Prove strong law of large numbers assuming bounded fourth moment

Question

Let $X_{1},X_{2},\dots$ be independent random variables satisfying $E(X_{n}^{4}) < B$ for some finite $B > 0$ and $E(X_{i})$ $\rightarrow\mu$.

a) Show that $Y_{n} = X_{n} -E(X_{n})$ are independent and $E(Y_{n}) = 0$, $E(Y_{n}^2)<\sqrt{B}$, $E(Y_{n}^4)<16B$.

b) Show that for $ \bar Y_{n} = (Y_{1}+\dots+Y_{n})/n$, $E(\bar Y_{n}^4) = \frac{1}{n^4}\sum_{i=1}^{n} E(Y_{i}^4) +\frac{6}{n^4}\sum_{i\neq j}E[Y_{i}^2 Y_{j}^2] \leq \frac{16B}{n^3} + \frac{6B}{n^2}$

c) Show that $\sum_{n=1}^{\infty} P(\bar Y_{n} > \epsilon) < \infty$ and conclude $\bar Y_{n} \rightarrow 0$ almost surely.

d) Show that $(\mu_{1}+\dots+\mu_{n})/n \rightarrow \mu$ and finish the proof of $\bar X_{n} \rightarrow \mu$ almost surely

I can handle question a) until proving $E(Y_{n})=0$ but have no idea about the rest of the whole problem. I think c) and d) might have something to do with law of large numbers but don't know how to proceed. Can someone give some detailed hint about the rest of a), b), c) and d) ?

Thanks!

Answer 1

No you are not supposed to assume SLLN, but to prove it in this case, and I assume you mean $\mathbb{E}X_i=\mu_i\to\mu$ instead of $\mathbb{E}(\lvert X_i\rvert)\to\mu$.

(a) Use $\mathbb{E}X_n^4<B$ to bound $\mathbb{E}X_n^2$ and hence $\operatorname{Var}X_n$. Similarly expand $\mathbb{E}Y_n^4$ and use e.g., Cauchy-Schwarz.

(b) Expand $\overline{Y_n}^4$ and use (a)

(c) Markov's inequality, and remember Borel-Cantelli.

(d) Cesaro summation in $\mathbb{R}$, together with (c).

Answer 2

Hint on b)

$\overline{Y}^{4}=n^{-4}\left(\sum_{i=1}^{n}Y_{i}\right)^{4}=n^{-4}\sum_{i_{1}=1}^{n}\sum_{i_{2}=1}^{n}\sum_{i_{3}=1}^{n}\sum_{i_{4}=1}^{n}Y_{i_{1}}Y_{i_{2}}Y_{i_{3}}Y_{i_{4}}$ so that

$\mathbb{E}\overline{Y}^{4}=n^{-4}\sum_{i_{1}=1}^{n}\sum_{i_{2}=1}^{n}\sum_{i_{3}=1}^{n}\sum_{i_{4}=1}^{n}\mathbb{E}\left[Y_{i_{1}}Y_{i_{2}}Y_{i_{3}}Y_{i_{4}}\right]$

If $i_{1}\notin\left\{ i_{2},i_{3},i_{4}\right\} $ then $\mathbb{E}\left[Y_{i_{1}}Y_{i_{2}}Y_{i_{3}}Y_{i_{4}}\right]=\mathbb{E}\left[Y_{i_{1}}\right]\mathbb{E}\left[Y_{i_{2}}Y_{i_{3}}Y_{i_{4}}\right]=0\mathbb{E}\left[Y_{i_{2}}Y_{i_{3}}Y_{i_{4}}\right]=0$.

The same story for $i_{2}\notin\left\{ i_{1},i_{3},i_{4}\right\} $ and $i_{3}\notin\left\{ i_{1},i_{2},i_{4}\right\} $ and $i_{4}\notin\left\{ i_{1},i_{2},i_{3}\right\} $.

So we have $\mathbb{E}\left[Y_{i_{1}}Y_{i_{2}}Y_{i_{3}}Y_{i_{4}}\right]=0$ if set $\left\{ i_{1},i_{2},i_{3},i_{4}\right\} $ has cardinality $3$ or $4$.