6
$\begingroup$

I have a data set $Y$, $Y$ is assumed to follow a poisson distribution with mean = $\lambda$ .

However, each element $y_i$ in $Y$ is accompanied with a covariate $x_i$. So here I want to use poisson regression to model this dataset. The link function is assumed to be log($\lambda(i)$)=$\beta_0$+$\beta_1x_i$.

I then want to use glm(Y~X,family='poisson') to fit the model and get $\beta_0$ and $\beta_1$.

After get $\beta_0$ and $\beta_1$, can I calculate the probability mass function given new $y$ and $x$.

I want to know if all of my above thought make sense?

$\endgroup$
  • 3
    $\begingroup$ Besides the fact that a conditional Poisson distribution (conditional on x) does not imply a marginal Poisson distribution, this would seem to be exactly the idea of Poisson regression, yes. $\endgroup$ – Nick Sabbe May 10 '13 at 15:51
3
$\begingroup$

If the probability model for $Y$ is this:

$$P(Y_i=y) = \exp(\lambda_i) {\lambda_i}^y / y!$$

and $i$-th observations rate parameter is in fact given by:

$$ \log(\lambda_i) = \beta_0 + \beta_1 x_i$$

(with no model misspecification per others' comments here)

Then the answer is yes you can calculate the PMF for a new $Y$ observation with a given $X$.

So if $X_i=x$, $$P(Y_i=y) = \exp(\exp(\beta_0 + \beta_1 x)) \exp(\beta_0 + \beta_1 x)^y / y!$$

If, however, the new $X$ observation is not known, then the marginal $Y$ distribution is a complex mixture of Poisson RVs.

$\endgroup$
2
$\begingroup$

Partially answered in comments:

Besides the fact that a conditional Poisson distribution (conditional on $x$) does not imply a marginal Poisson distribution, this would seem to be exactly the idea of Poisson regression, yes. – Nick Sabbe

For more information on Poisson regression, see Scaling vs Offsetting in Quasi-Poisson GLM

$\endgroup$
  • 1
    $\begingroup$ If you used quasilikelihood you would not in fact be able to get a pmf, it should be said. $\endgroup$ – AdamO Sep 28 '18 at 14:23

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.