Let $(X_n)_n$ be a sequence of independent random variables such that:
$\mathbb P(X_n=1)=\mathbb P(X_n=-1)=\frac{1}{2}(1-2^{-n})$
$\mathbb P(X_n=2^{\frac{n}{2}})=\mathbb P(X_n=-2^{\frac{n}{2}})=2^{-n-1}$
a) check whether $|X_n| \overset{\mathbb P}{\to} 1 $
b) check whether $(X_n)_n$ satisfies WLLN and SLLN
In general I can tackle these kind of exercises, but in this case I have my random variable defined by probabilities and none of them equals to 1 as $n \to \infty$ and I'm a bit lost.
EDIT|
My approach:
WLLN:
$E[X_n]=1 \cdot \frac{1}{2}(1-2^{-n})-1 \cdot \frac{1}{2}(1-2^{-n})+ 2^\frac{n}{2} \cdot 2^{-n-1}-2^\frac{n}{2} \cdot 2^{n-1}=0$
$E[X_n^2]=1 \cdot \frac{1}{2}(1-2^{-n})+1 \cdot \frac{1}{2}(1-2^{-n})+ 2^n \cdot 2^{-n-1}+2^n \cdot 2^{-n-1}=2-2^{-n}$
$Var[X_n] = 2-2^{-n}$
$\underset{n \to \infty}{\lim} \frac{Var[\sum X_k]}{n^2} \overset{IND.}{=} \underset{n \to \infty}{\lim} \frac{\sum Var[X_k]}{n^2} =\underset{n \to \infty}{\lim} \frac{\sum 2-2^{-k}}{n^2} \leq {\lim} \frac{n(2-2^{-n})}{n^2} =0$
SLLN:
$\sum \frac{2-2^{-n}}{n^2}$ converges so $X_n$ satisfies SLLN.
Is this alright?
Best Answer
a) We need to establish whether for any $\epsilon > 0$, $ \mathbb P( | |X_n| - 1 | > \epsilon) \to 0 $ as $n \to \infty$.
We have $\mathbb P ( | |X_n| - 1 |> \epsilon) = \mathbb P( |X_n| = 2^{\frac{n}{2}}) = \frac{1}{2^n}$, which clearly tends to $0$ as $n$ tends to $\infty$.
b) Define $Y_n$, such that $Y_n = 1$ when $X_n \ge 0$, $Y_n = -1$ otherwise.
Then $\mathbb P( X_n \neq Y_n) = \mathbb P( |X_n| = 2^{\frac{n}{2}} ) = \frac{1}{2^n}$, so by borel cantelli ( since $\sum_{n=1}^\infty \mathbb P(X_n \neq Y_n) < \infty $ ), we have that with probability $1$ only finitelly many events $X_n$ are different from $Y_n$. Call this set of measure $1$ by $\Omega_0$
But $(Y_n)$ is sequence of independent rademacher random variables, for which we have $\mathbb E[Y_k] = 0$, so by SLLN we have: $ \frac{Y_1 + ... + Y_n}{n} \to 0$ almost surely, so in probability, too.
Now, by observation above, take any $\omega \in \Omega_0$. Then we have $0 = \lim_{n \to \infty} \frac{\sum_{k=1}^n Y_k(\omega)}{n} = \lim_{n \to \infty} \frac{\sum_{k=1}^n X_k(\omega)}{n}$ and we get that $X_n$ also satisfies WLLN and SLLN