The question is:

" Let $ \{N(t) , t \geq 0 \} $ be a $PP(\lambda)$. Compute

$$P(N(t) = k | N(t = s) = k + m), $$

where k and m are non-negative integers and $ t, s \geq 0 $ are any real numbers".

So I did this:

Using the definition of joint probability, you get:

$$ P(N(t) = k | N(t = s) = k + m) = \frac{P(N(t) = k, N(t + s) = k + m)}{P(N(t + s) = k + m)} $$

Then I got a bit stuck, looking at the answersm, this happens:

From what I did, it goes to:

$$P(N(t) = k | N(t = s) = k + m) = \frac{P(N(t) = k, N(t + s) - N(t) =m)}{P(N(t + s) = k + m)} $$

Because a PP has independent increments:

$$= \frac{ \left( e^{ \lambda t} \frac{ (\lambda t)^k}{k!} \right) \cdot \left( e^{\lambda s} \frac{(\lambda s)^m}{m!} \right) } { \left( e^{ \lambda (t + s)} \frac{(\lambda (t + s))^{(k+m)}}{(k+m)!} \right) } $$

Because the increments have a Poisson distribution

$$= \begin{pmatrix} k + m & k \end{pmatrix} \frac{t^k s^m}{(t + s)^{k+m} } $$

The questions I have are these:

1) Why do you go from $ N(t + s) = k + m $to $ N(t + s) - N(t) = m$? Does it make any difference if you don't do this?

2) Why do you ignore the -N(t) when writing it in its "exponential form"?

3) What does the last nCr bit mean?


1) Think of each of those $k+m$ events as having occurred either before $t$ or after $t$, and you're trying to calculate the probability that exactly $k$ of them occurred before $t$, which leaves exactly $m$ of them occurring after $t$.

2) $N(t+s)-N(t)$ is just the number of events occurring during a period of length $s$; since this is a Poisson process with constant rate, the probability that $m$ occur in that interval is independent of $N(t)$, depending solely upon $s$, so we can just refer to $s$ instead.

3) nCr is a shorthand for $n$ choose $r$, often written $n \choose r$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.