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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 $\lim\limits_{h\to \infty}$ $\dfrac{dR(t)}{dt} = -\lambdaR(t)$ (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$$

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    @DilipSarwate, thanks for the quick response. To be honest, I have an exam coming up and I am just trying to solidify some concepts in my mind, some questions are from homework assignments, others are just questions I am asking in order to make sense of the concept of survival functions in my head. Sorry if you might have felt that I misled you in some way in order to get you to answer my question.2012-02-08

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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).

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    **exponential decay**, that's what I was looking for. Thanks!2012-02-08