# Solved:Derive the expression for survival function of $T$,

#### ByDr. Raju Chaudhari

Jul 12, 2020

Derive the expression for survival function of $T$, where $T$ is the survival time having Two parameter weibull distribution.

#### Solution

Let the survival time $T$ follows two parameter Weibull distribution. The pdf of $T$ is

 \begin{aligned} f(x;\alpha, \beta)&= \begin{cases} \frac{\alpha}{\beta} \big(\frac{x}{\beta}\big)^{\alpha-1}e^{-\big(\frac{x}{\beta}\big)^\alpha}, & x>0, \alpha, \beta>0; \\ 0, & Otherwise. \end{cases} \end{aligned}

Distribution function of Weibull Dist.

Let $T$ be the survival time and $T\sim W(0,\alpha,\beta)$. Then the pdf of $T$ is

 \begin{aligned} f(t;\alpha, \beta)&= \begin{cases} \frac{\alpha}{\beta} \big(\frac{t}{\beta}\big)^{\alpha-1}e^{-\big(\frac{t}{\beta}\big)^\alpha}, & t>0, \alpha, \beta>0; \\ 0, & Otherwise. \end{cases} \end{aligned}

The distribution function of Weibull distribution is

 \begin{aligned} F(t) &= P(T \leq t) \\ &= \int_0^t f(t)\; dt\\ &=\int_0^t \frac{\alpha}{\beta} \big(\frac{t}{\beta}\big)^{\alpha-1}e^{-\big(\frac{t}{\beta}\big)^\alpha}\; dt \end{aligned}

Let $Y = -e^{-\big(\frac{t}{\beta}\big)^\alpha}$. Then

 \begin{aligned} \frac{dY}{dt} & = -\bigg[- e^{-\big(\frac{t}{\beta}\big)^\alpha} \frac{\alpha}{\beta}\big(\frac{t}{\beta}\big)^{\alpha-1} \bigg]\\ &= \frac{\alpha}{\beta}\big(\frac{t}{\beta}\big)^{\alpha-1}e^{-\big(\frac{t}{\beta}\big)^\alpha}\\ &= f(t). \end{aligned}

Thus $\int f(t) \; dt = Y$. Using this, the distribution function of $T$ is

 \begin{aligned} F(t) &= \bigg[-e^{-\big(\frac{x}{\beta}\big)^\alpha} \bigg]_0^t \\ &= \bigg[-e^{-\big(\frac{t}{\beta}\big)^\alpha} + 1\bigg]\\ &= 1- e^{-\big(\frac{t}{\beta}\big)^\alpha}. \end{aligned}
Thus the survival function is

 \begin{aligned} S(x) &= 1- F(t)\\ &= e^{-\big(\frac{t}{\beta}\big)^\alpha}. \end{aligned}

The force of mortality is given by

 \begin{aligned} \mu(t) & = \frac{f(t)}{1-F(t)}\\ &= \frac{ \frac{\alpha}{\beta} \big(\frac{t}{\beta}\big)^{\alpha-1}e^{-\big(\frac{t}{\beta}\big)^\alpha}}{e^{-\big(\frac{t}{\beta}\big)^\alpha}}\\ &= \frac{\alpha}{\beta} \big(\frac{t}{\beta}\big)^{\alpha-1} \end{aligned}