# Almost Sure Convergence in $L^{p}$

Let $(X_n)_{n\geq 1}$ be a sequence of i.i.d. random variables, on the same probability space, with law given by $\displaystyle \mathbb P(X_1=(-1)^{m}m)=\frac{1}{(cm^2\log m)}$ for $m\geq 2$ where $c$ is the normalization constant $\displaystyle c=\sum_{m\geq 2}\frac{1}{m^2\log m}$. Prove that $\mathbb E(|X_1|)=\infty$, but there exists a constant $\mu \not\in \{\pm\infty\}$ such that $\displaystyle (n^{-1}\sum_{k=1}^{n}X_{k})_{n\geq 1}$ converges to $\mu$ in probability. Does it converge almost surely, and in $L^{p}$?

Integral test to show $$\mathbb E(|X_1|)=\infty$$

$$\mathbb E(|X_1|)=\sum_{m=1}^{\infty}|\left((-1)^mm\right)\left(\frac{1}{cm^2\log(m)}\right)|\propto\sum_{m=1}^{\infty}\frac{1}{m\log(m)}$$

Which is on the order of $\displaystyle \int_{1}^{\infty}\frac{1}{x\log(x)}dx=\int_{1}^{\infty}\frac{1}{u}=\infty$, $u=\log(x)$

Hence the expectation diverges.

We have the result:

Let $S_n=X_1+\ldots+X_n$, where $X_i$ are iid. Then $\frac{S_n}{n}-\mu_n$ converges in probability to 0 for some $\mu_n$ if and only if $\lim_{x\rightarrow\infty} x\mathbb P(|X_1|>x)\rightarrow 0$

Then:

$$\lim_{x\rightarrow \infty} x\mathbb P(|X_1|)>x\rightarrow 0=\lim_{k\rightarrow\infty}\left[k\left(\sum_{m=k}^{\infty}\frac{1}{m^2\log(m)}\right)\right]$$

$$\sim k\int_{x=k}^{\infty}\frac{1}{x^2\log(x)}dx$$

$$=k\int_{\log(k)}^{\infty}\frac{1}{u e^{u}}du,\hspace{5mm} u=\log(x)$$

$$\leq k\frac{1}{\log(k)}\int_{\log(k)}^{\infty}\frac{1}{e^{u}}du$$

$$=\frac{k}{log(k)}\times e^{-\log(k)}=\frac{k}{\log(k)}\frac{1}{k}$$

Where the $\frac{1}{k}$ cancels with the $k$,

And then $\displaystyle \lim_{k\rightarrow\infty}\frac{1}{\log(k)}\rightarrow 0$.

We do not have $L^{p}$ convergence because $\mathbb E(A_n)=\infty$.

Let $S_n=X_1+\ldots+X_n$, where $X_i$ are iid. Then $\frac{S_n}{n}-\mu_n$ converges in probability to 0 for some $\mu_n$ if and only if $\lim_{x\rightarrow\infty} xP(|X_1|>x)\rightarrow 0$
To investigate almost sure convergence, play around with Borel Cantelli Lemma looking at $|S_n/n-\mu_n|>\epsilon$ infinitely often. In particular, take a look at $X_n/n>\epsilon$ infinitely often.
As far as $L_p$ convergence goes, generally $L_p$ convergence implies convergence in probability from something like Chebyshev's inequality, which you don't have since there is no first (and therefore no second) moment. Yet, you do have convergence in probability. What are the ways that $L_p$ convergence can fail in this case, if any?
• Questions: How do I prove the statement in the highlighted box? Also how do we prove the Borel Cantelli Lemma for the statements? In the case of $\frac{X_n}{n}>\epsilon$, Borel Cantelli's convergence requirements don't seem to be met. Commented Apr 16, 2013 at 4:44