Timeline for The expected value of product of random variables which have the same distribution but are not independent
Current License: CC BY-SA 4.0
22 events
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| Feb 8 at 17:47 | comment | added | esg | @mathworker21: not much thought, really. I thought about constructing (highly dependent) uniform marginals using Dirichlet distributions (and was lucky at the first try). | |
| Feb 7 at 13:56 | comment | added | mathworker21 | @esg Sorry for the late response; thank you very much for your comment! Very nice. Don't know how you thought of it though... | |
| Jan 22 at 20:44 | comment | added | esg | @mathworker21: the following example shows that asymptotically the "truth" is near the lower bound given by Jensen. Let $(Y_1,\ldots,Y_k)$ be uniform on the $k-1$-dimensional simplex $\mathcal{S}_{k-1}:=\{(x_1,\ldots,x_k)\in [0,1]^k\,\mid \sum_{i=1}^k x_i=1\}$, and let $X_i:=(1-Y_i)^{k-1}$. Then each $X_i$ is uniform on $[0,1]$ and $$d_k:=\mathbb{E}\prod_{i=1}^k X_i=\mathbb{E}\prod_{i=1}^k (1-Y_i)^{k-1}\leq \big((1-\frac{1}{k})^k\big)^{k-1}\approx e^{\tfrac{1}{2}-k} $$ since by the AGM-inequality $\prod_{i=1}^k (1-Y_i)\leq (1-\frac{1}{k})^k$. | |
| Jan 22 at 7:11 | comment | added | Fedor Petrov | for improving a lower bound, we may try to find a pointwise estimate $x_1\ldots x_k\geqslant \sum_{i=1}^k f(x_i)$, $x_i\in [0,1]$, with maximal possible $\int f$. mike's argument corresponds to $f(x)=e^{-k}(1/k+(1+\log x))$. | |
| Jan 22 at 6:28 | comment | added | Fedor Petrov | for large $k$ I would try the distribution on the hypersurface $X_1\ldots X_k=c^{-k}$ (cut by $0<X_i<1$), its marginals are not uniform, but possibly not that not uniform? | |
| Jan 22 at 6:08 | comment | added | Fedor Petrov | @mathworker21 not a find really, I was well aware of this work, listening the talks and discussing the stuff with the authors. It is amazing that if we fix two-dimensional uniform distributions of $(X_i,X_j), 1\leqslant i <j \leqslant 3$, then the optimal measure for $\int X_1X_2X_3$ leaves on the "surface" $X_1\oplus X_2\oplus X_3=1$, where $\oplus$ is a binary bit-sum (=XOR). | |
| Jan 22 at 4:51 | comment | added | mathworker21 | @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? | |
| Jan 21 at 22:03 | history | edited | Michael Hardy | CC BY-SA 4.0 |
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| Jan 21 at 21:34 | comment | added | Fedor Petrov | 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 | |
| Jan 21 at 21:31 | comment | added | Stef | @ChristopheLeuridan Thanks. But the OP wrote "It is known that c(2) = 1/6", not "It is easy to prove c(2) = 1/6 using rearrangement inequalities", which suggests this problem has been discussed or studied somewhere before. | |
| Jan 21 at 21:16 | comment | added | Fedor Petrov | when $k=2$, you may simply integrate the inequality $2X_1X_2\geqslant X_1(1-X_1)+X_2(1-X_2)+(X_1+X_2-1)$ | |
| Jan 21 at 20:31 | comment | added | Christophe Leuridan | @Stef When $k=2$, it follows from rearrangement inequalities. View en.wikipedia.org/wiki/Rearrangement_inequality for a discrete analogue. | |
| Jan 21 at 18:27 | comment | added | Stef | "It's known that c(2) = 1/6" << Can you link to a reference? Is there a name for this function c? | |
| Jan 21 at 18:01 | history | became hot network question | |||
| Jan 21 at 15:50 | comment | added | mike | you get a bound by taking logs & using Jensen | |
| Jan 21 at 14:54 | history | edited | gmvh | CC BY-SA 4.0 |
MathJax, added tags
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| Jan 21 at 13:41 | history | edited | fengzju | CC BY-SA 4.0 |
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| Jan 21 at 13:39 | comment | added | fengzju | @JamesMartin Yes. The probability space is $[0,1]$, so the integral is the same as the expected value. The latter is a better notation for random variables. I will edit the question. | |
| Jan 21 at 11:53 | answer | added | Christophe Leuridan | timeline score: 7 | |
| Jan 21 at 10:16 | comment | added | James Martin | 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$? | |
| S Jan 21 at 10:00 | review | First questions | |||
| Jan 21 at 11:56 | |||||
| S Jan 21 at 10:00 | history | asked | fengzju | CC BY-SA 4.0 |