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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} = -\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:

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

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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 1

up vote 1 down vote accepted

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! –  Alistair Feb 8 '12 at 20:32

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