# Tom Chen

##### [cv][google scholar] Ph.D. Student, expected 2020

Biostatistics Department

Harvard T.H. Chan School of Public Health

## Publications

Selected ones. Full list here

## Question of the Day

More questions? here!

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 $\mathbb{P}(\sum_{i=1}^{\infty}X_i = 0) < e^{-1/3}$.

Harvard Biostatistics Qualifying Exam (Jan 2017)

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}