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

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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*}

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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.

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