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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 $c_1 \le x \le c_2$?

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Mohsen: Care to accept one of the answers below? –  Did Apr 25 '11 at 14:47

2 Answers 2

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.

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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? –  Mohsen 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. en.wikipedia.org/wiki/… –  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. –  Mohsen 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: $(K\;E(a))^L$

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