Finding Survival Function given hazard rate

Question

Let X be a random variable defined for 0 < x < 4 with hazard rate $$lambda(t)=1/(4-t)$$ for 0 <= t <= 4. find the survival function, S(x) = P(X>x).

Using the formula S(x) e^-integral from 0 to infinity \lambda(u)du, I plugged in for lambda and got t-4 after all of the calculations. I do not think that is the correct answer. I'm thinking the P(X>x) has to do something with it, but I don't really understand what that part of it means. My next idea is that the limits from 0 to 4 would have to do something with it. Any thoughts?

Answer 1

For a random variable which is inherently non-negative, we have $$S(x)=\exp\left(-\int_0^x \lambda(t)\,dt\right).\tag{1}$$ In our particular case, we therefore want to find $$\int_0^x \frac{1}{4-t}\,dt.$$ Integration gives (for $0\le x\lt 4$), that in our case $$\int_0^x \lambda(t)\,dt=\ln{4}-\ln(4-x)=-\ln\left(\frac{4-x}{4}\right).$$ Substituting in (1), we get that $S(x)=\frac{4-x}{4}$ for $0\le x\lt 4$.

Answer 2

Your formula for the survival function is incorrect, instead you want $$ S(t)=\exp\left(-\int_0^{\color{red} t} \lambda(s)\,\mathrm ds\right),\quad t>0. $$

Answer 3

By definition $\lambda(t)=-\frac{\bar{F}(t)'}{\bar{F}(t)}$. Therefore in your case $\frac{1}{4-t}=-\frac{\bar{F}(t)'}{\bar{F}(t)}$

Integrating both sides you get $\bar{F}(t)=C(4-t)$ for some constant $C$. Using the relation $\bar{F}(0)=1$ you get $C=1/4$. So, $\bar{F}(t)=\frac{4-t}{4}$ for $t \in [0,4]$ and $\bar{F}(t)=0$ for $t \ge 4$. From here $F(t)=0, t<0, F(t)=t/4, 0 \le t \le 4, F(t)=1, t >4$