Question about computing expected value of the limit of a geometric mean of random variables

Question

If I have the random variables $ X_{i} $ for $ i=1 \ldots N$ with the random variables being randomly selected integers from $1$ to $9$, how would I calculate the expected value of $$\lim_{N \to \infty} \left( \prod_{i = 1}^{N} X_{i} \right)^{\dfrac{1}{N}} . $$ This is obviously just the limit as $N$ goes to infinity of the geometric mean of N randomly selected integers from 1 to 9.

I'm wondering not only the answer but first how you would go about solve a problem like this. I haven't taken real higher-that-intro-level probability classes so my only knowledge of probability is a pretty dry and not very intensive 1st year college course on statistics and probability. Though I do have extensive math knowledge outside of that specific topic (Calculus, PDEs, Fourier Analysis, Abstract Algebra). Is there even a real number value that could be calculated from this question? I'm really just doing this out of my own curiosity (after reading this Wikipedia page on Khinchin's constant) I'm not cheating on math homework or something like that.

Answer 1

By the strong law of large numbers for i.i.d. integrable sequences, $$\frac1N\sum_{i=1}^N\log X_i\to\nu\quad\text{almost surely},$$ where $\nu=E(\log X_1)$. This is equivalent to the assertion that $$\lim_{N \to \infty} \left( \prod_{i = 1}^{N} X_{i} \right)^{1/N}=\mathrm e^\nu\quad\text{almost surely},$$ hence the expected value of the LHS exists and is $\mathrm e^\nu$. Numerically, $\mathrm e^\nu=(9!)^{1/9}\approx4.147166$.

A heuristics to understand this result is that, for every integer $1\leqslant k\leqslant9$, the number of times $T_N(k)$ that $X_i=k$ for $1\leqslant i\leqslant N$ is roughly $\frac19N$ in the sense that $T_N(k)=\frac19N+o(N)$ (this is again the strong law of large numbers, but applied once for each $k$). Thus, almost surely, $$\left( \prod_{i = 1}^{N} X_{i} \right)^{1/N}=\prod_{k=1}^9k^{T_N(k)/N}\approx\prod_{k=1}^9k^{1/9}.$$ Similarly, for every function $A$, $$\lim_{N \to \infty} \left( \prod_{i = 1}^{N} A(X_{i}) \right)^{1/N}=\left(\prod_{k=1}^9A(k)\right)^{1/9}\quad\text{almost surely}.$$

Answer 2

This is my crack at the question, please if anybody sees any flaws, point them out

I am assuming that $X_{i}$'s are independent. From this it is important to know that $$\left(\prod_{i=1}^{N}X_{i}\right)^{\frac{1}{N}}=e^{\frac{1}{N}\sum_{i=1}^{N}\ln(X_{i})}$$

Well notice that $\frac{1}{N}\sum_{i=1}^{N}\ln(X_{i})$ is just the average of the log of the $X_{i}$ and we know that as $N\rightarrow \infty$ that the average will resemble a normal distribution by Central Limit Theorem. Thus since $$E(\ln (X))=\frac{1}{9}\left(\sum_{i=1}^{9}\ln(i)\right)\approx1.422426$$ and $$Var\left(\frac{1}{N}\sum_{i=1}^{N}\ln (X_{i})\right)=\frac{V(\ln (X))}{N}=\frac{1}{N}\left(\frac{1}{9}\left(\sum_{i=1}^{9}\left(\ln(i)-1.422426\right)^{2}\right)\right)\approx\frac{0.4598559}{N}$$ Thus we have $\frac{1}{N}\sum_{i=1}^{N}\ln(X_{i})\sim N(1.422426,\frac{0.4598559}{N})$ approximately for large N. Thus this means that $e^{\frac{1}{N}\sum_{i=1}^{N}\ln(X_{i})}$ is lognormal where expected value of lognormal is $$E\left(e^{\frac{1}{N}\sum_{i=1}^{N}\ln(X_{i})}\right)=e^{E(X)+\frac{1}{2}Var(X)}=e^{1.422426+\frac{0.4598559}{2N}}$$ Now if you can prove that $$E\left(\lim_{N\rightarrow\infty}\left(\prod_{i=1}^{N}X_{i}\right)^{\frac{1}{N}}\right)=\lim_{N\rightarrow\infty}E\left(\left(\prod_{i=1}^{N}X_{i}\right)^{\frac{1}{N}}\right)$$ which is held true by dominated convergence theorem (http://web.stanford.edu/class/msande322/docs/app_B.pdf) since the geometric mean in this case will always be bounded by arithmetic mean which has a finite expectation which would just be $E(X)$. Thus finally the expectation should just be $$\lim_{N\rightarrow\infty}E\left(\left(\prod_{i=1}^{N}X_{i}\right)^{\frac{1}{N}}\right)=\lim_{N\rightarrow\infty}e^{1.422426+\frac{0.4598559}{2N}}=e^{1.422426}\approx4.147169$$