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I'm currently studying some probability and I'm stuck with this question.


Let $R_1, . . . , R_n$ be independent continuous uniform over [0, 1] random variables.

The geometric mean of $R_1, . . . , R_n$ is defined by $$G_n = 􏰠\sqrt[n]{R_1 ×...×R_n }=(R_1 ×...×R_n)^\frac{1}{n}.$$ Show that $G_n$ converges in probability as well as with probability 1 (i.e. almost surely) to a constant, and identify the limit. Make sure you state any theorem you use.


From what I gather, I'm pretty sure I'm supposed to use the Strong Law of Large Numbers. $$\mathbb{P}[\lim_{n \to \infty} M_n = \mu] = 1 $$ Where $M_n$ is the sample mean.

But I don't really understand how to show this, I'm completely stuck.

Any tips/help would be greatly appreciated

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    $\begingroup$ Have you tried taking a logarithm? (The logarithm of a geometric mean is the arithmetic mean of the logarithms.) $\endgroup$ – Ian Jan 19 '15 at 1:15
  • $\begingroup$ I can take a logarithm of $G_n$, but I don't know where to go from there. Can you explain a little bit more about the arithmetic mean of the logarithm? $\endgroup$ – Dr.Doofus Jan 19 '15 at 3:07
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    $\begingroup$ When you expand the logarithm of $G_n$ you get $\frac{1}{n} \sum_{i=1}^n \log(R_i)$. This is the arithmetic mean of the random variables $\log(R_i)$. These are trivially iid. You need to check that they have finite mean and variance, so that you can use the strong law. Then you can pass to the limit and invert the logarithm to find the limit. You probably did something similar in calculus to evaluate certain indeterminate forms like $0^0$ or $1^\infty$. In fact your limit is "in spirit" of the form $0^0$. $\endgroup$ – Ian Jan 19 '15 at 3:09
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    $\begingroup$ Sorry, there's an error in my previous comment: finite variance isn't required. $\endgroup$ – Ian Jan 19 '15 at 3:15
  • $\begingroup$ Ah, I see where you're coming from up until the last part. I haven't done anything like that (not to my knowledge anyway). How does inverting the logarithm help find the limit? Sorry to keep hassling you with questions! $\endgroup$ – Dr.Doofus Jan 19 '15 at 3:27
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Consider $F_n = \ln(G_n)$. Then by logarithm rules $F_n = \frac{1}{n} \sum_{i=1}^n \ln(R_i)$. This is the arithmetic mean of the iid random variables $\ln(R_i)$. Now prove that these have finite mean $\mu$. Then the strong law says that $F_n$ converges almost surely to $\mu$.

Now $G_n = \exp(F_n)$. Note that $\exp$ is continuous. So if $F_n(\omega)$ converges to $\mu$ then $G_n(\omega)$ converges to $\exp(\mu)$. Hence $G_n$ converges almost surely to $\exp(\mu)$.

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  • $\begingroup$ The more I read this, the more it makes sense to me (which is always a good thing). However, I am not sure I fully understand the notation of $F_n(ω)$ (math newbie here!) $\endgroup$ – Dr.Doofus Jan 19 '15 at 3:40
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    $\begingroup$ $\omega$ is an element of the sample space $\Omega$. The first paragraph tells us that there is a subset $A$ of $\Omega$, which has probability 1, and such that if $\omega \in A$ then $F_n(\omega) \to \mu$. The argument of the second paragraph is that $G_n \to \exp(\mu)$ on the same set $A$, and so it converges almost surely. $\endgroup$ – Ian Jan 19 '15 at 3:42
  • $\begingroup$ Ahhhhh, I see! Now when it says "identify the limit", what does it mean by that? Thanks so much for the help btw $\endgroup$ – Dr.Doofus Jan 19 '15 at 3:45
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    $\begingroup$ @Shotokan The limit is $\exp(\mu)$. You can find $\mu$ by direct calculation, since it's the expectation of $\ln(X)$ when $X$ is $U(0,1)$. $\endgroup$ – Ian Jan 19 '15 at 3:49
  • $\begingroup$ Oh, of course! Since $G_n(\omega)$ converges to $exp(\mu)$ (duh). Thanks so much for the help! You're an absolute legend. $\endgroup$ – Dr.Doofus Jan 19 '15 at 3:51

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