Convergence of Uniformly Distributed Random Variables (n-dimensional)

Question

Suppose that ${U_n} = ({U_{n1}},{U_{n2}},...,{U_{nn}})$ is uniformly distributed over the n-dimensional cube ${C_n}={[0,2]^n}$ for each $n=1,2,...$ That is, that the distribution of ${U_n}$ is ${Q_n}(dx) = {2^{-n}} {{\chi}_{C_{n}(x)}}{m_n}(dx)$, where ${m_n}$ is n-dimensional Lebesgue measure. Let ${X_n} = ({U_{n1}}{U_{n2}}...{U_{nn}}), n \geq 1$. Show that:

a) ${X_n} \rightarrow 0$ in probability as $n \rightarrow \infty$, and

b) $\{{X_n} : n > 1\}$ is not uniformly integrable.

Attempt at a solution:

We want to show that, for each $\epsilon > 0$, $P(|{X_n} - X|>\epsilon) \rightarrow 0$ as $n \rightarrow \infty$. Now, $$P(|{X_n}|>\epsilon) = 1-\int {\frac{1}{{{2^n}}}} {\chi _{[{X_n} < \varepsilon ]}}{m_n}(dx)$$ so then $$P(|{U_{n1}}{U_{n2}}...{U_{nn}}|>\epsilon) = 1- \int {\frac{1}{{{2^n}}}} {\chi _{[{X_n} < \varepsilon ]}}{m_n}(dx)$$ $$=1- \frac{{{\varepsilon ^n}}}{{{2^n}}}$$ which is where I'm stuck. Is the math up to here correct? I'm just beginning to study probability, so I know that there are some similar questions on this site but the answers involve martingales and other concepts that we haven't reached yet. I'm also pretty shaky on my measure theory, so there is probably a nuance somewhere in the integral calculation that I'm missing.

For part b, I'm really not even sure where to start.

Answer 1

A way to show convergence in probability to $0$ is to show that $\mathbb E\left[\sqrt{X_n}\right]\to 0$. This can be done in the following way: we observe that $$ \mathbb E\left[\sqrt{X_n}\right]=2^{-n}\int_{[0,2]^n}\prod_{i=1}^n\sqrt{x_i}\mathrm dx_1\dots\mathrm dx_n=\left(\frac 12\int_0^2\sqrt u\mathrm du\right)^n $$ and since $\frac 12\int_0^2\sqrt u\mathrm du=2^{3/2}/3\lt 1$, it follows that $\mathbb E\left[\sqrt{X_n}\right]\to 0$.

Observe that $\mathbb E[X_n]=1$. If the sequence $\left(X_n\right)_{n\geqslant 1}$ was uniformly integrable, then by item 6 of this answer, we would deduce from the convergence in probability to $0$ that $\mathbb E[X_n]\to 0$, which is not the case as $\mathbb E[X_n]=1$.