How to show two variables are asymptotically independent Let $X_1,...,X_n$ be iid from $Exp(\theta)$ with density function $f(x) = \frac{1}{\theta}e^{-x/\theta}$. Show that $M_n = X_{n:n} - \theta \ln(n)$ and $T_n = nX_{1:n}$ are asmyptoically independent with $X_{n:n} = max{(X_i)}$ and $X_{1:n}=min({X_i})$.
I know I first need to compute the limiting distributions of $M_n$ and $T_n$. Is this done by showing $F_{M_n}\xrightarrow{D}F$ and  $F_{T_n}\xrightarrow{D}F$? Because when I attempt to do this I get different answers for F. Also, is this F the CDF of the given PDF?
I was told that after I find the limiting distributions of  $M_n$ and $T_n$, to show they are asymptotically independent a good thing to show is that
$$\lim_{n\to\infty}F_{M_nT_n}(x,t)\propto\lim_{n\to\infty}F_{M_n}(x) * \lim_{n\to\infty}F_{T_n}(x)$$
But what exactly is $F_{M_nT_n}(x,t)$? I am unsure as to what this is.
Any help and pointer to help me solve this would be greatly appreciated. Thanks.
 A: In your second paragraph you write

Because when I attempt to do this I get different answers for $F$

and this is as it should be, you are led by your notation to think the limiting distributions should be the same. They are not.
Let us first look at the marginals:
$$ \DeclareMathOperator{\P}{\mathbb{P}} 
   \P(T_n \le t)=\P(n X_{1:n}\le t)=\P( X_{1:n} \le t/n) =\\
   1-\P(X_{1:n} > t/n)=1-\P(X_1>t/n)^n = 1- (e^{-\frac{t/n}{\theta}})^n=\\
   1-e^{-t/\theta}
$$ which does not depend on $n$ at all, so in this case the limit when $n\to\infty$ is very easy to compute! So this scaled minimum has itself an exponential distribution. Then
$$
\P(M_n \le m)=\P(X_{n:n} -\theta \log n \le m)=\\
P(X_{n:n} \le m+\theta \log n) =\P( X_1 \le m+\theta \log n)^n =\\
(1-e^{-\frac{m+\theta \log n}{\theta}})^n=
(1- \frac1n e^{-m/\theta})^n
$$ and we use the known limit: $\lim_{n \to\infty} (1-x/n)^n = e^{-x}$
to find the limit which is $e^{-e^{-m/\theta}}$, a Gumbel distribution with scale $\theta$.
But we want the joint distribution of this scaled min and max. Rather than the joint cdf, we start with
$$
\P( T_n > t, M_n \le m) = \\
\P(X_{1:n} > t/n, X_{n:n} \le m+\theta \log n) = \\
\P(X_1>t/n,X_2>t/n, \dotsc, X_n > t/n, X_1\le m+\theta \log n, \dotsc,
X_n \le m+\theta \log n) =\\
\P(t/n<X_1 \le m+\theta\log n, \dotsc, t/n<X_n \le m+\theta\log n)=\\
\P( t/n < X_1 \le m+\theta \log n)^n=\\
\left( (1-e^{-\frac{m+\theta \log n}{\theta}})-(1-e^{-\frac{t}{n\theta}})  \right)^n =\\
\left( e^{-\frac{t}{n\theta}} - e^{-m/\theta + \log(1/n)}  \right)^n =\\
\left( e^{-\frac{t}{n\theta}} - \frac1n e^{-m/\theta}  \right)^n =\\
(e^{-\frac{t}{n\theta}})^n \cdot (1-\frac1n e^{\frac{t}{n\theta}}\cdot e^{-m/\theta})^n =\\
e^{-t/\theta} \cdot (1 - \frac1n e^{-m/\theta  + t/(\theta n)})^n 
$$
and again taking the limit we get
$$
   e^{-t/\theta}  \cdot  e^{-e^{-m/\theta}}
$$
and since this has the separated product form, asymptotic independence is proved.
