# Convergence in p of bounded sequence of random variable implies convergence in rth mean

This is a homework question for a course in asymptotic statistics.

Definitions.

Let $X_1, X_2, \ldots$ be sequence of random variables with $|X_n| < B$ almost surely for all $n$.

Let $X_n \xrightarrow[]{p} X$ denote convergence in probability: $\lim_{n\to \infty} P(|X_n -X| > \epsilon ) = 0.$

Let $X_n \xrightarrow[]{r} X$ denote convergence in the rth mean: $\lim_{n\to \infty} E(|X_n -X|^r) = 0.$

Let $r > 0$.

Problem.

Show that $X_n \xrightarrow[]{p} X$ if and only if $X_n \xrightarrow[]{r}X$.

Work.

I have a proof for the if part, but not for the "only if" part: $$X_n \xrightarrow[]{r} X \Leftarrow X_n \xrightarrow[]{p} X.$$

I tried to show that $E|X_n - X|^r \leq P(|X_n-X| \geq \epsilon)$.

First: I tried expanding $|X_n - X|$ using the binomial theorem. Could only find a bound in terms of a sum over $B$.

Second: Tried writing the expectation in terms of a probability density integral. Again I can only find a bound in terms of $B$.

Third: I tried splitting the $E|X_n - X|^r$ using indicator functions. I ended up getting that $E|X_n - X|^r \geq E|X_n - X|^r$. I don't see how to do this better. I feel like it should involve $B$ somehow.

Status.

I am very stuck.

• When $|X_n-X|<\epsilon$ fails, which it might sometimes, $B^r$ bounds the difference $|X_n-X|^r$. The hit you take (in bounding $E|X_n-X|^r$) from this is bounded by $B^r$ times the chance that $|X_n-X|\ge \epsilon$ . – kimchi lover Sep 23 '17 at 21:08

First off, notice that we have $|X| \leq B$ almost surely. Let $\epsilon > 0$. We find \begin{align*} \mathbb{E}[|X_n - X|^r] &= \int_{|X_n -X| < \epsilon} |X_n -X|^r dP + \int_{|X_n -X| > \epsilon}|X_n -X|^r dP \\ &\leq \epsilon^r + \int_{|X_n -X| > \epsilon} (2B)^r dP = \epsilon^r + (2B)^r \cdot P(|X_n -X| > \epsilon) \end{align*} Taking the lim sup with respect to $n$ of both sides, we obtain $$\limsup_{n\to\infty} \mathbb{E}[|X_n -X|^r] \leq \epsilon^r$$ for any $\epsilon > 0$, which proves the claim.