Suppose $P(\lambda, i)$ is the probability that a Poisson random variable with average $\lambda$ is equal to $i$, i.e. $\frac{\lambda^i}{e^{\lambda}i!}$

I think the following system of equations always has solution in $x$ and $y$, non-negative real numbers, for any $\alpha>0$ and $k\in \mathbb{N}_+$

\begin{cases} \alpha=\sum_{i=0}^{\infty}P(x, i)\cdot P(y, k+i) \\ \alpha=\sum_{i=0}^{\infty}P(x, i)\cdot P(y, k+i+1) \end{cases}

where the necessary condition $\alpha\leq P(k+1, k+1)$ holds. It is easy to prove that this is indeed a necessary condition, equivalent to the condition that $\alpha=P(\lambda,k+1)$ has a solution. It is also easy to see that solution $y$ of the system is smaller or equal to $\lambda$, the largest solution to the equation $\alpha=P(\lambda,k+1)$. Experiments show that for each fixed $\alpha$ and $k$, there is a solution, but I did not manage to prove it analytically.

Is there any analogue for mean value theorem for multidimensional functions? Any suggestion for the proof directions will be appreciated.


Let $a:=\alpha$ and \begin{equation*} F_k(x,y):=\sum_{j=0}^\infty \frac{x^j}{j!}\frac{y^{k+j}}{(k+j)!}\,e^{-x-y}, \end{equation*} assuming the standard convention $0^0:=1$. We have to consider the existence of a solution in $x$ and $y$ of the system \begin{equation*} a=F_k(x,y)=F_{k+1}(x,y). \tag{1} \end{equation*} We shall prove the following.

Theorem 1. Take any natural $k$ and any \begin{equation*} a\in(0,a_k],\quad\text{where}\quad a_k:=\sup_{x,y\ge0}F_{k+1}(x,y). \tag{1.5} \end{equation*} Then the system (1) has a solution $x,y\ge0$.

Remark 1. Since $F_k>0$, the condition $a\in(0,a_k]$ is obviously necessary in Theorem 1.

Proof of Theorem 1. Note that $F_k(x,y)\ge0$ for any real $x,y\ge0$ and $F_k(x,y)$ is continuous in real $x,y\ge0$. The crucial observation is the identity \begin{equation*} \partial_y F_{k+1}(x,y)=F_k(x,y)-F_{k+1}(x,y) \tag{2} \end{equation*} for real $x,y$.

Next, fix for a moment any real $x\ge0$. Then $F_{k+1}(x,0)=0$ and, by dominated convergence, $F_{k+1}(x,\infty-)=0$. So, $F_{k+1}(x,y)$ attains its maximum in $y$ at some real point $y=y_x\ge0$. At this point, we have $\partial_y F_{k+1}(x,y)=0$. So, by (2), \begin{equation*} F_k(x,y_x)=F_{k+1}(x,y_x)=\max_{y\ge0}F_{k+1}(x,y)=:M_{k+1}(x), \tag{3} \end{equation*} for all real $x\ge0$.

Next, \begin{align*} M_k(x)&\le\sum_{j=0}^\infty \frac{x^j}{j!}\max_{y\ge0}\frac{y^{k+j}}{(k+j)!}\,e^{-x-y} \\ &=\sum_{j=0}^\infty \frac{x^j}{j!}e^{-x}\frac{(k+j)^{k+j}}{(k+j)!}\,e^{-k-j} \tag{4} \\ &\ll\sum_{j=0}^\infty \frac{x^j}{j!}e^{-x}\frac1{\sqrt{k+j}}=E\frac1{\sqrt{k+\Pi_x}} \underset{x\to\infty}\longrightarrow0 \end{align*} by dominated convergence and because $\Pi_x\underset{x\to\infty}\longrightarrow\infty$ in probability, where $\Pi_x$ is a Poisson random variable with parameter $x$. So, $M_k(\infty-)=0$. It is also not hard to see that $F_k(x,y)$ is continuous in real $x\ge0$ uniformly in real $y\ge0$ (see the Appendix), so that $M_k(x)$ is continuous in $x\ge0$. So, $M_{k+1}(x)$ attains its maximum in $x\ge0$ (equal $a_k$, by (1.5)) and takes all values in the interval $(0,a_k]$. Now Theorem 1 follows by (3). $\qquad\Box$

Appendix. Similarly to (2), \begin{equation*} \partial_x F_k(x,y)=F_{k+1}(x,y)-F_k(x,y). \end{equation*} for real $x,y$. Therefore and because $0\le F_k\le1$, we have $|\partial_x F_k(x,y)|\le1$ for real $x,y$, so that $F_k(x,y)$ is indeed continuous in real $x\ge0$ uniformly in real $y\ge0$.

Added: Let us now show that \begin{equation*} a_k=c_{k+1},\quad\text{where}\quad c_k:=\frac{k^k}{k!}\,e^{-k} \sim\frac1{\sqrt{2\pi k}} \end{equation*} as $k\to\infty$. To this end, note first that \begin{equation*} c_{k+1}/c_k=(1+1/k)^k/e<1, \end{equation*} and so, $c_k$ is decreasing in $k$. So, recalling (4), we have \begin{equation*} M_k(x)\le\sum_{j=0}^\infty \frac{x^j}{j!}e^{-x}\,c_{k+j} \le\sum_{j=0}^\infty \frac{x^j}{j!}e^{-x}\,c_k=c_k=M_k(0). \end{equation*} Thus, in view of (1.5) and (2), \begin{equation*} a_k=\max_{x\ge0}M_{k+1}(x)=M_{k+1}(0)=c_{k+1}, \end{equation*} as desired.

In particular, for $k=0,1,2,3$ the values of $a_k$ are $\approx0.367879, 0.270671, 0.224042$.

  • $\begingroup$ I have added Remark 2 concerning the values of $a_k$. $\endgroup$ – Iosif Pinelis Aug 5 '18 at 20:10
  • $\begingroup$ The proof of $a_k=P(k+1,k+1)$ is very easy, if $\alpha>P(k+1,k+1)$ then it implies $\alpha>P(\lambda, k+1)$ for each $\lambda$, since $P(\lambda,k+1)$ is maximized for $\lambda=k+1$. On the other hand, by induction we can prove that $\alpha>P(\lambda,k+1+i)$ for any $i>0$ and $\lambda$, therefore the second equation does not have a solution! $\endgroup$ – Tina Aug 5 '18 at 20:25
  • $\begingroup$ @Tina : I am afraid I don't understand your comment: (i) So, what if $\alpha>P(\lambda,k+1)$? What does this imply? (ii) What is your $\alpha$ in the inequality $\alpha>P(\lambda,k+1+i)$? (iii) How do you prove this inequality? (iv) What does this latter inequality imply? $\endgroup$ – Iosif Pinelis Aug 5 '18 at 20:45
  • $\begingroup$ (i) if $\alpha>P(\lambda, k+1)$ for any $\lambda$ this implies that $\alpha>P(\lambda,k+2)$ or for any $i>0$, $\alpha>P(\lambda,k+1+i)$. (ii) $\alpha$ is the parameter from the system of equations. (iii) proof is by induction, if $\alpha>P(\lambda,k+1)$, this means $\alpha\cdot e^{\lambda} - \frac{\lambda^{k+1}}{(k+1)!}>0$, then $\alpha\cdot e^{\lambda} - \frac{\lambda^{k+2}}{(k+2)!}>0$, since the latter holds for $\lambda=0$ and its derivative is the former, always positive. (iv) this implies that the second inequality of the system is not solvable, since $\sum_{i=0}^{\infty}P(x,i)=1$. $\endgroup$ – Tina Aug 5 '18 at 20:58
  • $\begingroup$ @Tina : I am sorry, I still don't understand your comment. $\endgroup$ – Iosif Pinelis Aug 5 '18 at 21:20

The sums over $P(\lambda,i)=\frac{\lambda^i}{e^{\lambda}i!}$ are evaluated in terms of a Bessel function as $$\sum_{i=0}^{\infty}P(x, i)\cdot P(y, k+i)=y^k e^{-x-y} \left(xy\right)^{-k/2} I_k\left(2 \sqrt{xy}\right)$$ $$\sum_{i=0}^{\infty}P(x, i)\cdot P(y, k+i+1)=(y/x)^{1/2}\,y^{k} e^{-x-y} \left(xy\right)^{-k/2} I_{k+1}\left(2 \sqrt{xy}\right)$$ for any positive integer $k$ these two expressions should be equal for some $x,y>0$. (For $x=y=0$ both expressions are identically zero.) So the function $$F_k(x,y)=\sqrt{x}\, I_k\left(2 \sqrt{xy}\right)-\sqrt{y} \,I_{k+1}\left(2 \sqrt{xy}\right)$$ should pass through zero in the quadrant $x,y>0$ for any positive integer $k$.

For large $z=2\sqrt{xy}$ both Bessel functions $I_k(z)$ and $I_{k+1}(z)$ grow as $(2\pi z)^{-1/2}e^z$, so by making $x$ much larger than $y$ the function $F_k(x,y)$ is positive and by making $y$ much larger than $x$ it is negative, hence it must go through zero when $x\approx y$.

I had not appreciated that $\alpha$ is fixed from the beginning like $k$, not a variable like $x$ and $y$. So we also need to show that $x\approx y\gg 1$ allows the sum to reach any $\alpha>0$, so $$\alpha=e^{-2x} I_k\left(2x\right)\approx (4\pi x)^{-1/2},\;\;x\gg 1.$$ This is possible only for $\alpha\ll 1$. The OP lists as necessary condition $\alpha\leq P(k+1,k+1)$, it is not clear to me this is sufficient.

  • $\begingroup$ Thank you! But I do not understand why both terms are equal to $\alpha$? $\endgroup$ – Tina Aug 5 '18 at 15:40
  • $\begingroup$ ah wait, $\alpha$ is fixed from the beginning like $k$ and not a variable like $x$ and $y$? $\endgroup$ – Carlo Beenakker Aug 5 '18 at 17:28
  • 2
    $\begingroup$ Yes, it is a fixed parameter of the system, like $k$. We should somehow use the necessary condition from the statement. $\endgroup$ – Tina Aug 5 '18 at 17:35

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