# Convergence in probability question clarification

Suppose you have a sequence of independent random variables {$$X_i, i \geq 1$$} where: $$\Bbb P(X_i = 1 )=1-\frac{1}{i}, \Bbb P(X_i =i )=\frac{1}{i}.$$ Let $$Y_n=\frac{1}{n}X_n.$$ How would you show that $$Y_n$$ converges to $$0$$ in probability? i.e $$\lim_{n \to \infty}\Bbb P(|Y_n| \gt \epsilon)=0$$ $$\forall \epsilon \gt0.$$

For a similar question, I had the distribution for $$X_i$$ [$$\Bbb P(X_i = \sqrt{i} )= \Bbb P(X_i = -\sqrt{i} )= \frac{1}{i+1}$$ and $$\Bbb P(X_i=0)=1-\frac{2}{i+1}$$ ] and where $$Y_n=\frac{1}{n}\sum_{i=1}^{n}X_i$$. I calculated $$E(Y_n)=0, Var(Y_n)=\frac{2i}{n(i+1)}$$. Therefore by Chebyshev's inequality, I obtained that the probability $$\Bbb P(|Y_n|\gt\epsilon)\leq$$Var($$X$$)/$$\epsilon^2$$= $$\frac{2i}{\epsilon^2n(i+1)}$$. Since Var($$X$$)/$$\epsilon^2$$ converges to $$0$$, the probability converges to $$0$$ and we say that $$Y_n \to 0$$ in probability.

However, that approach became particularly messy with the first question above.

## 1 Answer

Let $$\epsilon > 0$$ and $$n>1+\lfloor\frac{1}{\epsilon}\rfloor$$ ($$\implies n\epsilon >1$$). You note that $$\mathbb{P}(|Y_n|>\epsilon) = \mathbb{P}(X_n >n\epsilon) = \mathbb{P}(X_n = n) = \frac{1}{n}$$ for all $$n>1+\lfloor\frac{1}{\epsilon}\rfloor$$ and taking $$n\to\infty$$, you have done.