Reliability function, proving exponential distribution

Solution 1:

Below is the standard argument, which is undoubtedly done in essentially the same way in your textbook. After a small amount of probability, the rest is just calculus. We are told that $$P(X>s+t\,|\,X>t) = P(X>s). \qquad\qquad(\ast)$$ Let $A$ be the event $X>s+t$, and let $B$ be the event $X>t$. We know that $$P(A\,|\,B)=\frac{P(A\cap B)}{P(B)}.$$ In this case, $P(A\cap B)=P(A)$. So
$$P(X>s+t\,|\,X>t) = \frac{P(X>s+t)}{P(X>t)}.$$ Using $(\ast)$, we conclude that $$\frac{P(X>s+t)}{P(X>t)}=P(X>s).$$ This may be more compactly rewritten as $$R(s+t)=R(s)R(t).$$ Precisely this equation was given in a comment by Dilip Sarwate.

To put you on more familiar ground, we replace $s$ by $x$, and $t$ with $h$. So we have reached the equation $$R(x+h)=R(x)R(h).\qquad\qquad(\ast\ast)$$ Subtract $R(x)$ from both sides, and then divide by $h$. We arrive at $$\frac{R(x+h)-R(x)}{h}=R(x)\frac{R(h)-1}{h}.$$ Let $h$ approach $0$. As $h$ approaches $0$, the right-hand side, by definition, approaches $R'(0)$, which we were told is $-\lambda$. So the left-hand side has a limit, which by definition is $R'(x)$. (By only considering positive $h$ only, we are being a little dishonest. Don't worry about it too much.)

So after some calculation, we have reached the differential equation $$R'(x)=-\lambda R(x), \quad\text{or if you prefer}\quad\frac{dR}{dx}=-\lambda R.$$ This is the familiar differential equation for exponential decay. The general solution is $$R(x)=R(0)e^{-\lambda x}.$$ We were told that $R(0)=1$. It follows that $R(x)=e^{-\lambda x}$.

So the cumulative distribution function $F_X(x)$ of the random variable $X$ is $1-e^{-\lambda x}$ (for $x>0$). Differentiate with respect to $x$. We conclude that the probability density function $f_X(x)$ of $X$ is $\lambda e^{-\lambda x}$ (for $x>0$).