# Reliability function, proving exponential distribution

We are given $R(t)$ = $P(X>t)$ for all $x > 0$ and

$$R(0) = 1 - Fx(0) = 1\text{ and }\lim\limits_{t \to \infty} R(t) = 0$$

The random variable $X$ also satisfies the memoryless property:

$$P(X>s+t|X>t) = P(X>s)\text{ for }s>0\text{ and }t>0$$

Let $R'(0) = - \lambda\ where \ \lambda\ ,$ is some positive constant. I need to show that X must be exponentially distributed.

Given that $\dfrac{R(t + h) - R(t)}{h}$ = $R(t)\left[\dfrac{R(h) - 1}{h}\right]$

Show that by letting $\lim\limits_{h\to \infty}$ $\dfrac{dR(t)}{dt} = -\lambda$$R(t) (I think we should use Hopital's rule here I am not sure by differentiating \left[\dfrac{R(h) - 1}{h}\right] and letting h tend to 0, we will get -\lambda for this part but I got stuck afterwards). Also argue that X is an exponential random variable with rate parameter \lambda by solving the differential equation above respecting the conditions:$$R(0) = 1 - Fx(0) = 1\text{ and }\lim\limits_{t \to \infty} R(t) = 0$$- Using the definition of conditional probability, the memoryless property can be re-written as$$P\{X > s+t\} = P\{X>s\}P\{X>t\} \Rightarrow R(s+t)=R(s)R(t).$$Many calculus texts show that a continuous function R(t) with this property must be of the form a^t for some a, or identically 0. Maybe you can adapt their proofs to your homework problem. – Dilip Sarwate Feb 8 '12 at 15:31 Hi @Dilip Sarwate, one of the questions was actually to prove that property. Which I did. So now I'm left with these two questions. – Alistair Feb 8 '12 at 15:32 R(0) = a^0 = 1. R'(0) = \left.\frac{d}{dt}a^t\right|_{t=0} = -\lambda. I always have to look up the derivative of a^t with respect to t but if I pulled down the book off my shelf, I suspect I would find that \ln a = -\lambda and so a = e^{-\lambda}, and so R(t) = e^{-\lambda t} or maybe not.... – Dilip Sarwate Feb 8 '12 at 15:41 @DilipSarwate, the thing is we are trying to prove that this is an exponential distribution, for the first part I believe by using l'hopital's rule, with limit of h tending to 0 and differentiating the part in brackets we will get [\dfrac{R'(0)}{1}] which would equal -\lambda – Alistair Feb 8 '12 at 15:45 Please read my comment on your Failure time distribution question where you said all this was not homework. Once you prove that$$R(t) = P\{X > t\} = 1 - F(t) = \exp(-\lambda t),$$for which I more or less gave you a complete proof above, all that remains to be done is show that$$f(t) = \frac{d}{dt}F(t) = -\frac{d}{dt}R(t) = \lambda\exp(-\lambda t)\mathbf 1_{(0,\infty)}$$and claim that X is an exponential random variable with parameter \lambda. – Dilip Sarwate Feb 8 '12 at 16:01 ## 1 Answer Below is the standard argument, which is undoubtedly done in essentially the same way in your textbook. After a small amount of probability, the rest is just calculus. We are told that$$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\$).

-
exponential decay, that's what I was looking for. Thanks! – Alistair Feb 8 '12 at 20:32