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 (Choi 1994)
\[\lambda - \ln 2 \leq \nu < \lambda + \frac{1}{3}\]
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.