Let $X\_1$ and $X\_2$ be independent Poisson distributed random variables with parameters $\lambda\_1$ and $\lambda\_2$, respectively.


Let $a = P(X\_1 > X\_2)$ and $b = P(X\_1 = X\_2)$. 


**Question**: regarding $(a,b)$ as data, does it uniquely determine $\lambda\_1$ and $\lambda\_2$?


**Idea 1**: we have the expressions $a = a(\lambda\_1,\lambda\_2) = e^{-\lambda\_1-\lambda\_2} \sum\_{k=0}^\infty \sum\_{j > k}(\lambda\_1^j\lambda\_2^k)/(j!k!)$ and $b = b(\lambda\_1,\lambda\_2) = e^{-\lambda\_1-\lambda\_2} \sum\_{j = 0}^\infty (\lambda\_1\lambda\_2)^j/(j!)^2$, so perhaps one can use this to directly show that $a(\lambda\_1,\lambda\_2) = a(\mu\_1,\mu\_2)$ and $b(\lambda\_1,\lambda\_2) = b(\mu\_1,\mu\_2)$ implies $(\lambda\_1,\lambda\_2) = (\mu\_1,\mu\_2)$.


**Idea 2**: the random variable $Z = X\_1 - X\_2$ is Skellam distributed with parameters $(\lambda\_1,\lambda\_2)$. The conditions amount to specifying the probability mass to the right of $0$ (this is $a$), and the probability mass at $0$ (this is $b$). Perhaps there is a clever way to see that this determines the distribution of $Z$. 



Does $P(X_1>X_2)$ and $P(X_1=X_2)$, where $X_1$ and $X_2$ are independent and Poisson distributed, uniquely determine the parameters?
Consider the function $F: (\lambda\_1,\lambda\_2) \rightarrow (a,c)$, where $a = P(X\_1 > X\_2)$ and $c = 1 - a - b = P(X\_1 < X\_2)$. We claim that $F$ is invertible, considered as a map from $(0,\infty)^2$ to the open triangular set $T = \{(x,y) \in \mathbb{R}^2; x>0, y>0, x + y < 1\}$.


**$F$ is injective**


First, we compute the Jacobian. 
\begin{eqnarray}
\frac{da}{d\lambda\_1} &=&-a + e^{-\lambda\_1-\lambda\_2}\sum\_{k = 0}^\infty\sum\_{j > k}\frac{\lambda\_1^{j-1}\lambda\_2^k}{(j-1)!k!}\\&=&-a + e^{-\lambda\_1-\lambda\_2}\sum\_{k = 0}^\infty\sum\_{j \geq k}\frac{\lambda\_1^j\lambda\_2^k}{j!k!}\\
&=&-P(X\_1>X\_2) + P(X\_1\geq X\_2)\\
&=&P(X\_1 = X\_2).
\end{eqnarray}
Similar calculations for the other partial derivatives gives:
$$
J = \left(\begin{array}{cc}P(Z = 0) &-P(Z = 1)\\-P(Z = -1) &P(Z=0)\end{array}\right),
$$
where $Z = X\_1 - X\_2$ is the Skellam distributed random variable. The probability mass function for $Z$ is given by $P(Z=k) = e^{-\lambda\_1 -\lambda\_2}(\lambda\_1/\lambda\_2)^{k/2}I\_{|k|}(2\sqrt{\lambda\_1\lambda\_2})$, where $I\_{k}$ is the modified Bessel function of the first kind. 


Now, $J$ is a P-matrix, i.e. all its principal minors are positive. Indeed, $P(Z=0) > 0$, and the determinant $P(Z=0)^2 - P(Z=1)P(Z=-1)$ equals $e^{-2\lambda\_1 -2\lambda\_2}(I\_0(2\sqrt{\lambda\_1\lambda\_2})^2-I\_1(2\sqrt{\lambda\_1\lambda\_2})^2)$, which is positive since $I\_0 > I\_1$.


But now we have everything we need to apply the "Fundamental Global Univalence Theorem (Gale-Nikaido-Inada)" (see On Global Univalence Theorems, Lecture Notes in Mathematics Volume 977, 1983, pp 17-27), which gives us injectivity of $F$. 


**$F$ is surjective**


It is clear by construction that $F = (a,c)$ does not take values outside the closure of $T$: $a$ and $c$ are probabilities of disjoint events, so they are non-negative and sum to at most $1$. Moreover, the range of $F$ is simply connected, since $F$ is continuous and its domain is simply connected. Consider the three line segments that constitute the boundary of $T$: $R\_1 = \{(x,0); 0 < x < 1\}$, $R\_2 = \{(0,y); 0 < y < 1\}$, and $R\_3 = \{(x,y); x>0,y>0,x + y = 1\}$. We are done if we can show that $F$ attains values arbitrarily close to each point in $R\_1\cup R\_2\cup R\_3$, but never takes values in that set. (One should also consider the three corners, $(0,0)$, $(1,0)$ and $(1,0)$, but the arguments are similar for these cases.)


Consider a point $(x,0) \in R\_1$. It is clear that this value is never attained by $F$, since $P(X\_2 > X\_1) = 0$ would imply either vanishing $\lambda\_2$, or infinite $\lambda\_1$. However, values arbitrarily close to $(x,0)$ are attained: choose $\lambda\_1$ such that $P(X\_1 > 0) = x$. Then $F(\lambda\_1,\lambda\_2) \rightarrow (x,0)$ as $\lambda\_2 \rightarrow 0$. The set $R\_2$ is treated similarly.


Finally, consider a point $(x,y) \in R\_3$. By symmetry, we can assume that $x > y$. This point can not be in the range of $F$, since that would imply $P(X\_1 = X\_2) = 0$, which is impossible for finite $\lambda\_1$ and $\lambda\_2$. Now, $1 - a - c \rightarrow 0$ when $\lambda\_2 \leq \lambda\_1$ and $\lambda\_2\rightarrow \infty$, so we can choose $\lambda\_2$ sufficiently large and send $\lambda\_1$ from $\lambda\_1 = \lambda\_2$ towards infinity: $F(\lambda\_1,\lambda\_2)$ will describe a curve from a point in the epsilon ball $B\_\varepsilon(0.5,0.5)$ to a point in $B\_\varepsilon(1,0)$, always staying at most $\varepsilon$ away from $R\_3$, hence passing within $\varepsilon$ of $(x,y)$.


