Editing Exponential distribution (section)

===Joint moments of i.i.d. exponential order statistics===

Let <math> X_1, \dotsc, X_n </math> be <math> n </math> [[independent and identically distributed]] exponential random variables with rate parameter ''λ''.
Let <math> X_{(1)}, \dotsc, X_{(n)} </math> denote the corresponding [[order statistic]]s.
For <math> i < j </math> , the joint moment <math> \operatorname E\left[X_{(i)} X_{(j)}\right] </math> of the order statistics <math> X_{(i)} </math> and <math> X_{(j)} </math> is given by
<math display="block">\begin{align}
  \operatorname E\left[X_{(i)} X_{(j)}\right]
    &= \sum_{k=0}^{j-1}\frac{1}{(n - k)\lambda} \operatorname E\left[X_{(i)}\right] + \operatorname E\left[X_{(i)}^2\right] \\
    &= \sum_{k=0}^{j-1}\frac{1}{(n - k)\lambda}\sum_{k=0}^{i-1}\frac{1}{(n - k)\lambda} + \sum_{k=0}^{i-1}\frac{1}{((n - k)\lambda)^2} + \left(\sum_{k=0}^{i-1}\frac{1}{(n - k)\lambda}\right)^2.
\end{align}</math>

This can be seen by invoking the [[law of total expectation]] and the memoryless property:
<math display="block">\begin{align}
  \operatorname E\left[X_{(i)} X_{(j)}\right]
    &= \int_0^\infty \operatorname E\left[X_{(i)} X_{(j)} \mid X_{(i)}=x\right] f_{X_{(i)}}(x) \, dx \\
    &= \int_{x=0}^\infty x \operatorname E\left[X_{(j)} \mid X_{(j)} \geq x\right] f_{X_{(i)}}(x) \, dx
      &&\left(\textrm{since}~X_{(i)} = x \implies X_{(j)} \geq x\right) \\
    &= \int_{x=0}^\infty x \left[ \operatorname E\left[X_{(j)}\right] + x \right] f_{X_{(i)}}(x) \, dx
      &&\left(\text{by the memoryless property}\right) \\
    &= \sum_{k=0}^{j-1}\frac{1}{(n - k)\lambda} \operatorname E\left[X_{(i)}\right] + \operatorname E\left[X_{(i)}^2\right].
\end{align}</math>

The first equation follows from the [[law of total expectation]].
The second equation exploits the fact that once we condition on <math> X_{(i)} = x </math>, it must follow that <math> X_{(j)} \geq x </math>. The third equation relies on the memoryless property to replace <math>\operatorname E\left[ X_{(j)} \mid X_{(j)} \geq x\right]</math> with <math>\operatorname E\left[X_{(j)}\right] + x</math>.