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I'm working with the following Poisson difference distribution: $$\text{Prob}\{X_1-X_2 \geq 0\} $$ where $X_1 \sim$ Poisson $(\mu_1)$ is independent from $X_2 \sim$ Poisson $(\mu_2)$. I need to understand the behavior of: $$\frac{\partial}{\partial \mu_1}\text{Prob}\{X_1-X_2 \geq 0\} $$

My intuition is that the CDF is increasing in $\mu_1$. However, my intuition proved to be wrong in many occasions : )

The Poisson difference distribution (Skellam distribution) has not a convenient closed form distribution, in particular, its PMF includes a BesselI function and it is not straightforward to take the derivative.

Therefore, exploiting the fact that $X_1$ and $X_2$ are independent "we can write" (this is my claim):

$$ \text{Prob}\{X_1-X_2 \geq 0\} = \sum_{k=1}^{\infty}\text{Prob}\{X_1=k\}\text{Prob}\{X_2\leq k\} $$

$$=\sum_{k=0}^{\infty}\text{Prob}\{X_1=k\} \sum_{h=0}^k\text{Prob}\{X_2=h\} $$

For a generic $k$

$$e^{-\mu_1}\frac{\mu_1^k}{k!} e^{-\mu_2}\sum_{h=0}^k \frac{\mu_2^h}{h!} $$

In my particular case I have that $\mu_2 = v-\mu_1$. To be more specific, $\mu_1 = vx $ where $v >1$ and $x \in (0,1)$.

So I have $$\frac{e^{-v}}{k!} \mu_1^k \sum_{h=0}^k \frac{\mu_2^h}{h!} $$

Forgetting the constant term I can show that the expression is increasing in $x$. Taking the derivative with respect to $x$ and studying the sign I get

$$vk\mu_1^{k-1}\sum_{h=0}^k \frac{(1-\mu_1^h)}{h!} -v\mu_1^k\sum_{h=0}^k \frac{(1-\mu_1)^h}{h!} >0 $$

If and only if

$$\frac{k}{v}>x $$

or $k > \mu_1$ if we just consider the mean.

However, this contrasts the simulations I made using the "proper" CDF of the Skellam distribution; so I do not know how to approach the problem or where my fault is

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