# Using Binomial Distribution to evaluate this probability distribution

I am trying to find the value of a skewed distribution but can't make sense of what to plug in to evaluate the answer.

This is the given:

$$\text{Let X be Binomial(n, p). } \text{Using that, evaluate:}$$ $$\beta = \frac{E[(X-\mu)^3]}{\sigma^3}$$

Now, I expanded the numerator and got this: (wikipedia)

$$\beta = \frac{E[X^3] - 3\mu\sigma^2 - \mu^3}{\sigma^3}$$

and I know that $\mu=np$ and $\sigma=\sqrt{3np(1-p)}$ and this is what it simplifies to from what I did

$$\beta = \frac{E[X^3] - 3np(3np(1-p)) - (np)^3}{(\sqrt{3np(1-p)})^3}$$

The Problem

The issue is that I can't make sense of $E[X^3]$. I don't know how to evaluate that in order to get a numerical value of $\beta$ for arbitrary n and p values. The binomial distribution should take arguments x, n and p right? What is x here?

Would $E[X^3]$ be just $(np)^3$?

Thanks

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Isn't $\sigma = \sqrt{np(1-p)}$? – Braindead Nov 12 '12 at 2:31
Sorry, yes thats correct. I fixed it but I still dont know what $E[X^3]$ should be – Kay Nov 12 '12 at 2:36
$E[X^3]$ is definitely not just $(np)^3$. I got an expression for $E[X^3]$ using a method similar to Michael Hardy's work, but then, I'm not really sure what level of mathematics your class is using. – Braindead Nov 12 '12 at 2:58
$(np)^3=(\mathbf{E}X)^3$ if $np=\mathbf{E}X$ – Alex Nov 12 '12 at 3:01
@Alex - Except (E[X])^3 is not the same as E[X^3]. – Braindead Nov 12 '12 at 3:03

If we are interested in higher moments, the moment generating function $M_X(t)$ of $X$ can be useful. This is given by $$M_X(t)=E(e^{tX}=\sum_{x=0}^n e{tx}\binom{n}{x}p^xq^{n-x},$$ where $q=1-p$. The moment generating function can be rewritten as $$\sum_{x=0}^n \binom{n}{x}(pe^t)^xq^{n-x},$$ which we recognize as $(pe^t+q)^n$.

Expanding $E(e^{tX})$ as a power series, we get $$M_X(t)=1+tE(X)+\frac{t^2}{2!}E(X^2)+\frac{t^3}{3!}E(X^3)+\cdots.$$ Differentiate $M_X(t)$ three times, and set $t=0$. We get $$E(X^3)=M_X'''(0).$$ The differentiation three times of $(pe^t+q)^n$ is straightforward.

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The Pochhammer symbol $(x)_k$ is used by some people, perhaps especially combinatorialists, to represent the falling factorial: $$(x)_k = x(x-1)(x-2)\cdots(x-k+1)$$ and by others, perhaps especially those who work with special functions, to represent the rising factorial $$(x)_k = x(x+1)(x+2)\cdots(x+k-1).$$

I will follow the former convention. One way to find $E(X^3)$ is first to write $X^3$ as a linear combination of falling factorials: $$X^3 = (X)_3 + 3(X)_2 + X.$$

Then $$E((X)_3) = \sum_{x=0}^n (x)_3 \binom n x p^x (1-p)^{n-x} = \sum_{x=0}^n x(x-1)(x-2)\frac{n!}{(n-x)!x!} p^x (1-p)^{n-x}$$ $$= \sum_{x=3}^n x(x-1)(x-2)\frac{n!}{(n-x)!x!} p^x (1-p)^{n-x}$$ Notice we're now starting at $x=3$ rather than $x=0$, since the terms for $0,1,2$ are $0$. Now do a substitution: let $y=x-3$ and notice that as $x$ goes from $3$ to $n$ then $y$ goes from $0$ to $n-3$: $$\sum_{y=0}^{n-3} (y+3)(y+3)(y+1) \binom{n!}{((n-3)-y)!(y+3)!} p^{y+3} (1-p)^{(n-3)-y}$$ $$= \sum_{y=0}^{n-3} n(n-1)(n-3) \frac{(n-3)!}{((n-3)-y)!y!} p^{y+3} (1-p)^{(n-3)-y}$$ $$= (n)_3 p^3 \sum_{y=0}^{n-3} \binom{n-3}{y} p^y (1-p)^{(n-3)-y}$$ $$= (n)_3 p^3 \cdot 1.$$

Then do similar things with $E((X)_2)$, and so on.

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What you did definitely makes sense but my concern here is that, this approach seems out of scope of this introductory course. Would it be possible for you to confirm that the simplification I performed is correct? It seems to me that there should be an easy (semi obvious) way of simplifying that expression (aka a few assumptions can be made and not a lot of rigor is required). – Kay Nov 12 '12 at 2:48

Recall the definition of the expectation: $$\mathbb{E}\left(X^3\right) = \sum_{k=0}^n k^3 \mathbb{P}\left(X=k\right) = \sum_{k=0}^n k^3 \binom{n}{k} p^k (1-p)^{n-k} \tag{1}$$ In order to evaluate $\mathbb{E}\left(X^3\right)$ it is useful to consider a different sum instead. Consider $$\mathcal{P}_X(z) = \sum_{k=0}^n z^k \mathbb{P}\left(X=k\right)$$ known as the probability generating function. Then $$\mathbb{E}\left(X\right) = \left.z \frac{\mathrm{d}}{\mathrm{d}z} \mathcal{P}_X(z)\right|_{z=1}, \quad \mathbb{E}\left(X^2\right) = \left. z \frac{\mathrm{d}}{\mathrm{d}z} z \frac{\mathrm{d}}{\mathrm{d}z} \mathcal{P}_X(z)\right|_{z=1}, \quad \mathbb{E}\left(X^3\right) = \left. z \frac{\mathrm{d}}{\mathrm{d}z} z \frac{\mathrm{d}}{\mathrm{d}z} z \frac{\mathrm{d}}{\mathrm{d}z} \mathcal{P}_X(z)\right|_{z=1}$$ Evaluation of $\mathcal{P}_X(z)$ for the binomial distribution is especially easy: $$\begin{eqnarray} \mathcal{P}_X(z) &=& \sum_{k=0}^n z^k \mathbb{P}\left(X=k\right) = \sum_{k=0}^n z^k \binom{n}{k} p^k (1-p)^{n-k} = \sum_{k=0}^n \binom{n}{k} (z p)^k (1-p)^{n-k} \\ &=& \left(1-p + p z\right)^n \end{eqnarray}$$ Instead of computing $\mathbb{E}(X^3)$, it is easier to evaluate $$\left. \frac{\mathrm{d}^3}{\mathrm{d}z^3} \mathcal{P}_X(z)\right|_{z=1} = \mathbb{E}\left(X(X-1)(X-2)\right) = n(n-1)(n-2) p^3$$ Now, after some algebra we get: $$\mathbb{E}\left(X^3\right) = n p \left( 1 + 3 (n-1)p + (n-1)(n-2) p^2\right)$$

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Makes more sense. Thank you! – Kay Nov 12 '12 at 5:07