Suppose an uniform distribution, with 0 and 80 as parameters, that explains the ‘future time of life’ of a person with age 20. Find:
- $f_T(30)$
- $F_T(30)$
- $S_T(30)$
- $μ_T(30)$
I think he meant 'a person with age 30' instead of 20. I'm not really sure where this 20 should fit. *PS: future time of life, probably is a mistranslation, but I think you got the point
You are given the future lifetime random variable $$T(20) \sim \operatorname{Uniform}(0,80),$$ meaning that $$\Pr[T(20) \le x] = \frac{x}{80}, \quad 0 \le x \le 80.$$
However, your question needs additional clarification. You write for instance $f_T(30)$, but in actuarial practice, $T$ is notation for the future lifetime random variable of a newborn; i.e., $T = T(0)$. If this is the case, we cannot answer the question because we do not have information about the probability ${}_{20} p_{0} = \Pr[T(0) > 20]$.
One alternative interpretation is that $T = T(20)$, so that for instance $f_T(30) = f_{T(20)}(30)$, the density of the future lifetime random variable evaluated at time $T(20) = 30$; i.e., at age $50$. This of course is simply $1/80$.
However, there is yet another interpretation: $f_T(30) = f_{T(30)}(x)$, meaning the question wants the probability density function for a life aged $30$. This is tractable because given that the future lifetime of $(20)$ is uniform, we immediately know that $$T(30) \sim \operatorname{Uniform}(0,70),$$ hence $$f_{T(30)}(x) = \frac{1}{70}, \quad 0 \le x \le 70,$$ and the other functions are easily determined from this. The reason is that given $(20)$ survives $10$ years to $(30)$, we see that $$\Pr[T(30) > x] = \Pr[T(20) > x + 10 \mid T(20) > 10] = \frac{\Pr[T(20) > x + 10]}{\Pr[T(20) > 10]}.$$ In actuarial notation, the above takes on the more familiar form $$({}_x p_{30})( {}_{10} p_{20} ) = {}_{x+10} p_{20}.$$