Compare difference of probability between two Poisson distributions, evaluate at certain point.

Question

$X,Y$ are two independent random variables such that $X\sim \mathop{\mathrm{Po}}(\lambda_1), Y\sim \mathop{\mathrm{Po}}(\lambda_2)$, where $\lambda_2>\lambda_1$. Is there a conclusion about the size relationship between $\mathop{\mathrm{Po}}(\lambda_2)\{k\}$ and $\mathop{\mathrm{Po}}(\lambda_1)\{k\}$, if $k>\lambda_2$?

Simulations show that $\mathop{\mathrm{Po}}(\lambda_2)\{k\}>\mathop{\mathrm{Po}}(\lambda_1)\{k\}$. Is there proof, or what other conditions should I add?

Answer 1

Let's show the following equivalent statement according to Andrew:

For $\lambda_3>\lambda_2>\lambda_1$, we have $\lambda_3\log(\frac{\lambda_2}{\lambda_1})>\lambda_2-\lambda_1$

Proof:

Define $\delta=\lambda_2/\lambda_1>1$, $\lambda_3\log(\delta)>\lambda_2\log(\delta)=\lambda_1\delta\log(\delta)$, we want to show $\lambda_1\delta\log(\delta)>(\delta-1)\lambda_1$, which is equivalent to show $\delta\log(\delta)>(\delta-1)$. Let $f(x)=x\log(x)-x+1,x>1$ $$f^{\prime}(x)=\log(x)+1-1>0$$ which imply $f(\delta)>f(1)=0$. #

Answer 2

Let $\lambda_1 <\lambda_2$. Then $$e^{-\lambda_1}\frac{\lambda_1^n}{n!}<e^{-\lambda_2}\frac{\lambda_2^n}{n!}$$ if and only if $$e^{\lambda_2-\lambda_1}< \left(\frac{\lambda_2}{\lambda_1}\right)^n$$ if and only if $$ n\log(\lambda_2/\lambda_1)>\lambda_2-\lambda_1$$