Cumulative distribution function of the median

Question

If X is the sample median from a poisson distribution, is there a way for me to find the cdf of the median i.e $F(y) = P(X <= y)$. I have never seen the cdf in terms of the median, always as a random variable and hence I am stumped.

Answer 1

Let $X_1,\ldots,X_n$ be i.i.d. with $\mathrm{Pois}(\lambda)$ distribution. Then the probability mass function of the $k^{\mathrm{th}}$ order statistic is given by \begin{align} \mathbb P(X_{(k)}=m) &= \mathbb P(X_{(k)}\leqslant m) - \mathbb P(X_{(k)}<m)\\ &= \sum_{j=0}^{n-k}\binom nj((1-F(m))^j(F(m))^{n-j} - (1-F(m)+f(m))^j(F(m)-f(m))^{n-j}), \end{align} where $F(m) = \sum_{k=0}^m \frac{\lambda^k}{k!}e^{-\lambda}$ is the distribution function of $X_1$ and $f(x) = \frac{\lambda^m}{m!}e^{-\lambda}$ is the probability mass function of $X_1$. Letting $n=3$ and $k=2$ we have \begin{align} \mathbb P(X_{(k)}=m) &= \sum_{j=0}^{1}\binom 3j\left(\left(\sum_{k=m+1}^\infty \frac{\lambda ^k}{k!}e^{-\lambda}\right)^j\left(\sum_{k=0}^m \frac{\lambda^k}{k!}e^{-\lambda}\right)^{3-j} - \left(\sum_{k=,}^\infty \frac{\lambda^k}{k!}e^{-\lambda}\right)^j\left(\sum_{k=0}^{m-1} \frac{\lambda^k}{k!}e^{-\lambda}\right)^{3-j}\right)\\ &=\frac{\Gamma (m+1,\lambda )^2 (3 m!-3 m \Gamma (m,\lambda )+\Gamma (m+1,\lambda ))}{(m!)^3}+\frac{2 \Gamma (m,\lambda )^3}{\Gamma (m)^3}-\frac{3 \Gamma (m,\lambda )^2}{\Gamma (m)^2} \end{align} I don't know how to simplify this further, unfortunately.