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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.

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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.

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