# Convergence in Probability (Analytical Solution Verification)

Problem: Let $$X_1,X_2,\cdots$$ be independent random variables that are uniformly distributed over $$[-1,1]$$. Show that the sequence $$Y_1,Y_2,\cdots$$ converges in probability to some limit, and identify the limit for

$$\begin{equation*} Y_n=\dfrac{X_n}{n} \end{equation*}$$

My-Solution I: $$X_n \sim \text{Uni}[-1,1]$$

Conjecture: As $$n \longrightarrow \infty$$ then $$\frac{X_n}{n} \longrightarrow 0$$ therefore $$Y_n$$ converges to $$0$$ in probability.

Definition Convergence in Probability: Let $$Y_1,Y_2,\cdots$$ be a sequence of random variables (not necessarily independent), and let $$a$$ be a real number. We say that the sequence of $$Y_n$$ converges to $$a$$ in probability, if for every $$\epsilon > 0$$, we have $$\begin{equation*} \displaystyle \lim_{n \to \infty} \mathbb{P}(|Y_n - a|) \geq \epsilon)=0 \end{equation*}$$

In our case $$a=0$$ and for $$\epsilon >0$$ we have $$\mathbb{P}(|Y_n - 0|) \geq \epsilon)$$

Using derived distributions

\begin{align*} F_{Y_n}(y_n)&=\mathbb{P}(Y_n \leq y_n)\\ &=\mathbb{P}\bigg(\frac{X_n}{n} \leq y_n\bigg)=\mathbb{P}(X_n \leq ny_n)\\ &= \frac{ny_n}{2}-\frac{1}{2} \end{align*}

differentiating with respect to $$y_n$$, we get

$$\begin{equation*} f_{Y_n}(y_n)= \begin{cases} \frac{n}{2}, & \ \text{if}\quad y_n \in [-\frac{1}{n},\frac{1}{n}]\\ 0, & \ \text{otherwise} \\ \end{cases} \end{equation*}$$

From the PDF of $$Y_n$$ it is clear when $$\epsilon > \frac{1}{n}$$

\begin{align*} \mathbb{P}(|Y_n-0| \geq \epsilon)&=\mathbb{P}(|Y_n| \geq \epsilon) \\ &= \mathbb{P}(Y_n \geq \epsilon)+\mathbb{P}(Y_n \leq -\epsilon)\\ &= 0+0=0 \end{align*}

for all $$n$$ with $$\epsilon > 0$$ and $$\epsilon > \frac{1}{n}$$, so $$\mathbb{P}(|Y_n|\geq \epsilon) \longrightarrow 0$$. Hence $$Y_n$$ converges to zero as per definition.

Solution II:

\begin{align*} \mathbb{P}(|Y_n-0| \geq \epsilon) &= \mathbb{P}(|Y_n| \geq \epsilon)\\ &= \mathbb{P}\bigg(\Bigg\vert\frac{X_n}{n}\Bigg\vert \geq \epsilon\bigg)\\ &= \mathbb{P}(|X_n| \geq n\epsilon)\\ &= \mathbb{P}(X_n \geq n\epsilon)+\mathbb{P}(X_n \leq -n\epsilon)\\ &= 0+0=0 \quad \text{as}\: \epsilon > 0 \:\text{with}\: n \rightarrow \infty \end{align*} So $$\mathbb{P}(|Y_n| \geq \epsilon) \longrightarrow 0$$ with $$\epsilon > 0$$ and $$\epsilon > \frac{1}{n}$$ for all $$n$$.

Manual Solution: For any $$\epsilon > 0$$, we have $$\begin{equation*} \mathbb{P}(|Y_n| \geq \epsilon) = 0, \end{equation*}$$ for all $$n$$ with $$\epsilon > \frac{1}{n}$$, so $$\mathbb{P}(|Y_n| \geq \epsilon) \longrightarrow 0$$.

Does my solutions make sense or not. Kindly validate the solutions.