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I'm struggling with the following problem:

Let $(\Omega, \mathcal{A}, \mathbb{P})$ be a probability space, $(X_n)_{n \in \mathbb{N}} \subseteq \mathcal{L}^0(\mathbb{P};\mathbb{R})$ i.i.d with: $$ \forall F \in \mathcal{B}(\mathbb{R}): \; \mathbb{P}(X_n \in F) = \int_F 1_{[1,e]} \frac{1}{z} \lambda^1(dz) $$ (Where $ 1_{[1,e]}$ is the indicator function). For all $n \in \mathbb{N}$ define $$ Y_n := ( \prod\limits_{k=1}^n X_k )^{1/n} $$

a) Show that for all $n \in \mathbb{N}$ $Y_n$ is $ \mathcal{A}- \mathcal{B}(\mathbb{R})$-measurable

I didn't have any issues solving this problem.

b) Show that $(Y_n)_{n \in \mathbb{N}}$ converges almost surely against 1.648721....

I tried using the law of large numbers. I calculated $$ \log((Y_n)^n) = \sum\limits_{k=1}^n \log(X_k) $$ The sequence $(\log(X_k))_{k \in \mathbb{N}}$ is still i.i.d. , thus for $k \in \mathbb{N}$ I tried to calculate the expected value $$ \mathbb{E}[\log(X_k)] = \int_\mathbb{R} \log\left(1_{[1,e]} \frac{1}{z}\right) \lambda^1(dz) $$ But when I tried to calculate this integral, it is undefined. Where did I go wrong? Another approach I thought about was trying to use the almost completely convergence, as in $$ \sum\limits_{n=1}^\infty \mathbb{P}( |Y_n - Y| > \epsilon) < \infty $$ for any $\epsilon \in (0,\infty)$, in order to imply almost sure convergence. I thought about this idea because the limit is given incompletely as a decimal representation, and not as (most likely) $\sqrt{e}$. But I can't make any significant progress with that idea

c) Show that $(Y_n)_{n \in \mathbb{N}}$ is uniformly integrable.

I tried using $L^1$-convergence to conclude that $(Y_n)_{n \in \mathbb{N}}$ is uniformly integrable. But here I encounter the same problem as in b), that I can't seem to solve the integral at hand and can't calculate the expectation.

I would be very thankful for any advice or progress to the solution!

vitamin d
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Bro Hman
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1 Answers1

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Actually, $$\mathbb{E}[\log(X_k)] = \int_\mathbb{R}1_{[1,e]} \log\left(z\right)\frac 1z \lambda^1(dz)$$ and the integrand is the primitive of $(\log z)^2/2$ hence $\mathbb{E}[\log(X_k)]=1/2 $. Then the law of large numbers show that $\log Y_n\to 1/2$ almost surely.

For uniform integrability, it is a good idea to show convergence in $\mathbb L^1$. Since $Y_n\to \sqrt e$ almost surely and $Y_n$ is non-negative, it suffices to show that $\mathbb E[Y_n]\to \sqrt e$. As $\mathbb E[Y_n]$ can be explicitely computed (indeed, $\mathbb E[Y_n]=\left(n\left(e^{1/n}-1\right)\right)^n$, this becomes now a calculus problem.

Or in a simpler way: $1\leqslant X_i\leqslant e$ for each $i$ hence $1\leqslant Y_n\leqslant e$.

Davide Giraudo
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