In something I'm working on, the expression $$\lim_{n\rightarrow\infty}\exp\left(-\frac{1}{n}\log\left(\prod_{i=1}^{n}X_{i}+\prod_{i=1}^{n}Y_{i}\right)\right)$$ came up, in which all $X_{i}$ $iid$ and all $Y_{i}$ $iid$. Is there any way to simplify or rewrite this nicely, hopefully with respect to expectations/central moments?

For example, in another location, the expression $$\lim_{n\rightarrow\infty}\exp\left(-\frac{1}{n}\log\left(\prod_{i=1}^{n}X_{i}\right)\right)$$ came up, and I was able to use the properties of log and the law of large numbers to get $$\exp\left(E[\log(X)]\right)$$ Then I substituted in $E[X]+\delta$ for $X$, and used a Taylor expansion, which can be truncated to get $$\exp\left(-\frac{\sigma^2_{X}}{2\mu^{2}_{X}}\right)$$ This is the kind of thing I'm hoping can be done with the first expression listed in this question.

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    $\begingroup$ Are the $X_i$ independent of the $Y_i$? $\endgroup$
    – whuber
    Aug 17, 2015 at 14:49
  • $\begingroup$ I'm most interested in not making that assumption, but still interested in the special case where they are. Does making that assumption allow for a simplification? $\endgroup$
    – Biomath
    Aug 17, 2015 at 15:03
  • $\begingroup$ are you studying the sum of geometric means? $\endgroup$
    – Aksakal
    Aug 17, 2015 at 16:00
  • $\begingroup$ Not quite, but close. This expression came up when I was looking at a geometric mean, in which there are two sequences in which at the same time point, the total value (for lack of better term) is a constant. I can elaborate more if you'd like. $\endgroup$
    – Biomath
    Aug 17, 2015 at 16:07
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    $\begingroup$ It seems to me that maybe you should get back to where you started, to the equation of interest before you took the logs and exps. Maybe there's intuition from that point to a different path $\endgroup$
    – Aksakal
    Aug 17, 2015 at 16:10

1 Answer 1


If you're analyzing the sum of geometric means then, I'm afraid you can't go too far. You may try applying geometric mean inequality to get an upper bounds:

$$ \left(a_1 a_2 \cdots a_n\right)^{1/n} \le \frac{1}{n} \sum_{k=1}^n a_k$$

Particularly, Poyla's proof looks promising.

Your starting expression is: $$\left(\prod_{i=1}^{n}X_{i}+\prod_{i=1}^{n}Y_{i}\right)^{\frac{1}{n}}$$

It's like similar to a geometric mean, so it's got to have similar bounds.

For instance, this must have some relation to the geometric mean of a sum $X_i+Y_i$, i.e. $\prod_{i=1}^n\left(X_i+Y_i\right)^\frac{1}{n}$

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    $\begingroup$ I don't think this is relevant to the problem at hand. Am I supposed to click down arrow if it doesn't help me? I feel weird about doing that. $\endgroup$
    – Biomath
    Aug 17, 2015 at 16:13
  • $\begingroup$ You can down vote if you wish so, but I usually give an answer a few days to stay before ruling that it's irrelevant. $\endgroup$
    – Aksakal
    Aug 17, 2015 at 16:28
  • $\begingroup$ @Alexis, thanks, it was typo. All this says is that geom mean is smaller than arithmetic. It's an often used fact in finance for geom and arith returns $\endgroup$
    – Aksakal
    Aug 17, 2015 at 16:34

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