Let $X_1, … , X_n$, ($n\ge 2$) be a sample from exponential distribution with parameter $\lambda=1$ and $X_{(1)}={\rm{min}}(X_1,…,X_n)$. Then we need to find the conditional expectation $E(X_1X_2\mid X_{(1)})$.
Is this correct:
\begin{align}
&E(X_1X_2\mid X_{(1)}=a)\cdot P(X_{(1)}=a)\\&=a\cdot\int_a^\infty f_{X_2}(x_2)x_2dx_2\cdot \prod_{i=3}^n \int_a^\infty f_{X_i}(x_i)dx_i\\
&+\int_a^\infty f_{X_1}(x_1)x_1dx_1\cdot a \cdot \prod_{i=3}^n \int_a^\infty f_{X_i}(x_i)dx_i\\
&+\sum_{k=3}^n \int_a^\infty f_{X_1}(x_1)x_1dx_1\cdot\int_a^\infty f_{X_2}(x_2)x_2dx_2\cdot \prod_{\substack{i=3\\ i\neq k}}^n \int_a^\infty f_{X_i}(x_i)dx_i,
\end{align}
where $P(X_{(1)})=ne^{-(n-1)a}$.
The formula above is self-explanatory: we consider $n$ cases where $X_{(1)}=X_k$, $k=1,2,…,n$ and take the sum of expectations in each case.
This seems intuitively correct to me, but I'm not sure if this is so. Especially I can't understand how do we use known properties of conditional mean to prove this?
By calculating this expression with $f_X(x)=e^{-x}, x>0$ I find
$$
E(X_1X_2\mid X_{(1)}=a)=(1+a)\left(a+1-\frac2n\right).
$$
EDIT. Whoever can give a definitive answer to this question will be given 200 pt bonus. Thanks.
Best Answer
It is necessary to suppose that the n exponential rv are independent so $\mathbb{E}[X_1 X_2]=\mathbb{E}[X_1]\mathbb{E}[X_2]=1$
Now, reminding the memoryless property of Exp Neg Law, when the minimum $X_{(1)}=a$ means that in $a$ the n rv's are "good as new" so their conditional expected value is $a+1$
EDIT: the previous solution is NOT CORRECT. Finally I found
$\mathbb{E}[X_1 X_2 |X_{(1)}=a]=\frac{2}{n}a(a+1)+\frac{n-2}{n}(a+1)^2$
that is equivalent to the solution found by the OP
I explain my brainstorming:
First of all, observe that $f(x|x>a)=e^a e^{-x}\mathbb{1}_{[a;+\infty)}(x)$
and so $$\mathbb{E}[X|X>a]=e^a \int_a^{+\infty}xe^{-x}dx=a+1$$
Now, if the minimun is NOT $X_1$ or $X_2$ the requested probability is like the following
$\mathbb{E}[X|X>a]\mathbb{E}[Y|Y>a]=(a+1)^2$
For similar reason, if $X_1$ or $X_2$ is the minimum, the expected value will be $a(a+1)$
As the probability of one in $n$ independent rv's to be the minimum is constant $=\frac{1}{n}$ the solution is what I showed