Question 1
When do the bounds on the median $\nu$ of a Poisson distribution equal (or nearly equal) the mean of the distribution, $\frac{1}{\lambda}$?
The bounds of the median are given as
\[\lambda - \ln 2 \leq \nu < \lambda + \frac{1}{3}\]Lower Bound
Let \(\lambda - \ln 2 = \frac{1}{\lambda}\) then we can obtain the polynomial
\(\lambda^2 - \lambda \ln 2 - 1 = 0\) for which we can then use the quadratic equation to obtain
\[\lambda = \frac{\ln2 \pm \sqrt{\left( \ln 2 \right)^2 +4}}{2}.\]We obtain two approximate solutions for lambda:
\[\lambda_- \approx -0.7117804400329233\] \[\lambda_+ \approx 1.4049276205928687\]However, since we already know that $\lambda > 0$ we can exclude $\lambda_-$ and take $\lambda_+$ as the unique solution.
Upper Bound
Let
\[\frac{1}{\lambda} = \lambda + \frac{1}{3}\]then we can arrange to see this is a quadratic polynomial
\(\lambda^2 + \frac{\lambda}{3} - 1 = 0\) and apply the quadratic formula:
\[\lambda = \frac{-\frac{1}{3} \pm \sqrt{\left( \frac{1}{3} \right)^2 + 4}}{2}\]This gives two candidate solutions:
\[\lambda_- \approx -1.18046042171637\] \[\lambda_+ \approx 0.8471270883830365\]Like with the lower bound, we can only take the upper bound to be $\lambda_+$. But unlike the lower bound, we can only say that the median is strictly less than $\lambda_+$ even if it is really really close.
Question 2
Assuming $\lambda$ is known, how wide a range of values can the median take?
\[\lambda + \frac{1}{3} - \lambda + \ln 2 = \frac{1}{3} + \ln 2\]Interestingly, no matter what value $\lambda$ takes there is a constant-width interval for the values of the median.