I need to find the model over a period of length t. This is what I've done:

Based on the Bayes' theorem, the relationship between the prior, the posterior, and the likelihood function is $p(\theta|x) = \frac{p(x|\theta)p(\theta)}{\int p(x|\theta^{`})p(\theta^{`})}$.

Before computing the posterior $p(\lambda|x)$ with prior $g(\lambda;r,\alpha) = \frac{\alpha^{r} \lambda^{r-1} e^{-\lambda\alpha}}{\Gamma(r)}$ and Poisson pmf $p(x|\lambda(t))= \frac{e^{{-\lambda}t}(\lambda*t)^x }{x!}$

After canceling some terms in the numerator and denominator, $p(\lambda(t)|x)= \frac{ (\alpha+t)^{r+x} \lambda^{r+x-1} e^{-\lambda(\alpha+t)}}{\Gamma(r+x)} = Gamma(\lambda; r + x, \alpha+t)$.

Now we know that $E[X(t)] = \frac{r+x}{\alpha+t}$. However I am trying to calculate: $E[X(t^∗+t)−X(t)|X(t)]$ and not sure how to approach this problem.

  • $\begingroup$ How long is the period from $t$ to $t^*+t$? See if you can figure out how that applies if you were to rewrite your first sentence. Also, you should put the "self-study" tag on for homework or self-study problems. $\endgroup$ – jbowman Feb 10 '18 at 2:23
  • $\begingroup$ The period from $t$ to $t^*+t$ is $t*$. Do I just re-derive posterior? $\endgroup$ – conums Feb 10 '18 at 2:48
  • $\begingroup$ You don't have to rederive it... $t$ is just a variable indicating the length of the period. If the length of the period is $t^*$, just use $t^*$ instead if $t$. $\endgroup$ – jbowman Feb 10 '18 at 2:52
  • $\begingroup$ @jbowman I posted what I think may be the solution below - would love your thoughts on it. $\endgroup$ – conums Feb 10 '18 at 2:54

I am thinking of doing this: $g(\lambda;r,\alpha) = A= \frac{\alpha^{r} \lambda^{r-1} e^{-\lambda\alpha}}{\Gamma(r)}$ and Poisson pmf $p(x|\lambda(t))= B = \frac{e^{{-\lambda}t}(\lambda t^*)^x }{x!}$ and using $p(\lambda(t)|x)= C= \frac{ (\alpha+t)^{r+x} \lambda^{r+x-1} e^{-\lambda(\alpha+t)}}{\Gamma(r+x)} = Gamma(\lambda; r + x, \alpha+t)$. and now recalculating/solving for $p(\theta|x) = \frac{p(x|\theta)p(\theta)}{\int p(x|\theta^{`})p(\theta^{`})}$.

So using the simplified notation, I would obtain: $\frac{AB}C$ which would be another Gamma at which point the $E[X=t^*]$ is $r/\alpha$

| cite | improve this answer | |
  • 1
    $\begingroup$ You are overthinking it. You have an expression for E[x(t)] and you want to find E[x(t*)] . Just substitute t* everywhere you see t. t is a parameter, it's not fixed. If it takes on the value 2, substitute 2 everywhere. If it takes on the value t*, just substitute t* everywhere. $\endgroup$ – jbowman Feb 10 '18 at 3:01
  • $\begingroup$ But doesn't seeing the first $t$ give info about the true value of $\lambda$ so we have to use the posterior distribution of $\lambda$ $\endgroup$ – conums Feb 10 '18 at 3:13
  • $\begingroup$ Please ignore my confusing initial comment here. To clear up a little confusion in your original post, you have $E[X(t)] = \frac{r+x}{\alpha+t}$. This isn't quite right, it's $E[\lambda|X,t]$. So, after you've seen the first $(x,t)$, you have the $x$ and $t$ necessary to plug into $\frac{r+x}{\alpha+t}$ to get the posterior mean of $\lambda$. Now, for $X(t^*+t)-X(t)$, that's the same (distributionally) as $X(t^*)$, so will have mean $\lambda t^*$. You can plug the expression for $E[\lambda|x,t]$ right in to find $E[X(t+t^*)-X(t)]$. $\endgroup$ – jbowman Feb 10 '18 at 16:49
  • $\begingroup$ @jbowman what do you mean by plug in? $\endgroup$ – conums Feb 13 '18 at 22:20
  • $\begingroup$ put into, substitute into. $\endgroup$ – jbowman Feb 13 '18 at 22:23

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.