1
$\begingroup$

Let's say I have a Poisson point process on $\left[0,T\right]$ with rate $\lambda\left(t\right)=2t^2$. Suppose I attach a mark $m_t$ to each point $t$ of the process such that $m_t\sim N\left(t,1\right)$. Then do I calculate the mean measure of the marked process $\left(m_t\right)$ by taking this?

\begin{equation} \mu\left(\left[0,T\right]\right)=\int_0^T 2t\times\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}\left(m-t\right)^2}\,dt \end{equation}

What if $m_t$ were a discrete random variable? For example, the mark associated with the point $t$ is now either $t+1$ or $t-1$ with probability $1/2$ each.

$\endgroup$

migrated from math.stackexchange.com Apr 14 '16 at 5:55

This question came from our site for people studying math at any level and professionals in related fields.

  • $\begingroup$ What do you mean by $m_t\sim N(t,1)$? $\endgroup$ – Math1000 Apr 9 '16 at 14:49
  • $\begingroup$ @user38584 marked process $\endgroup$ – Neil G Apr 14 '16 at 6:09
0
$\begingroup$

For the normally distributed marks case, I think you can forget about the variance since it increases the mean measure by as much as it decreases it. So, set it to zero to make things simple. Then, you have a process whose intensity is $2t^2$ and whose marks are $t$. As far as the mean is concerned more events is the same as bigger marks, so I think it's the same as having intensity $2t^3$. That makes the expectation $\frac12T^4$.

For the Bernoulli case, I think the same logic applies.

$\endgroup$

Your Answer

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