Lately, I am doing an investigation on Stirling's formula and its applications. So I thought I could use it to prove that the mode of the Poisson model is approximately equal to the mean. Of course, you do that by considering the curve that is formed by connecting the points of the probabilities of occurrence and the different values of the discrete random variable. Then you differentiate the p.d.f. where for $x!$ you use Stirling's formula $x!\approx \sqrt{2\pi x}~x^xe^{-x}$. The result is $\lnλ-1/(2x)-\ln x$ whose roots cannot be found analytically, but by iterative methods we find that as λ is larger and larger, the mode~mean.
Problem is, I found the following paper online, which seems to be the solution from a Harvard's undergraduate problem set.
http://www.physics.harvard.edu/academics/undergrad/probweek/sol84.pdf
It reads "You can also show this by taking the derivative of eq. (2), with Stirling’s expression in place of the $x!$. Furthermore, you can show that $x = a[=λ ~\text{in my case}~]-1/2$ leads to a maximum $P(x)$ value of $P_\max\approx1/\sqrt{2\pi a}$."
Does this puzzle you as much as it puzzles me? My main concern is over the "=" sign: how does this hold? The derivative=0 equation cannot have such an exact solution. Furthermore at such x, how does $P(X=a-1/2)$ give $1/\sqrt{2\pi a}$?
Am I (and my professor) missing something rather obvious or is the solution wrong?
Discuss!
PS: This sort of question might have been asked before, but still, I am really curious that somebody reads the paper in the link above, so that I can figure out what's going on.
Best Answer
To find the mode of the Poisson distribution, for $k > 0$, consider the ratio $$ \frac{P\{X = k\}}{P\{X = k-1\}} = \frac{e^{-\lambda}\frac{\lambda^k}{k!}}{e^{-\lambda}\frac{\lambda^{k-1}}{(k-1)!}} = \frac{\lambda}{k}$$ which is larger than $1$ for $k < \lambda$ and smaller than $1$ for $k > \lambda$.
If $\lambda < 1$, then $P\{X = 0\} > P\{X = 1\} > P\{X > 2\} \cdots$ and so the mode is $0$.
If $\lambda > 1$ is not an integer, then the mode is $\lfloor\lambda\rfloor$ since $P\{X = \lceil\lambda\rceil\} < P\{X = \lfloor\lambda\rfloor\}$.
If $\lambda$ is an integer $m$, then $P\{X = m\} = P\{X = m-1\}$ and so either $m$ or $m-1$ can be taken to be the mode.
In all cases, the mode and the mean differ by less than $1$. You do not need to use Stirling's approximation at all.