Prove the convergence in probability of $\bar{X}_{n}=\frac{1}{n} \sum_{i=1}^{n} X_{i}$, with iid X_i

Question

I need to show that: $\bar{X}_{n} \rightarrow^{p} 1+\theta$ and that $min(X_i) = X_{(1)} \longrightarrow^{p} \theta$.

Here $X_{1}, X_{2} \ldots$ is a sequence of iid random variables with p.d.f $f(x)=e^{-x+\theta}, x \geq \theta,\theta>0$, and $\bar{X}_{n}=\frac{1}{n} \sum_{i=1}^{n} X_{i}$.

I tried to apply the definition of convergence in probability, but I didn't know how to move forward.

Answer 1

First observe that the expectation of your rv is

$$\mathbb{E}[X]=e^{\theta}\int_{\theta}^{\infty}xe^{-x}dx=\dots=\theta+1$$

Using the Strong Law of Large Numbers you get that

$$\overline{X}_n\xrightarrow{a.s.}\theta+1$$

Thus also

$$\overline{X}_n\xrightarrow{\mathcal{P}}\theta+1$$

Second, the CDF of $U=X_{(1)}$ is the following

$$F_U(u)=1-e^{n(\theta-u)}$$

Thus

$$\lim\limits_{n \to \infty}\mathbb{P}[\theta<U<\theta+\epsilon]=1-e^{-n\epsilon}=1$$

this proves that

$$X_{(1)}\xrightarrow{\mathcal{P}}\theta$$

---

Using the facts that the sequence is monotonic (non increasing) and lower bounded to $\theta$ you can also prove that

$$X_{(1)}\xrightarrow{a.s.}\theta$$