I want to formulate the likelihood function in a linear regression problem in which I have censored observations. Considering the dataset $(x_i,y_i)$, I suppose that the dependent variable is normally distributed with mean $y=ax+b$ and standard deviation $\sigma$. The model parameter vector is $\theta=(a,b,\sigma)$
In case there is no censored observation, the likelihood function can be written as
$L(\theta;data)=\prod_{I=1}^N f(y_i|\theta)$
where $f(y_i|\theta)$ is the density function of $y$ given the parameters $\theta$.
In case of right censored data, the likelihood is
$L(\theta;data)=\prod_{I=1}^N [f(y_i|\theta)]^{\delta_i} [1-F(y_i|\theta)]^{1-\delta_i},$
where the first term is evaluated for non censored observations and the second term deals with censored data, since the value of the right censored indicator $\delta$ is 1 for uncensored data and 0 for censored data. Moreover, $F$ is the cdf of y and the term in the second parenthesis is the survival function.
Thus, considering the pdf of the Normal distribution having mean $\mu$ and standard deviation $\sigma$
$f(y)=\frac{1}{\sigma \sqrt(2 \pi)} \exp(-0.5 (\frac{y-\mu}{\sigma})^2) $
its cdf
$F(y)=\int_{-\infty}^{y} \frac{1}{\sigma \sqrt(2 \pi)} \exp(-0.5 (\frac{y-\mu}{\sigma})^2) dy $
and the linear regression model
$y=ax+b$
the likelihood function can be expressed, in case of homoskedasticity, as
$$L(\theta;data)=\prod_{I=1}^N [\frac{1}{\sigma \sqrt(2 \pi)} \exp(-0.5 (\frac{y_i-(ax_i+b)}{\sigma})^2) ]^{\delta_i} [1-\int_{-\infty}^{y} \frac{1}{\sigma \sqrt(2 \pi)} \exp(-0.5 (\frac{y-(ax_i+b)}{\sigma})^2) dy ]^{1-\delta_i}.$$
Then, for simplicity I can modify the likelihood in the loglikelihood function obtaining:
$$\log(L(\theta;data))=\sum_{I=1}^N (\delta_i) \log (\frac{1}{\sigma \sqrt(2 \pi)} \exp(-0.5 (\frac{y_i-(ax_i+b)}{\sigma})^2) ) + (1-\delta_i) \log (1-\int_{-\infty}^{y} \frac{1}{\sigma \sqrt(2 \pi)} \exp(-0.5 (\frac{y-(ax_i+b)}{\sigma})^2) dy )$$
Do you agree with this formulation of the likelihood function?
Does somebody have some good reference about this topic?
