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Suppose that $N_t$ is a Poisson process with rate $\lambda$.

  1. Find $\mathbb E\left[\frac{N^2_1}{N_3}\right]$
  2. Find $\mathbb E\left[\frac{N^2_3}{N_1}\right]$

Now, I know that if $N_t$ is a Poisson process then $N_t$ represents the number of arrivals up to time $t$. In fact, we can derive that if $N_t$ is a Poisson process then (surprise!) $N_t$ is Poisson-distributed with parameter $\lambda t$. That is, \begin{equation*} \mathbb P(N_t = k) = \frac{e^{-\lambda t}}{k!} (\lambda t)^k, \quad t \geq 0 \end{equation*}

Additionally, an important result is that the inter-arrival times of each arrival is exponentially distributed with rate $\lambda$. That is, if $T_1$ is the time until the first arrival of our Poisson process $N_t$ then \begin{equation*} \mathbb P(T_1 \leq t) = 1 - e^{\lambda t}, \quad t \geq 0 \end{equation*}

Up until now I've basically listed just a bunch of (probably useful) facts/results about Poisson processes. However, I've yet to manage to weave them into something that'll help me with a solution.

One problem I encountered is that, taking (1) as an example, $N_3 = 0$ with a nonzero probability, blowing up the expectation? Clearly this is just a heuristic argument, but we can illustrate it more clearly by attempting to solve the expectation by simply computing it:

\begin{equation*} \mathbb E\left[\frac{N^2_1}{N_3}\right] = \sum^\infty_{n = 0} \sum^\infty_{m = 0} \frac{n^2}{m} \mathbb P(N_1 = n, N_3 = m) \end{equation*}

Clearly any term with $m = 0$ will give us a singularity, and so I must have approach this poorly. It is also not obvious to me how to find the joint probability

\begin{equation*} \mathbb P(N_1 = n, N_3 = m) \end{equation*}

Thank you for any input!

Edit. I've thought about this a bit and looked through some older course material (+ the help of Henry's comment!). I forgot a nice property of Poisson processes: If $N_t$ is the number of arrivals up to time $t$ and $N_{t+s}$ the arrivals up to time $t + s$ $(t, s > 0)$ then \begin{equation*} \mathbb P(N_{t + s} - N_t = k) = e^{-\lambda t} \frac{(\lambda t)^k}{k!}, \quad k = 0, 1, 2, ... \end{equation*}

That is, $N_{t + s} - N_t \sim \text{Pois}(\lambda s)$. This property permits us to write $N_3$ in terms of $N_1$ and an independent Poisson random variable with parameter $2\lambda$: \begin{equation*} N_3 = N_1 + (N_3 - N_1) \end{equation*}

where \begin{align*} N_1 &\sim \text{Pois}(\lambda) \\ N_3 - N_1 &\sim \text{Pois}(2\lambda) \end{align*}

and the independence of $N_1$ and $N_3 - N_1$ is gained via the fact that Poisson processes have independent (non-overlapping) increments.

For brevity, denote \begin{align*} X &\stackrel{\text{def}}{=} N_1 \\ Y &\stackrel{\text{def}}{=} N_3 - N_1 \end{align*}

then $X \sim \text{Pois}(\lambda)$ and $Y \sim \text{Pois}(2\lambda)$, and $X$ and $Y$ are independent of each other. Therefore, our expectation of $N_1$ and $N_3$ becomes the expectation of independent $X$ and $Y$ \begin{align*} \mathbb E\left[\frac{N^2_1}{N_3}\right] &= \mathbb E\left[\frac{X^2}{X + Y} \right] \\ &= \sum_x \sum_y \frac{x^2}{x + y} \mathbb P(X = x, Y = y) \\ &= \sum_x \sum_y \frac{x^2}{x + y} \mathbb P(X = x)\mathbb P(Y = y) \quad \text{($X$ and $Y$ indep.)} \\ &= \sum_x \sum_y \frac{x^2}{x + y} e^{-\lambda} \frac{\lambda^x}{x!} e^{-2\lambda} \frac{(2\lambda)^y}{y!} \end{align*}

Now, I've played a bit around with these sums and I've yet to yield anything conclusive. Not to mention the issue with 0 in the denominator is still present.

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  • $\begingroup$ I would agree that $0$ in the denominator with non-zero probability blows up the expectation. $\endgroup$ – Henry Feb 16 '17 at 23:52
  • $\begingroup$ For the first problem, you can fudge it by noting that if $N_3=0$ then of necessity $N_1=0$ as well, and can treat the $0^2/0$ expression as zero. Still hate that. For the second problem, even that fudge is impossible. $\endgroup$ – Mark Fischler Feb 16 '17 at 23:56
  • $\begingroup$ Given memorylessness of a Poisson process $\mathbb P(N_1 = n, N_3 = m)$ is the probability that a Poisson distributed random variable with parameter $\lambda$ takes the value $n$ multiplied by the probability that a Poisson distributed random variable with parameter $2\lambda$ takes the value $m-n$ $\endgroup$ – Henry Feb 17 '17 at 1:27
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The counts of events in disjoint regions are independently distributed.

$$\mathsf P(N_1=n, N_3=m)~=~\mathsf P(N_1=n)~\mathsf P(N_3-N_1=m-n)$$

The count of arrival after time 1 until time 3 is Poisson distributed rate $(3-1)\lambda$.

$$N_3-N_1\sim\mathcal{Poiss}(2\lambda)$$

So putting it all together:

$$\mathsf P(N_1=n, N_3=m)~=~\dfrac{\lambda^m 2^{m-n}e^{-3\lambda}}{n!(m-n)!} \;\mathbf 1_{(n,m)\in\Bbb N^2, 0\leq n\leq m, }$$

(NB: the support is $\{(n,m)\in\Bbb N^2: 0\leq n \leq m\}$ because there cannot be less arrivals in the whole three minutes as there were in the first one minute.)

And in your equation

$$\begin{align}\mathsf E(N_1^2/N_3) & = \sum_{n=0}^\infty\sum_{m=n}^\infty \dfrac{n^2}{m}\dfrac{\lambda^m2^{m-n}e^{-3\lambda}}{n!(m-n)!} \\ & = \sum_{n=1}^\infty\sum_{m=n}^\infty \dfrac{n^2}{m}\dfrac{\lambda^m2^{m-n}e^{-3\lambda}}{n!(m-n)!} \tag{$\star$}\end{align}$$

Which works if we allow, as mentioned in the post comments, that $\lim_{n\to 0}n^2/n = 0$

However, yes, unfortunately the same convenience does not help avoid blowout for the other expectation.

$$\begin{align}\mathsf E(N_3^2/N_1) & = \sum_{m=0}^\infty\sum_{n=0}^m \dfrac{m^2}{n}\dfrac{\lambda^m2^{m-n}e^{-3\lambda}}{n!(m-n)!} \\ & = \sum_{m=1}^\infty\sum_{n=0}^m \dfrac{m^2}{n}\dfrac{\lambda^m2^{m-n}e^{-3\lambda}}{n!(m-n)!} \tag{$\star$}\end{align}$$

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  • $\begingroup$ Thank you! It seems I forgot to take care about the indices of our inner sum (starting at $m = n$). How do we deal with division by zero for $n = m = 0$? $\endgroup$ – JonMann Feb 17 '17 at 1:33
  • $\begingroup$ We don't. The second expectation is indefinite. $\endgroup$ – Graham Kemp Feb 17 '17 at 1:43
  • $\begingroup$ Thank you! All is clear after having gone through your edit. Is there an obvious way to evaluate these sums? I'm satisfied leaving it in such a form since the question is essentially answered at this point. However, wouldn't it be nice if it was indeed convergent to something elegant? $\endgroup$ – JonMann Feb 17 '17 at 1:43
  • $\begingroup$ Just to wrap up a few lingering questions: Regarding your final point, that we can't do the same trick to avoid blowout. Can we not swap the sums (via Fubini's Theorem) $\sum^\infty_{m = 0} \sum^m_{n = 0} = \sum^\infty_{n = 0} \sum^\infty_{m = n}$, and then perform the same trick as before? Also, is it just a typo that we have written $e^{-m\lambda}$ instead of $e^{-3\lambda}$? $\endgroup$ – JonMann Feb 17 '17 at 2:16
  • $\begingroup$ We still get an infinitude of non-zero numbers divided by zero. $$\sum_{m=0}^\infty m^2 \sum_{n=0}^m \frac{1}{n}\times\text{stuff} = \sum_{n=0}^\infty \frac 1n \sum_{m=n}^\infty m^2\times\text{stuff}$$ If the "stuff" does not cancel the $n$ in the denominator, the bloat cannot be eliminated (that is: the series will not converge to a finite value). $\endgroup$ – Graham Kemp Feb 17 '17 at 3:15

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