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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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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?2011-04-22
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    Please ask your question using **exactly** the notations of your post so that I can see what step causes a problem.2011-04-22
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    @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 $Y$, $E(XY) = E(X)E(Y)$?2011-04-22
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    @Carl: Yes. http://en.wikipedia.org/wiki/Independence_%28probability_theory%29#Independent_random_variables2011-04-22
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    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.2011-04-22
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    If $KL$ is big enough, we can say $E[a_{ij}] = ((\sum_{i, j}{a_{ij}})/KL)$, *i.e.*, the average of $a_{ij}$'s. Your answer means that the expected value equals the upper bound while we know the expected value should be less than the optimal value (which is hard to find).2011-04-22
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    Let $P$ denote the random variable in your post and $S=\sum_{ij}a_{ij}/KL$. You note that $P\le(KS)^L$ and that $S\to\alpha$ almost surely when $KL\to\infty$. And $E(P)=(K\alpha)^L$ for every $K$ and $L$. To summarize, $R=P/K^L$ is such that $R\le S^L$ and $E(R)=\alpha^L$ with $\alpha=E(S)$. Where is the contradiction?2011-04-22
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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: $(K\;E(a))^L$