The median of random variables

Question

Suppose we have $n$ independent exponential random variables say $X_1,X_2,...,X_n$ all with rate $\lambda$.
The max and min of these r.v's can be found easily:
if $Y= min\{X_i\}$, then $P(Y\geq k)=e^{-n\lambda k}$ then finding the cdf and pdf from that.
also if $Z=max\{X_i\}, P(Z\leq d)=(1-e^{-\lambda d})^n $ which is the cdf itself
how to find the median in a similar way?
the median is the cdf $F(m)=P(X \leq m)=1/2$, but to find its distribution for the $X_i$ r.v's?

Answer 1

Let's suppose $n = 2k+1$ is odd, so the median is the $(k+1)$'th order statistic (if $n$ is even, there is no universal agreement on how to define the median). The number of $X_i \ge m$ is a binomial random variable $Y(m)$ with parameters $n$ and $p = e^{-\lambda m}$. Thus $$\mathbb P(Median \le m) = \mathbb P(Y(m) \ge k+1) = \sum_{j=k+1}^n {n \choose j} e^{-\lambda j m} (1-e^{-\lambda m})^{n-j}$$