Find $a$ such that $X_n \xrightarrow[n\rightarrow \infty]{ P } a $, where $X_n = n^{-1} \sum^n_{i=1} Y_i $, where the $Y_i \sim Exp(2)$ are independent random variables.
I tried: \begin{equation*}\begin{split} \lim _{n \rightarrow \infty} P(|X_n - a| \ge \varepsilon) = & \, 1 - \lim _{n \rightarrow \infty} P\left(|X_n - a| \leq \varepsilon \right) \\[8pt] = & \, 1 - \lim _{n \rightarrow \infty} P\left(\left|\frac{\sum Y_i}{n} - a\right| \leq \varepsilon \right) \\[8pt] = & \, 1 - \lim _{n \rightarrow \infty} P\left(-\varepsilon \leq\frac{\sum Y_i}{n} - a \leq \varepsilon \right) \\[8pt] = & \, 1 - \lim _{n \rightarrow \infty} P\left(n(a-\varepsilon) \leq \sum Y_i \leq n(a + \varepsilon) \right) \\[8pt] = & \, 1 - \lim _{n \rightarrow \infty} P\left( \sum Y_i \leq n(a + \varepsilon) \right) % + \lim _{n \rightarrow \infty} P\left( \sum Y_i \leq n(a-\varepsilon)\right) \\[8pt] \end{split}\end{equation*} here we have iid, thus $$P\left( \sum Y_i \leq n(a+\varepsilon) \right) = P( Y_1 \leq n(a+\varepsilon), Y_2 \leq n(a+\varepsilon),...)= [P(Y \leq n(a+\varepsilon) )]^n $$ similarly, $$P\left( \sum Y_i \leq n(a-\varepsilon) \right) = P( Y_1 \leq n(a-\varepsilon), Y_2 \leq n(a-\varepsilon),...)= [P(Y \leq n(a-\varepsilon) )]^n $$ returning to first equation, \begin{equation*}\begin{split} =& \, 1-\lim _{n \rightarrow \infty} [P(Y \leq -2n(a+\varepsilon) )]^n + \lim _{n \rightarrow \infty} [P(Y \leq -2n(a-\varepsilon) )]^n\\[8pt] =& \, 1-\lim _{n \rightarrow \infty} \left[e^{-2n(a+\varepsilon)}\right]^n + \lim _{n \rightarrow \infty} \left[e^{-2n(a-\varepsilon)}\right]^n\\[8pt] =& \, 1+\lim _{n \rightarrow \infty} e^{-2n^2(a+\varepsilon)} - \lim _{n \rightarrow \infty} e^{-2n^2(a-\varepsilon)}\\[8pt] =& \, 1+\lim _{n \rightarrow \infty} e^{-2n^2(a+\varepsilon)} - \lim _{n \rightarrow \infty} e^{-2n^2(a-\varepsilon)}\\[8pt] =& \, 1+0 -0\\[8pt] =& \, 1\\[8pt] \end{split}\end{equation*} Where did I go wrong?