I need some help with the following problem related to convergence of a sequence of random variables:
Suppose we have a biased coin with probability $p$ of coming up heads and probability $q = 1 − p$ of tails. Let $X_{n+1}=X_n \pm 1$ with "$+$" in case of "heads" and "$−$" in case of "tails". Let $Y_n=(q/p)^{X_n}$
How do I show that:
- $Y_n$ converges almost surely to $0$
- Determine whether $Y_n$ converges to $0$ in $L^1$.
- Determine whether the sequence $|Y_n - Y_{n-1}|$ converges in probability
Thanks!
Let $(Z_n)_{n\in\Bbb N}$ i.i.d with $$P(Z_n = 1) = p = 1-q = 1-P(Z_n = -1)$$
Then $E[Z_n] = p-q, X_n = \sum\limits_{k=1}^n Z_k$ and the strong law of large numbers gives us $$\lim_{n\to\infty} \frac{X_n}{n} = \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n Z_n = E[Z_1] = p-q$$
1.)
If $q<p$ this implies $$\lim_{n\to\infty} X_n = +\infty \quad \text{and} \quad \frac{q}{p} < 1$$ hence $$\lim_{n\to\infty} Y_n = 0$$
If $p<q$ this implies $$\lim_{n\to\infty} X_n = -\infty \quad \text{and} \quad \frac{p}{q} < 1$$ hence $$\lim_{n\to\infty} Y_n = 0$$
If $p=q=\frac{1}{2}$ it holds $Y_n \equiv 1$ so $Y_n \not\to 0$ but $Y_n \to 1$
2.)
In all 3 cases above $Y_n$ is dominated by $1$ so dominated convergence theorem gives us also $L^1$ convergence
3.)
$\Bbb P$-a.s convergence and $L^1$ convergence imply convergence in probability.