{S_n}(t)=E(e^{tX_1})\ldots E(e^{tX_n})$. Now $M_{S_n}(t)={E(e^{tX_1})}^n$ because the $X_i$ have the same distribution. It is easy to get $E(e^{tX_1})=1-p+pe^t$ and finally $\boxed{M_{S_n}(t)={(1-p+pe^t)}^n}$.

Now it remains to calculate $M_{S_n}'$, $M_{S_n}''$, $M_{S_n}'''$, to take the values at $0$, and to inject the results in the expansion of $E\bigl[(S_n-np)^3\bigr]$. I agree this is not really funny.

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  • $\begingroup$ As Stephane notes, we are just seeking the $3^{rd}$ central moment of a Binomial random variable. That is straight forwards with a computer (or just look it up). If one is calculating by hand, then, along the same lines – but perhaps simpler – might be to derive the central mgf. i.e. $\text{cmgf} = E\left[e^{t \left(S_n-\mu \right)}\right]=e^{-t \mu} M_{S_n}(t)$ where $\mu = n p $. Taking the third derivative etc will then yield the $3^{rd}$ central moment directly, side-stepping the substitution. $\endgroup$ – wolfies Apr 1 '16 at 12:55

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