Law of Large Numbers and Bernoulli Distribution

Question

Let ${X_i}$ be a sequence of random variables which follows the Bernoulli distribution with $Bernoulli(\frac1i)$, for each $i = 1,...,n$. In other words, $P(X_i=1) = \frac 1i$ and $P(X_i=0) = 1 - \frac 1i$ for each $i = 1,...,n$. Show that $X_n$ tends in probability to $0$.

I have no idea how to even get started on this, so any tips would help greatly.

My lecturer has just covered the Law of Large Numbers, convergence in probability and what makes a good estimator (unbiased, efficient and consistent) but I think I speak for the whole class when I say that we understood little of what she was teaching.

EDIT

I would first like to thank the kind souls who gave me hints on how I should go about solving this and also vetting what I had come up with.

This is my answer:

To show $X_n \to 0$, we should aim to show $P(|X_n - 0| > \epsilon) = 0$ for any positive $\epsilon$. Then, since $X_n$ follows a Bernoulli Distribution, so $X_n$ can only take on non-negative integer values. Thus, $P(|X_n - 0| > \epsilon) = P(X_n > \epsilon)$. Now, I make use of Markov's Inequality, so $P(X_n > \epsilon) \leqslant \frac 1\epsilon E(X_n) = \frac 1\epsilon \frac 1n$. As $n \to \infty$, $\frac 1n \to 0$, so $P(|X_n - 0| > \epsilon) = 0$ and we can conclude that $X_n$ tends in probability to $0$.

Any further comments on how I can improve this answer are welcome :) or if anyone has any other more elegant ways to answer the question, they are welcome too!

Answer 1

As a hint, follow through your comment where you said

this is how my lecturer described "convergence in probability" - for any positive $ϵ$, the probability that $Z_n$ differs from $c$ by more than $ϵ$ will eventually decrease to $0$ as $n$ tends to infinity.

In this case $c=0$ and $Z_n=X_n$. So

• For any $ϵ$ with $0 \lt ϵ \lt 1$, what is the probability that $X_n$ differs from $0$ by more than $ϵ$?
• What happens to that probability as $n$ tends to infinity?