## Qualifying Exam Day I

Consider an infinite sequence of independent Bernoulli trials $X_1, X_2, \cdots.$ Suppose $\mathbb{P}(X_i=1)=1/4^i$, show:

1. $\mathbb{P}(\sum_{i=n}^{\infty}X_i > 0) \le \frac{1}{3\times 4^{n-1}}$.
2. Use the previous result to show $\mathbb{P}(\sum_{i=1}^{\infty}X_i < n) \ge 1-\frac{1}{3\times 4^{n-1}}$.
3. Find $\mathbb{P}(\sum_{i=1}^{\infty}X_i < \infty)$.
4. Show that $\mathbb{P}(\sum_{i=1}^{\infty}X_i = 0) < e^{-1/3}$.

solution

1. By Markov Inequality, we have
\begin{align}
LHS &= \mathbb{P}(\sum_{i=n}^{\infty}X_i \ge 1) \le \mathbb{E}(\sum_{i=n}^{\infty}X_i) = \sum_{i=n}^{\infty} \frac{1}{4^i} \\
&= \lim_{m\rightarrow \infty} \frac{1}{4^n} \frac{1-(1/4)^{m-n-1}}{1-(1/4)} = RHS
\end{align}
2. It follows from:
$$\mathbb{P}(\sum_{i=1}^{\infty}X_i < n) \ge \mathbb{P}(\sum_{i=n}^{\infty}X_i = 0)$$
3. Sending $n$ to $\infty$, we obtain that the probability is 1. Alternatively, we can use Borel Cantelli. Since the probabilities are summable, we have $\mathbb{P}(\{X_i=1 \; i.o.\}) = 0$. And it follows that $\exists \, \Omega_0$ with $\mathbb{P}(\Omega_0)=1$, and $\forall \, \omega \in \Omega_0$, we can find a large number $M$ such that $X_i(\omega)=0$ for all $i \ge M$, implying $\sum_{i=1}^{\infty}X_i(\omega) < M < \infty$.
4. Using the fact $1-x < e^{-x}$ for $x>0$, we have
\begin{align}
LHS &= \prod_{i=1}^\infty \mathbb{P}(X_i=0) = \prod_{i=1}^{\infty} \big( 1-\frac{1}{4^i} \big) \\
&< \prod_{i=1}^{\infty} \big( e^{-1/4^i} \big) = e^{\sum_{i=1}^\infty -1/4^i} =RHS
\end{align}