I am trying to understand a proof for the Kurtosis of a sum of independent random variables, however, there is one part where I am quite stuck:


for $X_1, X_2, ..., X_n$ independent random variables with means $ \mu_1, ... , \mu_n$ and variance $\sigma_1^2, ... , \sigma_n^2 $ $E(X_i^4) < \infty$

Define $S_n = X_1 + ... + X_n$ and $S_n$ will be appropriately normal

$ kurt(S_n) - 3 = (\sum_{i=1}^n\sigma_i^n)^{-2}\sum_{i=1}^n\sigma_i^4(kurt(X_i)-3)$

Proof (first part):

assume WLOG that $E[X_i] = 0$ for all $i$

$ kurt(Sn) = \frac{E[(X_1 + ... + X_n )^4]}{(\sigma_1^2 + ... +\sigma_n^2)^2 } $

$ E[(X_1 + ... + X_n )^4] = \sum_{i=1}^n\sum_{j=1}^n\sum_{k=1}^n\sum_{l=1}^n E(X_iX_jX_kX_l) = \sum_{i=1}^n E(X_i^4) +6 \sum_{i<j}^n\sigma_i^2\sigma_j^2$

NOTE: $E(X_iX_jX_kX_l) = 0 $ unless $ i=j=k=l $ or if combinations of two pairs.

Question: why is the NOTE: true? From my understanding the expected value of INDEPENDENT random variables is equal to the product of the expected values of the random variables. However I intuitively understand this case does not apply, or else, we would not get very far! So what is going on?

Assuming this hickup true, i understand the rest of the proof, which I will not write. (as i am improvising the syntax)


Consider for example the case of $i=j=k=l$ where you are then taking the expectation of $X_i^4$. Since this real number raised to an even power is never negative, and is sometimes positive, its expectation cannot be zero.

On the other hand, consider $X_i^3 X_j$ with $i\neq j$. Now just because $E(X_i) = 0$ that does not allow you to say that $E<X_i^3>$ = 0; consider a discrete random that is $-1$ with probability $\frac23$ and $1$ with probability $\frac13$. But by the multiplication of independent expectations, you get zero since the expectation of $X_j$ is zero.

Only if all the independent variates are raised to powers greater than $1$ can the expectation be non-zero. For four variables, the only such cases are four of a kind and two pairs.

  • $\begingroup$ Alright I get it! Discrete example helped! $\endgroup$ – rannoudanames Sep 28 '16 at 23:44

From my understanding the expected value of [the product of] INDEPENDENT random variables is equal to the product of the expected values of the random variables.

This is true. Moreover, you assumed "(..) WLOG that E[Xi]=0 for all i".

  • $\begingroup$ Yes, that is why I am confused about the NOTE:, I would have expected E(X_i * X_j * X_k * X_l) to be 0 under all circumstances, however that is not the case. $\endgroup$ – rannoudanames Sep 28 '16 at 23:08
  • $\begingroup$ I see. The case i=j=k=l gives the first term, while (i=j, k=l, i!=k) gives the second term. $\endgroup$ – LinAlg Sep 28 '16 at 23:10
  • $\begingroup$ Yes, I see that, but why would those expected values, not be equal zero, due the independence of the random variables, and the assumption that E(X_i) = 0 $\endgroup$ – rannoudanames Sep 28 '16 at 23:14
  • $\begingroup$ For i=j=k=l, you end up with $E(X_i^4)$. For i=j, k=l, i!=k you have $E(X_i X_i X_k X_k) = E(X_iX_i)E(X_kX_k)$ (due to independence), which in turn equals $\sigma_i^2 \sigma_k^2$. $\endgroup$ – LinAlg Sep 28 '16 at 23:22

Suppose some index appears exactly once in the product $X_i X_j X_k X_l$, say, $i$ is different from all of $j,k,l$. Then $E(X_i X_j X_k X_l) = E(X_i) E(X_j X_k X_l) = 0$, since $E(X_i) = 0.$

There are only two ways that no index appears exactly once: if they are all the same, or if you have two indices that each appear twice.


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