Find $E(X_1X_2 \mid X_{(1)})$ where $X_i$'s are i.i.d Exponential. Is my solution correct?

Question

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.

Answer 1

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

Answer 2

I would do the computations as follows.

Due to the symmetry $P(X_{(1)}=X_i)=1/n$. Now we can fix $X_{(1)}$ and compute the expectation by definition. Assume that $X_{(1)}=X_3$ (with probability $1/n$), then $$ A= E(X_1X_2|X_{(1)}=X_3)= \int_0^\infty e^{-x_3}dx_3 \int_{x_4=x_3}^\infty e^{-x_4}dx_4 \dots \int_{x_n=x_3}^\infty e^{-x_n}dx_n \int_{x_1=x_3}^\infty \int_{x_2=x_3}^\infty e^{-x_1} e^{-x_2} x_1 x_2 dx_1 dx_2 = {2+2n+n^2 \over n^3}. $$

Under assumption $X_{(1)}=X_i$, $i \ge 4$, the answer will be the same and equal to $A$.

Now assume that $X_{(1)}=X_2$ (with probability $1/n$), then $$ B= E(X_1X_2|X_{(1)}=X_1)= \int_0^\infty e^{-x_2} x_2 dx_2 \int_{x_1=x_2}^\infty e^{-x_1} x_1 dx_1 \int_{x_3=x_2}^\infty e^{-x_3}dx_3 \dots \int_{x_n=x_2}^\infty e^{-x_n}dx_n = {n+2\over n^3}. $$

Under assumption $X_{(1)}=X_1$ the answer will be the same and equal to $B$.

Puting everyting together by the law of total probaiblity, we have ${(n-2)\over n}A+{2 \over n}B={1 \over n}$.

By the way, if $n$ is very large, then the amswer to your question must be $\approx 0$. This is another check for the candidate answer.