# Variance-Covariance matrix of Weibull Distribution for right-censored data

The probability distribution function, cumulative distribution function and survival function of Weibull distribution are given by respectively,

$$f(t;\alpha, \beta)= \dfrac{\beta}{\alpha^{\beta}} t^{\beta - 1} exp\left[- \left(\dfrac{t}{\alpha} \right)^{\beta} \right]$$

$$F(t;\alpha, \beta)= 1 - exp\left[- \left(\dfrac{t}{\alpha} \right)^{\beta} \right]$$

$$S(t;\alpha, \beta)= exp\left[- \left(\dfrac{t}{\alpha} \right)^{\beta} \right]$$

where $\alpha$ is a scale parameter and $\beta$ is a shape parameter.

Here, we are interested in reparameterization of Weibull distribution in terms of scale parameter $\sigma=\dfrac{1}{\beta}$ and location parameter $\mu = \log (\alpha)$. Thus, we substitute those in to the equations above, we will get $f(t)$, $F(t)$ and $S(t)$ like,

$$f(t;\sigma, \mu)= \dfrac{1}{\sigma \left[\exp(\mu)\right]^{\frac{1}{\sigma}}} t^{\frac{1}{\sigma} - 1} exp\left[- \left(\dfrac{t}{\exp(\mu)} \right)^{\frac{1}{\sigma}} \right]$$

$$F(t;\sigma, \mu)= 1 - exp\left[ - \left(\dfrac{t}{\exp(\mu)} \right)^{\frac{1}{\sigma}} \right]$$

$$S(t;\sigma, \mu)= exp\left[- \left(\dfrac{t}{\exp(\mu)} \right)^{\frac{1}{\sigma}} \right]$$

Now, we are trying to derive variance-covariance matrix of these parameters, i.e. $\sigma$ and $\mu$ when we have right-censored data. The definition and how to find likelihood function of right-censored data are given below:\

Let $f(t)$ and $S(t)$ denote the pdf and survivor functions, each of which is a function of a parameter vector $\theta$ and individual covariate information $x_i$ ($i= 1, \ldots, n$). Assume independent observations. Let $T_i$ denote the lifetime for the $i^{th}$ subject and let $C_i$ denote the time from the date of entry to the end of the study data. Thus we observe $t_i = \min\{T_i, C_i\}$. Finally, let $Y_i$ be an indicator variable, where 1 indicates the observation is right censored, i.e., $T_i > C_i$ and 0 indicates the observation is not censored. Then

$$L(\theta) = \prod_{i=1}^n f(t_i|\theta)^{1-Y_i} S(t_i|\theta)^{Y_i}.$$

Each observation contributes one factor in the likelihood, either the value of the density or of the survivor probability.

If we take the natural logarithm of this, we will get log-likelihood function which is

$$\begin{split} \log L(\theta) &= \sum_{i=1}^n (1-Y_i) f(t_i|\theta) + \sum_{i=1}^n (Y_i) S(t_i|\theta)\\ &=\sum_{i=1}^n (1-Y_i) \left\{-\log (\sigma) - \frac{\mu}{\sigma} + \frac{\log(t)}{\sigma} - \log(t) - \left[\dfrac{t}{\exp(\mu)}\right]^{\frac{1}{\sigma}} \right\} - \sum_{i=1}^n Y_i \left[\dfrac{t}{\exp(\mu)}\right]^{\frac{1}{\sigma}} \end{split}$$

From then on, I know what I need to do. I need to take the first derivaties with respect to $\sigma$ and $\mu$ and find score matrix. And then I have to take second derivatives and get Hessian matrix so I can easily get observed Fisher Information matrix.

but I cannot do it because it is too complicated. Can someone help me about it?