Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How can we find the value of the following term,

$$ E[\prod_{i = 1}^{L}{\sum_{j = 1}^{K}{a_{ij}}}] $$

i.e., the expected value of the product of the sum of $a_{ij}$'s where $a_{ij}$ is a random variable drawn from a probability distribution $f(x)$. How can I compute the value for a general $f(.)$? What if $f(x) = \frac{1}{\sqrt{x}}$ and $c_1 \le x \le c_2$?

share|cite|improve this question
Mohsen: Care to accept one of the answers below? – Did Apr 25 '11 at 14:47

If the $a_{ij}$ are not only identically distributed but also independent, your expectaton is $(K\alpha)^L$ where $\alpha=E(a_{ij})$.

Since the independence assumption is only needed to disentangle the sums $b_i=\displaystyle\sum_{j=1}^Ka_{ij}$ but not to compute $E(b_i)=K\alpha$, this assumption can be relaxed to the $b_i$s being $L$ independent random variables.

share|cite|improve this answer
This is true that $E[\sum_{i = 1}^{k}a_i] = k E[a_i]$, but I can't see why $E[\prod_{i = 1}^{k}a_i] = E[a_i]^k$. $E[a_i]^k$ is the upper bound of the products, not the expected value, right? – Helium Apr 22 '11 at 0:08
Please ask your question using exactly the notations of your post so that I can see what step causes a problem. – Did Apr 22 '11 at 0:19
@Didier; I have the same instinct you do, but my attempts at counterexamples all fail. Is it the case that for independent variables $X$ and $Y$, $E(XY) = E(X)E(Y)$? – Carl Brannen Apr 22 '11 at 0:36
@Carl: Yes.… – Did Apr 22 '11 at 5:52
Let me explain my concern a bit more. Assume we have $KL$ random numbers $a_{ij}$ drawn from $f(.)$. An upper bound for the product of the sum of $a_{ij}$'s is $((\sum_{i, j}{a_{ij}})/L)^L$ (The product is maximized when the numbers are equally distributed). It is known that finding the optimal assignment of $a_{ij}$'s is NP-hard. However, the product of the sum for the optimal assignment is always less than or equal to the upper bound I mentioned. – Helium Apr 22 '11 at 8:12

Since $E(XY) = E(X)E(Y)$ for random and independent variables as can be seen by: $$\int_x\int_y\;xy\;f(x)g(y)\;dx\;dy = \int_x xf(x)\;dx\int_y yf(y)\;dy$$ Didier Pau's answer is correct: $(K\;E(a))^L$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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