Survival Functions and PDFs I have a general question about survival functions and their associated PDFs (probability density functions).
Background. A survival function $s(x)$ is the probability that an individual will survive $x$ more years. In particular $s(x) = P(X > x)$. Also $\mu(x)$ is the "force of mortality." In particular, $\mu(x) \ dx$ gives the probability that an individual aged $x$ will die in the interval $(x, x+dx)$. 

Question. Why is the pdf of $X$ the following: $$f_{X}(x) = s(x) \cdot \mu(x)$$ where $\mu(x)$ is the "force of mortality"? 

My Answer. The random amount of years $X$ a person will live is conditional upon the fact that he survived up to the current time? So we have to weight the fact that the person has probability of dying or living at some time? 
 A: What is the probability that a person dies in the time interval $(x,x+dx)$?  By definition, it's $f(x)dx$.  But in order for that person to die in that interval, he must have survived until time $x$.  So another way to express the probability is 
$P$ (person survives until time $x) \times P$ (person dies in $(x,x+dx)$|person survived until time $x$);
i.e., $s(x) \mu(x) dx$.  So we get $f(x) = s(x) \mu(x)$.
A: We give a somewhat formal derivation of the result.  
Let the continuous random variable $X$ have density function $f_X(x)$.  Then the survival function $s(x)$ is $P(X>x)$.  Thus
$$s(x)=\int_x^\infty f_X(t)\,dt=1-F_X(x),$$
where $F_X(x)$ is the cumulative distribution function of $x$.
Now we look at definition of the force of mortality $\mu(x)$.  The idea is that we consider the probability of death in the interval $(x,x+h)$, given that we have survived to time $x$, divide by $h$, and take the limit as $h$ approaches $0$ from the right.  So the force of mortality measures the death rate in the cohort of people who are age exactly $x$.
In symbols, 
$$\mu(x)=\lim_{h\to 0^+} \frac{P((x <X \le x+h)|X>x)}{h}.$$
Using the ordinary formula for conditional probability, we find that 
$$\mu(x)=\lim_{h \to 0^+}\frac{F(x+h)-F(x)}{hs(x)}.$$
But the derivative of the cumulative distribution function $F_X(x)$ is $f_X(x)$, so we get
$$\mu(x)=\frac{f_X(x)}{s(x)}.$$
It follows that $f_X(x)=\mu(x)s(x)$.  
