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Regarding the asymptotic behavior when $\lambda \to \infty$:

To get an estimate one simply finds that the dominating element among $$A_k= \mathbb{P}(Y=k+1)\mathbb{P}(X=k) = e^{-1-\lambda} \frac{\lambda^k}{k!(k+1)!}$$ is $A_{\sqrt{\lambda}}$ which gives roughly $e^{2\sqrt{\lambda}-\lambda}$. Probably there's also a Poly($\lambda$) factor here.

Oh, and the $F'(\lambda)$ is simply $\mathbb{P}(X=Y)$ since the probability of adding 1 when increasing the intensity of a Poisson RV by $\epsilon$ is $\epsilon$. In the special case $\lambda=1$ we get $$F'(1)=e^{-2} \sum_{k=0}^\infty \frac{1}{(k!)^2}$$

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Regarding the asymptotic behavior when $\lambda \to \infty$:

To get an estimate one simply finds that the dominating element among $$A_k= \mathbb{P}(Y=k+1)\mathbb{P}(X=k) = e^{-1-\lambda} \frac{\lambda^k}{k!(k+1)!}$$ is $A_{\sqrt{\lambda}}$ which gives roughly $e^{2\sqrt{\lambda}-\lambda}$. Probably there's also a Poly($\lambda$) factor here.

Oh, and the $F'(\lambda)$ is simply $\mathbb{P}(X=Y)$ since the probability of adding 1 when increasing the intensity of a Poisson RV by $\epsilon$ is $\epsilon$. In the special case $\lambda=1$ we get $$F'(1)=e^{-2} \sum_{k=0}^\infty \frac{1}{(k!)^2}$$

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Regarding the asymptotic behavior when $\lambda \to \infty$:

To get an estimate one simply finds that the dominating element among $$A_k= \mathbb{P}(Y=k+1)\mathbb{P}(X=k) = e^{-1-\lambda} \frac{\lambda^k}{k!(k+1)!}$$ is $A_{\sqrt{\lambda}}$ which gives roughly $e^{2\sqrt{\lambda}-\lambda}$. Probably there's also a Poly($\lambda$) factor here.