If $Y_i$ has a $\textrm{Poisson}(\lambda_i)$ distribution then it has density function,
$$p(Y_i|\lambda_i) = \frac{\lambda_i^{Y_i}}{Y_i!}e^{−\lambda_i}$$
Suppose we think the variables $Y_1, \dots , Y_n$ follow a Poisson distribution. For each variable, we have an predictor $X_i$ and would like to model $Y$ as a function of X. So we get the poisson regression model with:
$$λ_i = e^{\beta_0+\beta_1X_i}$$
write the log likelihood function in terms of $x,y,\beta_1$ and $\beta_0$.
I am a little stuck with this and think that the likelihood function here would be equal to $$-\sum e^{\beta_0+\beta_1 x_i}-\sum \log(y_i!)+\sum y_i (\beta_0+\beta_1 x_i)$$
Is this correct and if not how would I go about finding this log likelihood function?
Best Answer
The likelihood function is $$ L(\beta_0,\beta_1\mid(Y_i)_{i=1}^n)=\prod_{i=1}^n p(Y_i\mid e^{\beta_0+\beta_1 X_i})=\prod_{i=1}^n\frac{(-e^{\beta_0+\beta_1 X_i})^{Y_i}}{Y_i!}\exp(-e^{\beta_0+\beta_1 X_i}) $$ and so the log-likelihood is $$ \begin{align*} l(\beta_0,\beta_1\mid(Y_i)_{i=1}^n)&=\log L(\beta_0,\beta_1\mid(Y_i)_{i=1}^n)\\ &=\sum_{i=1}^n\log\left[\frac{(e^{\beta_0+\beta_1 X_i})^{Y_i}}{Y_i!}\exp(-e^{\beta_0+\beta_1 X_i})\right]\\ &=\sum_{i=1}^n\log(e^{\beta_0+\beta_1 X_i})^{Y_i}-\sum_{i=1}^n\log(Y_i!)-\sum_{i=1}^ne^{\beta_0+\beta_1 X_i}\\ &=\sum_{i=1}^n(\beta_0+\beta_1 X_i)Y_i-\sum_{i=1}^n\log(Y_i!)-\sum_{i=1}^ne^{\beta_0+\beta_1 X_i}. \end{align*} $$