# Show $P(X > x) = e^{-\lambda x} \forall x > 0$ and some $\lambda > 0$ [duplicate]

Given that $$X$$ is a non-negative random variable, satisfies $$P(X = x) = 0$$ and

$$P(X > x + y\mid X > x) = P(X > y)$$ $$\forall x, y \in \mathbb{R^{+}}$$. Prove that $$P(X > x) = e^{-\lambda x}$$ $$\forall x > 0$$ and some $$\lambda > 0$$

By applying the definition of conditional probability, for now I have $$P(X > x) = \frac{P(X > x + y)}{P(X > y)}$$. I guess we are trying to prove something like $$P(X > x) > 0$$, but I dont know how to proceed.

• Also see this answer. – StubbornAtom Dec 6 '19 at 9:52

If $$f(x)=P(X>x)$$ then $$f(x+y)=f(x)f(y)$$ and $$f$$ is right-continuous. It is well know that the only measurable solutions of this functional equation are of the type $$f(x)=e^{cx}$$.

Note that $$f(x) \leq 1$$ which forces $$c$$ to be $$\leq 0$$. Since $$f(x) \to 0$$ as $$x \to \infty$$, $$c$$ cannot be $$0$$. Hence $$c <0$$. Take $$\lambda =-c$$.

Sketch of proof of the fact that the solutions of the functional equation are of the type $$e^{cx}$$: Let $$g(x)=\log (f(x))$$. Then $$g(nx)=ng(x)$$. This implies $$g(rx)=rg(x)$$ for all positive rational numbers $$r$$. Using right continuity it follows that $$g(rx)=rg(x)$$ for al $$r,x>)$$. Put $$r=\frac 1 x$$ to get $$g(1)=\frac {g{(x)}} x$$ or $$g(x)=cx$$ where $$c=g(1)$$. Hence $$f(x)=e^{cx}$$.

• I see... But just wondering why does the question emphasize $x, y > 0$ and $\lambda > 0$? @Kabo Murphy – TrueWarrior09 Dec 6 '19 at 6:51
• Since $X$ is a non-negative random variable $P(X>x)=1$ for all $x \leq 0$. So only $x>0$ is relevant. @TrueWarrior09 – Kavi Rama Murthy Dec 6 '19 at 7:16
• @TrueWarrior09 As for why $\lambda>0$, note that the cumulative distribution function of $X$ is $F_X(x)=P(X\le x)=1-P(X>x)=1-e^{-\lambda x}, x>0$ and $0$ otherwise, which must converge to $1$ as $x\to\infty$. – Shubham Johri Dec 6 '19 at 7:31
• @TrueWarrior09 I have provided more details now. – Kavi Rama Murthy Dec 6 '19 at 7:41

\begin{align} \Pr(Y>2\mid Y>1) & = \Pr(Y>1) \\ \Pr(Y>3\mid Y>2) & = \Pr(Y>1) \\ \Pr(Y>4\mid Y>3) & = \Pr(Y>1) \\ \Pr(Y>5\mid Y>4) & = \Pr(Y>1) \\ & \,\,\,\vdots \end{align} \begin{align} \Pr(X>5) & = \Pr(X>5\mid X>4)\cdot\Pr(X>4) \\[10pt] & = \Pr(X>1)\cdot\Pr(X>4) \\[10pt] & = \Pr(X>1)\cdot \Pr(X>4\mid X>3)\cdot\Pr(X>3) \\[10pt] & = \Pr(X>1)^2 \cdot\Pr(X>3) \\[10pt] & = \Pr(X>1)^2 \cdot\Pr(X>3\mid Y>2)\cdot\Pr(X>2) \\[10pt] & = \Pr(X>1)^3 \cdot\Pr(X>2) \\[10pt] & = \Pr(X>1)^3 \cdot\Pr(X>2\mid Y>1)\cdot\Pr(X>1) \\[10pt] & = \Pr(X>1)^4 \cdot\Pr(X>1) \\[10pt] & = \Pr(X>1)^5. \\[10pt] \Pr(Y>n) & = \Pr(X>1)^n. \end{align} So $$\Pr(X>x)$$ is an exponential function of $$x$$ as long as $$x$$ is an integer.

But now suppose instead of increments of $$1,$$ you use increments of $$0.001.$$ Then the same argument shows $$\Pr(X>x)$$ is an exponential function of $$x$$ as long as $$x$$ is an integer multiple of $$0.001.$$

And moreover, $$\Pr(X>x)$$ is a decreasing function of $$x$$ even without that restriction, since if $$x_1 then $$\Pr(X>x_1) = \Pr(x_1x_2).$$

And then take increments of $$0.00001,$$ etc.

The only functions of $$x$$ that are thus squeezed between decreasing functions that are exponential functions when restricted to integer multiples of $$0.000\ldots001,$$ and that continued to be so no matter how many $$0$$ you put in there, are exponential functions of $$x.$$

And every exponential function of $$x$$ is of the form $$a^2 = e^{-\lambda x}$$ where $$\lambda = -\log_e a.$$