# Likelihood formulation in linear regression with right censored data [duplicate]

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?

• The $f(y)$ you give is not a normal density. Fix what you have there (and what follows from it) – Glen_b -Reinstate Monica Nov 23 '16 at 11:30