I am a bit stuck on the following question and need some help with the following question:
"For a certain experiment, the Poisson distribution with parameter $λ = m$ has been assigned. Show that a most probable outcome for the experiment is the integer value $k$ such that $m − 1 ≤ k ≤ m$. Under what conditions will there be two most probable values? Hint: Consider the ratio of successive probabilities."
The solution says that $$\frac{P(\text{outcome is } j+1)}{P(\text{outcome is } j)} = \frac{\frac{m^{j+1}e^{-m}}{(j+1)!}}{\frac{m^je^{-m}}{j!}} = \frac{m}{j+1}$$
I'm not quite sure how it came up with this, would someone mind explaining please ?
Thank you
ok, I will go out on a bit of a limb and guess that you are familiar with the fact that $$\Pr(\text{outcome is }j) = \frac{m^j e^{-m}}{j!}$$ (mainly because it doesn't make sense to work on problems involving the Poisson distribution unless you know that), and that the part you're having trouble with is showing that $$ \frac{\left(\dfrac{m^{j+1}e^{-m}}{(j+1)!}\right)}{\left(\dfrac{m^je^{-m}}{j!}\right)} = \frac{m}{j+1}. $$
We have $$ \frac{\left(\dfrac{m^{j+1}e^{-m}}{(j+1)!}\right)}{\left(\dfrac{m^je^{-m}}{j!}\right)} = \frac{m^{j+1} e^{-m}}{(j+1)!}\cdot\frac{j!}{m^j e^{-m}} = \frac{m\cdot 1\cdot2\cdot3\cdots j}{1\cdot2\cdot3\cdots j\cdot(j+1)} = \frac m {j+1}. $$