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I just derived it by using the generation function to first get raw moments. The result is


It was merely brutal force calculation, nothing interesting. So I was wondering, if there any one knows tricks that could simplify the process a bit.

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When the probability mass function involves factorials in the denominator (as in binomial and Poisson pmf), it is often easier to find $E[X(X-1)(X-2)(X-3)]$ rather than $E[X^4]$ etc. Of course, using the moment generating function (if you know what it is) is even easier. –  Dilip Sarwate May 8 '12 at 1:11

3 Answers 3

up vote 4 down vote accepted

Let $X$ be binomial with $n$ trials and success probability $p$. I assume by fourth central moment you mean $E[(X - np)^4]$. Write $X = \sum_{i = 1} ^ n X_i$ where the $X_i$ are iid Bernoulli with success probability $p$. Then we want $$E\left[\left(\sum_{i = 1} ^ n (X_i - p)\right)^4 \right] = E\left[\sum_{1 \le i, j, k, l \le n} (X_i - p)(X_j - p)(X_k - p)(X_l - p) \right] = \sum_{1 \le i, j, k, l \le n}E[(X_i - p)(X_j - p)(X_k - p)(X_l - p)]$$ which effectively turns this into a counting problem. The terms in the sum on the right hand side are $0$ unless either $i = j = k = l$, which happens $n$ times, or when there are two pairs of matching indicies, which happens $\binom n 2 \binom 4 2 = 3n(n - 1)$ times. This gives us $$E[(X - np)^4] = n\mu_4 + 3n(n - 1)\sigma^4 \tag{$\dagger$}$$ as the fourth central moment, when $\sigma^2 = p(1 - p)$ is the variance $X_1$ and $\mu_4 = p(1 - p)^4 + p^4 (1 - p)$ is the fourth central moment of $X_1$. Note that in deriving ($\dagger$) we actually derived a general formula for the fourth central moment of a sum of iid random variables.

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+1 I was going exactly this way. –  leonbloy May 8 '12 at 1:31
It's strange, I did these exact same calculations on a completely unrelated problem. I could almost copy and paste this answer from there. –  guy May 8 '12 at 1:48

An alternative approach would be to use cumulants. The fourth central moment of a random variable $X$ can be expressed in terms of cumulants as follows: $$\mu_4(X)=\kappa_4(X)+3\kappa^2_2(X).$$

Now, cumulants add over independent random variables and the second cumulant is just the variance, i.e., $\kappa_2=\mu_2$.

Writing $Y=\sum_{i=1}^n Z_i$, where the $Z_i\,$s are i.i.d. random variables, we have

\begin{eqnarray*} \mu_4(Y)&=&\kappa_4(Y)+3\kappa^2_2(Y)\\ &=&n\kappa_4(Z)+3[n\kappa_2(Z)]^2\\ &=&n\left[\mu_4(Z)-3\kappa_2^2(Z)\right]+3[n\kappa_2(Z)]^2\\ &=&n\, \mu_4(Z) +3n(n-1)\,\mu_2^2(Z). \end{eqnarray*}

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If you decide to pursue Dilips' strategy: For $k\geq 0$, define $$m_k=E\!\left[{X\choose k}\right]=\sum_{x\geq 0} {x\choose k} {n\choose x} p^x (1-p)^{n-x}.$$

We may not know what these numbers are, but we do know that \begin{eqnarray*} \sum_{k\geq0} m_k y^k &=& \sum_{k\geq0} \sum_{x\geq 0} {x\choose k} {n\choose x} p^x (1-p)^{n-x} y^k \\[5pt] & =& \sum_{x\geq 0} (1+y)^x {n\choose x} p^x (1-p)^{n-x}\\[5pt] &=& (py+1)^n. \end{eqnarray*}

Extracting the coefficient of $y^k$ on both sides gives $$m_k=E\!\left[{X\choose k}\right]= {n\choose k} p^k.$$

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