Compare difference of probability between two Poisson distributions, evaluate at certain point. $X,Y$ 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?

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 $f(\delta)>f(1)=0$. #
