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 $Z$.

  • $\begingroup$ I've previously seen the word "intensities" used only when speaking of a process in which one associates with each measurable subset of some space a Poisson random variable with expected value proportional to the measure of the subset, and where the Poisson random variables are independent iff the subsets are essentially disjoint. That's not a reason why the term couldn't be used in this somewhat simpler situation, but the fact that these "intensities" are also expected values would make it seem simpler to just call them expected values or expectations. $\endgroup$ Aug 8, 2013 at 1:56
  • $\begingroup$ I agree @MichaelHardy, have edited to simply mention "parameters". $\endgroup$
    – svangen
    Aug 8, 2013 at 7:49

1 Answer 1


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 $(x,y)$.


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