Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A study of patients’ survival was classified by sex (female or male) with follow-up of patients until the patient died or the study ended. We have the following information: $y_1$ - Number of deaths of female patients; $t_1$ - Females’ time at risk/person years; $y_2$ - Number of deaths of male patients; and $t_2$ - Males’ time at risk/person years.

The model with only one covariate (sex) is: $$ \log\left(\frac{E[Y]}{t}\right) = \log \left(\frac \mu t\right) = \alpha + \log(t) + \beta 1_{\mathrm{Sex}}. $$

Assuming normal distribution approximation, i.e. if $Y \sim \mathrm{Poisson}(\lambda t)$, then $Y\sim N (\lambda t, \lambda t)$ and based on the maximum likelihood theory, derive $\alpha$ and $\beta$ as: $$ \hat{\alpha} = \log \left(\frac{y_1}{t_1}\right),\quad \hat{\beta}= \log\left(\frac{y_2/t_2}{y_1/t_1}\right) = \log\left(\frac{y_2}{t_2}\right) – \log\left(\frac{y_1}{t_1}\right). $$

The likelihood function with offset takes the following form-- P(Y=y) = [exp^(-lambda.t).(lambda.t)^y]/y!

If lambda1= y1/t1; and lambda2= y2/t2; what are the values for t and y; and how to consolidate two lambdas into one for plugging it into the above function in order to solve for alpha and beta hats?

Table 9.11 at page 386 of Agresti's book available at http://xa.yimg.com/kq/groups/20006164/692104409/name/agresti-second-edition.pdf may give further insight to solve this problem, as this is a simpler adaptation of the problem discussed in the book.

I have no idea how to solve this. Could anyone help please?

share|improve this question

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.