Probabilities of two random variables with different Poisson distribution

Question

Assume that random variables $X\sim\operatorname{Poisson}(\mu)$ and $Y\sim\operatorname{Poisson}(\lambda)$, $\mu > \lambda > 0$. How should I approach to prove that for every $k > 0$: $P(X > k) > P(Y > k)$?

Answer 1

Hint: Let $Z \sim Poisson(\mu-\lambda)$, independent of $Y$. Then $X$ has the same distribution as $Y+Z$, moreover $Y>k$ implies $Y+Z>k$.