finding $P(Y_i=X_n)$ in exponential distribution Suppose $X_1,X_2,\cdots,X_n$ are independent random variables and  we show order statistics of this random variables with $Y_1,Y_2,\cdots,Y_n$.
$X_1,X_2,\cdots,X_{n-1}$ have exponential distribution with mean $1$ and $X_n$ has exponential distribution with mean $\theta$.
How can find $P(Y_i=X_n)$ for ($i=1,2,\cdots,n$)?
 A: One has
$$
\mathrm P(Y_i=X_n) =  \int_{0}^{\infty} {n-1 \choose i-1} (1-\mathrm e^{-x})^{i-1} \mathrm e^{-(n-i)x}  \frac1\theta \mathrm e^{-x/\theta} \, \mathrm dx.
$$
The change of variable $\mathrm e^{-x}=s$ yields
$$
\mathrm P(Y_i=X_n)= \frac1\theta{n-1 \choose i-1}\int_{0}^{1}  (1-s)^{i-1} s^{n-i}   s^{1/\theta} \, s^{-1}\,\mathrm ds,
$$
that is,
$$
\mathrm P(Y_i=X_n)= \frac1\theta{n-1 \choose i-1}\mathrm{B}(i,n-i+1/\theta)=\frac1\theta\,\frac{\Gamma(n)}{\Gamma(n+1/\theta)}\,\frac{\Gamma(n-i+1/\theta)}{\Gamma(n-i+1)}. 
$$
Sanity checks: 
(i) When $\theta=1$, the distribution is uniform on $\{1,2,\ldots,n\}$. 
(ii) When $\theta\to0$, the distribution concentrates on $i=1$. 
(iii) When $\theta\to+\infty$, the distribution concentrates on $i=n$. 
(iv) (More involved) For every $n\geqslant1$ and $a\gt0$,
$$
\sum_{k=0}^{n-1}\frac{\Gamma(k+a)}{\Gamma(k+1)}=\frac{\Gamma(n+a)}{a\Gamma(n)}.
$$
Using this identity for $a=1/\theta$, one sees that the sum of $\mathrm P(Y_i=X_n)$ over $i$ is $1$... as it should.
A: You might start with $$P(Y_i=X_n) = \int_{x=0}^{\infty} P(Y_i=X_n | X_n =x)\,  p( X_n =x)  \, dx  $$ 
$$ = \int_{x=0}^{\infty} {n-1 \choose i-1} P(X_1 \lt x)^{i-1}   P(X_1 \gt x)^{n-i} p(X_n = x) \, dx $$ 
$$ = \int_{x=0}^{\infty} {n-1 \choose i-1} (1-e^{-x})^{i-1} (e^{-x})^{n-i}    e^{- x/\theta} /\theta\, dx $$ 
A: Since I had trouble grasping the integral expression for the probability $\mathbb{P}(Y_k = X_n)$ given by @Henry and @did, I am writing up my own derivation of it.
Consider first $n-1$ iid samples $X_1, \ldots, X_{n-1}$ from exponential distribution with unit rate, and let $Z_k = X_{n-1:k}$ be its order statistics. $X_n$ and $Z_k$ are independent random variables. 
We now insert $X_n$ into the ordered sequence $Z_1, \ldots, Z_{n-1}$. The event $\{Y_k = X_n\}$ is equivalent to $\{Z_{k-1} < X_n <Z_k \}$:
$$\begin{eqnarray}
  \mathbb{P}\left(Y_k = X_n\right) &=& \mathbb{P}\left(Z_{k-1} < X_n <Z_k\right) \\ 
  &=& \mathbb{E}\left(\mathbb{P}\left(Z_{k-1} < X_n <Z_k|Z_{k-1},Z_k\right)\right) 
 \\ &=& \mathbb{E}\left(\mathbb{P}\left(X_n > Z_{k-1}\right) - \mathbb{P}\left(X_n > Z_{k}\right)|Z_{k-1},Z_k\right) \\ 
 &=& \mathbb{E}\left( \mathbb{P}\left(X_n > Z_{k-1}|Z_{k-1}\right)\right) -  \mathbb{E}\left(\mathbb{P}\left(X_n > Z_k |Z_k\right) \right) \\
  &=&\mathbb{P}\left(X_n > Z_{k-1}\right)  - \mathbb{P}\left(X_n > Z_k \right)
\end{eqnarray}
$$
This readily produces the integral (using the pdf of the single order statistics):
$$ \begin{eqnarray}
 \mathbb{P}\left(X_n > Z_k\right) &=& \frac{(n-1)!}{(k-1)!(n-1-k)!} \int_0^\infty \mathrm{e}^{-(n-1-k)z} \left(1-\mathrm{e}^{-z}\right)^{k-1} \mathrm{e}^{-z/\theta} \mathrm{d}z \\
    &=& \frac{\Gamma(n)}{\Gamma\left(n+\frac{1}{\theta}\right)} \cdot \frac{\Gamma\left(n-k+\frac{1}{\theta} \right)}{\Gamma\left(n-k\right)}
\end{eqnarray}
$$
The order integral is obtained replacing $k$ with $k-1$. Then subtracting and using recurrence relation for $\Gamma$ function yields the result.
