Let $(X_n)_{n}∈_{\mathbb{N}_+}$ be a sequence of independent random variables such that Xn ∼ Exp{n} and let
$Y_n :=\frac{1}{n} \sum_{i=1}^{n}X_{i}$ for n$\in \mathbb{N}_+$
Does the sequence ($Y_n$) converge in probability?
My intuition is that the sequence converge in probability since the exponential distribution has a finite mean but I do not know how to prove this, the exercise give me only this information
Firstly the Weak Law of Large numbers will give you the immediate answer "yes". I'd recommend reading the wikipedia page briefly. In fact, the Strong Law of Large numbers is even stronger and will give you almost sure convergence.
To prove it, note that $EY_n = 1$ for all $n$ and $Var(Y_n) = \frac{1}{n}$ for all $n$. You can show this using linearity of expecation and linearity of variance (given independent variables).
Hence:
$$\lim_{n\to \infty} P(|Y_n -1|>\epsilon) = \lim_{n\to \infty} P((Y_n -1)^2>\epsilon^2)\leq \lim_{n\to \infty} \frac{Var(Y_n)}{\epsilon^2} = \lim_{n\to \infty} \frac{1}{n \epsilon^2} = 0$$
The inequality is an application of Markov's inequality (or Chebyshev's inequality, which is a special case involving the variance).
Since that probability goes to $0$ for all $\epsilon>0$, $Y_n$ converges in probability to $1$.