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What is the expectation of $\prod\limits_{k=1}^\infty \left(1+\prod\limits_{i=1}^k u_i\right)$, where the $u$'s are i.i.d. $\text{Uniform}(0,1)$-variables ?

My first thought was to try replacing $u_i$ with its mean value $\frac12$, giving $\prod\limits_{k=1}^\infty \left(1+\left(\frac12\right)^k\right)=$ 2.38423102903..., but this approach did not seem justifiable.

Taking the log of the products didn't seem to help.

An Excel simulation with half a million trials gave an average (arithmetic mean) of $2.7201$, suggesting that the answer is just $e$.

Context: This question was inspired by a related question asking for the probability that $\sum\limits_{k=1}^\infty \left(\prod\limits_{i=1}^k u_i\right)>1$.

Dan
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1 Answers1

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Write

$$ f(x)=\mathsf E_{u_1,u_2,u_3,\ldots}[(1+xu_1)(1+xu_1u_2)(1+xu_1u_2u_3)\cdots]\;. $$

You want $f(1)$. We have

\begin{eqnarray} f(x) &=& \int_0^1\mathsf E_{u_2,u_3,\ldots}[(1+xu_1)(1+xu_1u_2)(1+xu_1u_2u_3)\cdots]\,\mathrm du_1 \\ &=& \int_0^1(1+xu_1)f(xu_1)\,\mathrm du_1 \\ &=& \frac1x\int_0^x(1+t)f(t)\,\mathrm dt \end{eqnarray}

(with the substitution $t=xu_1$). Multiplying by $x$ and then differentiating with respect to $x$ yields

$$ f(x)+xf'(x)=(1+x)f(x)\;, $$

and then cancelling $f(x)$ and dividing by $x$ yields

$$ f'(x)=f(x)\;, $$

with the solution $f(x)=c\,\mathrm e^x$. Since $f(0)=1$, we have $c=1$ and thus $f(1)=\mathrm e$.

joriki
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  • I don't quite understand the line with $\int_0^1(1+xu_1)f(xu_1),\mathrm du_1$. I guess you're using $f(xu_1)=\mathsf E_{u_1,u_2,u_3,\ldots}[(1+xu_1u_1)(1+xu_1u_1u_2)(1+xu_1u_1u_2u_3)\cdots];.$ But $u_1$ and $u_2$ are independent, so I don't quite understand why $xu_1 u_2$ can be replaced by $xu_1 u_1$. – Dan Feb 22 '23 at 06:36
  • @Dan: Rename $u_1$ to something else and rename $u_2$ to $u_1$, $u_3$ to $u_2$ etc. – joriki Feb 22 '23 at 06:57
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    OK, thanks. Very nice solution! – Dan Feb 22 '23 at 09:17
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    Could you (try to) explain how you came up with this solution? :) – PhoemueX Feb 22 '23 at 16:11
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    @PhoemueX: I'll give it a try, though I don't remember exactly. I wondered what would happen if I tried to integrate over $u_1$, and I noticed that the expression that results if I plug in a value for $u_1$ looks a lot like the original expression, except for an additional factor, so I figured it might be useful to consider the expression more generally as a function of this factor, because then I can relate its value with $u_1$ plugged in to its other values. – joriki Feb 22 '23 at 17:02
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    There are quite a number of answers on this site that solve similar problems using integral equations transformed into differential equations by differentiation (including my answers here, here, here and here), and the fact that the value seemed to be $\mathrm e$ made it seem even more likely that that sort of approach might work, since a differential equation might yield an exponential function. – joriki Feb 22 '23 at 17:02
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    @joriki: Thanks! – PhoemueX Feb 22 '23 at 17:48