## Survival Function

Assume $T$ is a continuous random variable indicates the death occurrence time, we have:

$$F(t) = P\lbrace T < t\rbrace = \int_0^t f(t) dt \tag{1.1}$$

Then the Survival Function should be:

$$S(t) = P\lbrace T > t\rbrace = 1 - F(t) = \int_t^\infty f(t) dt \tag{1.2}$$

## Harzard Function

An alternative way to characterization the distribution is given by harzard function, or instantaneous rate of occurrence of the event:

\begin{align} \lambda(t) &= \lim_{dt \to 0} \frac{P\lbrace t \le T < t + dt | T \ge t\rbrace}{dt} \\ &= \lim_{dt \to 0} \frac{P\lbrace t \le T < t + dt \rbrace}{P \lbrace T \ge t\rbrace dt} \\ &= \lim_{dt \to 0} \frac{f(t)dt}{S(t) dt} \\ &= \frac{f(t)}{S(t)} \end{align} \tag{2.1}

Given $(1.2)$ we have $\frac{d}{dt} S(t) = -f(t)$, so $(2.1)$ has another form

$$\lambda(t) = -\frac{d}{dt} log S(t) \tag{2.2}$$

We could derive survival function from harzard function as well:

$$S(t) = exp\lbrace - \int_0^t \lambda(x)dx \rbrace = exp\lbrace -\Lambda(t) \rbrace \tag{2.3}$$

In which $\Lambda(t) = \int_0^t \lambda(x)dx$, called cumulative hazard

### Example 2.1

Here we’re modeling a constant risk over time:
$$\lambda(t) = \lambda$$
From $(2.2)$, we could solve corresponding survival function and pdf
\begin{align} S(t) &= exp\lbrace - \int_0^t \lambda(x)dx \rbrace = e^{-\lambda t} \\ f(t) &= \lambda e^{-\lambda t} \end{align}
That is exactly an exponential distribution

### Expectation of Life

Given $S(t)$ or $\lambda(t)$, it’s easy to denote expected value of $T$
$$\mu = \int_0^\infty tf(t)dt =\int_0^\infty S(t)dt$$

## Censoring and the likelihood function

### Censoring Type

1. Type I
Typically 2 types of observatioin:

• A sample of $n$ units is followed for a fixed time $\tau$
• Generalization, fixed censoring: each unit has a fixed time $\tau_i$

In cases above, number of deaths is a random variable.

2. Type II

• A sample of $n$ units is followed as long as necessary until $d$ units have experienced the event
• Generalization, random censoring: Each unit has:
• Censoring time $C_i$
• Potential lifetime $T_i$
• Observe time $Y_i = min\lbrace C_i, T_i\rbrace$
• Indicator $d_i, \delta_i$ tells us whether the observation is terminated by death or censoring

### Likelihood of censoring model

1. Unit died at $t_i$. Since we know it is dead while survives till $t_i$, we have:
$$L_i = f(t_i) = S(t_i)\lambda(t_i) \tag{3.1}$$

2. Unit still alive at $t_i$. We only know it survives till $t_i$
$$L_i = f(t_i) = S(t_i) \tag{3.2}$$

Given 2 conditions above, we have:
$$L = \prod\limits_{i=1}^{n}L_i = \prod\limits_{i} \lambda(t_i)^{d_i}S(t_i) \tag{3.3}$$
Taking logs, considering $(2.3)$, we have:
$$log L = \sum\limits_{i=1}^{n} \lbrace d_ilog\lambda(t_i) - \Lambda(t_i) \rbrace \tag{3.4}$$

### Example 3.1

Considering exponential distribution $\lambda(t) = \lambda$, from$(3.4)$, we have
$$log L = \sum\limits_{i=1}^{n} \lbrace d_ilog\lambda - \lambda t_i \rbrace$$

We could estimate $\lambda$ using MLE:

Let $D=\sum d_i$ denotes the total number of deaths, $T = \sum t_i$ denotes total number of observation time:

\begin{align} log L &= Dlog\lambda - T\lambda \\ \frac{\partial}{\partial \lambda} L &= \frac{D}{\lambda} - T \end{align}

Letting $\frac{\partial}{\partial \lambda} L = 0$ we get the estimation of $\lambda$

$$\hat \lambda = \frac{D}{T}$$

Reference