Let $X$ and $Y$ be two independent poisson random variables with parameters $\mu$ and $\lambda$, respectively. Assuming that $\mu\geq\lambda$ , is it true that $P\left(X=Y-k\right)$ is decreasing in $\mu$ (holding $\lambda$ fixed) for any integer $k\geq0$? In other words, does the probability that the realization of $X$ being less than the realization of $Y$ by exactly $k$ become smaller as $\mu$ becomes larger (or equivalently, as $\lambda$ becomes smaller)? Intuitively, I believe this is true but cannot prove it.
One fact that I think is potentially helpful is that a Poisson r.v. with parameter $\mu$ first order stochastically dominates another Poisson r.v. with parameter $\mu'$ if $\mu\geq\mu'$. One consequence is that $P(X\leq Y)$ is decreasing in $\mu$, but this does not directly imply that $P(X=Y-k)$ is necessarily decreasing in $\mu$ for all $k$.
This is not true
$X \sim \text{Poi}(\mu)$ $Y \sim \text{Poi}(\lambda)$ $$P(X = Y - k) = \sum_{j = k}^\infty \frac{e^{-\lambda} \lambda^j}{j!} \frac{e^{-\mu} \mu^{j-k}}{(j-k)!} = f(\mu)$$
$$ f'(\mu)= \frac{e^{-\lambda} \lambda^k}{k!} (-e^{-\mu}) + \sum_{j = k+1}^\infty \frac{e^{-\lambda} \lambda^j}{j!}\bigg( \frac{-e^{-\mu} \mu^{j-k}}{(j-k)!}+ \frac{e^{-\mu} \mu^{j-k -1}}{(j-k - 1)!} \bigg) $$
Now call $$\bigg( \frac{-e^{-\mu} \mu^{j-k}}{(j-k)!}+ \frac{e^{-\mu} \mu^{j-k -1}}{(j-k - 1)!} \bigg) = \alpha (\mu,j-k)$$
note that $\alpha (\mu,j-k) \to -\infty $ as $\mu \to \infty$.
Now since for fixed $k$ $\frac{e^{-\lambda} \lambda^k}{k!} \to 0$ as $\lambda \to \infty$ you just need to take $\lambda$ large and $\mu > \lambda$ to obtain that
$f'(\mu) \leq 0$ so the function itself is decreasing
remark: to derive insede the sum just use that $f_n(\mu) \to f(\mu)$ uniformly and that $f_n'(\mu)\to g(\mu)$ uniformly implies $f'(\mu)= g\mu$ with $$f_n(\mu) = \sum_{j = k}^n \frac{e^{-\lambda} \lambda^j}{j!} \frac{e^{-\mu} \mu^{j-k}}{(j-k)!}$$