Is my proof for $Y_n \overset{\text{$\Bbb P$}}\longrightarrow Y$ correct ? Expected Value and Variance are given

Let $Y$ be a real random variable and $Y_n:=Y+X_n$, with $(X_n)_{n \in \Bbb N}$ a sequence of real random variables with $$\Bbb E[X_n]=\frac{1}{n}, Var[X_n]=\frac{\sigma^2}{n} $$
with $\sigma>0.$

Proof: $ Y_n \overset{\text{$\Bbb P$}}\longrightarrow Y$

Now this is my attempt, I look for advice, for corrections, etc. thanks in advance:

Let $\epsilon >0$. Than $$\begin{align}\Bbb P(|Y_n-Y|\geq \epsilon)&=\Bbb P(|Y_n-(Y_n-X_n)|\geq \epsilon)\\ &=\Bbb P(|X_n|\geq \epsilon)\\ &=\Bbb P(|X_n-0|\geq \epsilon)\\ &=1-\Bbb P(X_n\in (0,\epsilon))\\ \end{align}$$ Assume that $\sigma<\infty$, than Chebychev gives us $$\Bbb P(|X_n-\frac{1}{n}|\geq \epsilon)\leq\frac{Var(X_n)}{\epsilon^2}=\frac{\sigma^2}{n\epsilon^2} \overset{\text{$n \to \infty$}}\longrightarrow 0$$

Solution 1:

I think you are missing some nuances. I would argue that, \begin{align*} \lim_{n\rightarrow \infty}\mathbb{P}(\lvert X_n\rvert \geq \epsilon) &= \lim_{n\rightarrow \infty}\mathbb{P}(\lvert X_n\rvert - \tfrac{1}{n} + \tfrac{1}{n} \geq \epsilon),\\ &\leq \lim_{n\rightarrow \infty}\mathbb{P}(\lvert X_n - \tfrac{1}{n}\rvert \geq \epsilon - \tfrac{1}{n}),\\ &\stackrel{cc}{\leq} \lim_{n\rightarrow \infty}\left(\frac{\sigma^2}{n\left(\epsilon - \tfrac{1}{n}\right)^2}\right) = 0. \end{align*}

Since you already showed that $\mathbb{P}(\lvert Y_n - Y\rvert \geq \epsilon) = \mathbb{P}(\lvert X_n\rvert \geq \epsilon)$, we have by definition convergence in probability.

Also Your last line of the first block, i.e. $1-\mathbb{P}(X_n \in (0,\epsilon))$ is, I believe, not necessarily correct. Since, it is not stated that $X_n \geq 0$.