Given data for $x$ and $t$. $X_i|p_i ~ Bin(2,p_i)$ where $p_i=\frac{\exp{(\beta_0+\beta_1t_i)}}{1+\exp{(\beta_0+\beta_1t_i)}}$, I am trying to estimate the MLE $\beta_0$ and $\beta_1$.
The log-likelihood function I got is: $l(\theta)=c+\Sigma_i(x_i\eta_i-2\log(1+\exp\eta_i))$ where $\eta_i=\beta_0+\beta_1t_i$.
Solving for the first derivative with respect to $\beta_0$ and $\beta_1$ I get:
$$0= \frac{\partial l}{\partial \beta_0}=\Sigma_i x_i - \Sigma_i \frac{2 \exp(\beta_0+\beta_1t_i)}{1+\exp(\beta_0+\beta_1t_i)}$$
$$0=\frac{\partial l}{\partial \beta_1} = \Sigma_i t_i x_i - \Sigma_i \frac{2 t_i \exp(\beta_0+\beta_1t_i)}{1+\exp(\beta_0+\beta_1t_i)} $$
I think that this is correct so far. Now I am having trouble setting up the problem to get the estimators using the Newton method. For our Newton method we give initial parameters to $\beta_0$ and $\beta_1$, how do we handle the $x$ and $t$?
Thanks
I am not sure that this answers the question but if the model is $$p_i=\frac{\exp{(\beta_0+\beta_1t_i)}}{1+\exp{(\beta_0+\beta_1t_i)}}$$you can rewrite it as $$\frac{p_i}{1-p_i}=\exp{(\beta_0+\beta_1t_i)}$$ and taking logarithms $$\log\big(\frac{p_i}{1-p_i}\big)=\beta_0+\beta_1t_i$$ So, a linear regression $$z_i=\beta_0+\beta_1t_i$$ will give you good estimates of the parameters if you define for this preliminary step $z_i=\log\big(\frac{p_i}{1-p_i}\big)$