Given a positive integer $k$, is there a positive real number $c(k)$ such that $\mathbb{E}\left(\prod_{i=1}^k X_i\right)\geq c(k)$ for any $k$-random variables $X_1,X_2,\ldots,X_k$ which all have the uniform distribution over $[0,1]$? What's the optimal value of $c(k)$? It's known that $c(2)=\frac{1}{6}$.
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6$\begingroup$ What do you mean by $\int_{[0,1]} \Pi_{i=1}^k X_i dx$? Do you perhaps mean the expectation of $ \Pi_{i=1}^k X_i$? $\endgroup$– James MartinCommented Jan 21 at 10:16
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4$\begingroup$ you get a bound by taking logs & using Jensen $\endgroup$– mikeCommented Jan 21 at 15:50
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1$\begingroup$ when $k=2$, you may simply integrate the inequality $2X_1X_2\geqslant X_1(1-X_1)+X_2(1-X_2)+(X_1+X_2-1)$ $\endgroup$– Fedor PetrovCommented Jan 21 at 21:16
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3$\begingroup$ for $k=3$ the problem is addressed in the paper by Alexander Zimin and Nikita Gladkov , and the things which happen in this case do not suggest that the answer for general $k$ has any chances to be explicit arxiv.org/pdf/1809.08554.pdf - but of course we may think on the reasonable bounds $\endgroup$– Fedor PetrovCommented Jan 21 at 21:34
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1$\begingroup$ @FedorPetrov Nice find! The asymptotic question would be the base of the exponent: Jensen gives a lower bound of $e^{-k}$, and taking $X_i$'s independent already gives an upper bound of $2^{-k}$. One can improve the upper bound by blowing up examples for fixed $k$ (such as $k=2$ or $k=3$), but I don't see how to improve (or match) the lower bound. Any ideas? $\endgroup$– mathworker21Commented Jan 22 at 4:51
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1 Answer
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The answer to the first question is positive, and the lower bound is achieved, since the set of all probability measures on $[0,1]^k$ with uniform marginals is compact and since the integral of the bounded continuous function $(x_1,\ldots,x_k) \mapsto x_1 \cdots x_k$ on $[0,1]^k$ depends continuously of the probability measure.
Moreover, given such a probability measure $\pi$ on $[0,1]^k$, the integral of $x_1 \cdots x_k$ with regard to $\pi$ is strictly positive since $x_1 \cdots x_k>0$ for $\pi$-almost every $(x_1,\ldots,x_k)$.
Yet, finding the minimum is not obvious. For all $i<j$, the conditional distribution of $(X_i,X_j)$ given $X_1,\ldots,X_{i-1},X_{i+1},\ldots,X_{j-1},X_{j+1},\ldots,X_k$ should be the decreasing coupling.
answered Jan 21 at 11:53
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1$\begingroup$ I feel like the argument in the first paragraph shows that $c(k)$ exists and is nonnegative; why is it strictly positive? $\endgroup$ Commented Jan 21 at 20:13
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2$\begingroup$ @Greg Martin I completed the answer to explain why. $\endgroup$ Commented Jan 21 at 20:34