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!


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 $1\leqslant Y_n\leqslant e$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.