2
$\begingroup$

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:

Theorem:

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)

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Alright I get it! Discrete example helped! $\endgroup$ – rannoudanames Sep 28 '16 at 23:44
0
$\begingroup$

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".

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.