Order Statistics from a sum of exponential distributions

Let $$X_i$$ $$(X_1, \dots, X_n)$$ and $$Y_i (Y_1, \dots,Y_n)$$ be i.i.d. exponential r.vs with rate $$\lambda$$. Let $$Z_i= X_i+Y_i$$.

How to write the pdf of the k-th order statistics of the $$Z_i$$ random variable ?

• Please use MathJax whenever you type a formula on this site. It’s pretty easy. You just enclose your formulae into the \$ signs, for the most part.
– Aig
Commented Apr 25 at 9:36

If $$X_i, Y_i\sim \exp(\lambda)$$ then $$Z_i = X_i + Y_i$$ has pdf of

$$f_{Z_i}(z) = \int_{-\infty}^\infty f_{X_i}(x)f_{Y_i}(z-x) dx = \int_0^z \lambda^2 e^{-\lambda x}e^{-\lambda(z-x)} dx = \lambda^2 e^{-\lambda z}z$$

And thus its CDF is

$$F_{Z_i}(z) = \int_0^z \lambda^2 e^{-\lambda x}x dx = 1-e^{-\lambda z} (\lambda z+1)$$

In general, the probability density function of the $$k$$-th order statistic $$Z_{(k)}$$ of $$n$$ i.i.d random variables from $$Z$$ is

$$f_{Z_{(k)}}(z) = \frac{n!}{(k-1)!(n-k)!} F_Z(z)^{k-1} (1-F_Z(z))^{n-k} f_Z(z)$$

Plugging in, we get

$$f_{Z_{(k)}}(z) = \frac{n!}{(k-1)!(n-k)!} \cdot \left(1-e^{-\lambda z} (\lambda z+1)\right)^{k-1} \cdot \left(e^{-\lambda z} (\lambda z+1) \right)^{n-k} \cdot \lambda^2 e^{-\lambda z}z$$

Where you expecting a simple closed form?