Convergence In Probability and Limiting Distribution Let $X_{1}, X_{2}$, . . . be i.i.d. with the following shifted exponential distribution:$f(x)=ae^{-a(x-1)}, x>1$.Let $Y_{n}=X_{(1)},Z_{n}=X_{(n)}$
Show that $Y_{n}$ converges to 1 in probability
Derive the limiting distribution of $T_{n}=Z_{n}-1-\frac{logn}{a}$
I have find $f_{Y_{n}}(x)=nae^{-a(x-1)n}$, but how can I prove $Y_{n}$ converges to 1 in probability?
$F_{Z_{n}}(z)=(1-e^{-a(z-1)})^{n}$
Any idea to the second questions? Thanks.

According to the definitions $Y_n$ is the minimum of the $n$ i.i.d random variables $X_i$. Thus $$F_{Y_n}(y)=P(\min\{X_1,X_2,\ldots,X_n\}\le y)=\ldots=1-\left(1-F_X(y)\right)^n$$ and and therefore (by substituting $F_X$) we obtain that $$F_{Y_n}(y)=1-\left(1-1+e^{-a(x-1)}\right)^n=1-e^{-na(x-1)}$$ (Observe that the power of $e$ is negative for every $x>1$ since $n>0, a>0$ and $x-1>0$. Thus as $n$ grows to infinity you will find that $$P(|Y_n-1|\ge ε)\overset{Y_n>1}=P(Y_n\ge 1+ε)=1-F_{Y_n}(1+ε)=e^{-na(x-1)}\longrightarrow 0$$ as $n\to \infty$, which proves the convergence of $Y_n$ to $1$ in probability.
Similarly, $Z_n$ is the maximum of the $n$ i.i.d random variables $X_i$. Thus $$F_{Z_n}(z)=P(\max\{X_1,X_2,\ldots,X_n\}\le y)=\ldots=\left(F_X(z)\right)^n$$ and therefore (as you correctly have) 
$$F_{Z_n}(z)=\left(1-e^{-a(z-1)}\right)^n$$ for $z>1$. From this last expression we can derive the CDF of the random variable $T_n$ as follows $$\begin{align*}F_{T_n}(t)&=P(T_n\le t)=P\left(Z_n-1-\frac{\ln n}{a}\le t\right)=P\left(Z_n\le 1+\frac{\ln n}{a}+t\right)\\[0.2cm]&=F_{Z_n}\left(1+\frac{\ln n}{a}+t\right)=\left(1-e^{-a\left(1+t+\frac{\ln n}{a}-1\right)}\right)^n\\[0.2cm]&=\left(1-e^{-at}e^{-\ln n}\right)^n=\ldots\end{align*}$$ where $e^{-\ln n}=\not e^{\not \ln n^{-1}}=n^{-1}=\frac{1}{n}$ and therefore $$\begin{align*}\ldots &=\left(1-\frac{1}{n}e^{-at}\right)^n=\left(1+\frac{\left(-e^{-at}\right)}{n}\right)^n\longrightarrow e^{-e^{-at}}\phantom{sssssssssssss}\end{align*}$$ as $n \to \infty$ (where we used that $\left(1+\frac{x}{n}\right)^n \to e^x$ as $n \to \infty$). Thus the limiting distribution of $T_n$ as $n \to \infty$ is $$F_{T}(t)=\lim_{n\to \infty}F_{T_n}(t)=e^{-e^{-at}}$$ for every $t \in \mathbb R$ (why?). Taking the derivative of $F_{T}(t)$ you can obtain the pdf of $T$.
