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.

  • $\begingroup$ Also see this answer. $\endgroup$ – 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 $f(x)=e^{cx}$.

  • $\begingroup$ I see... But just wondering why does the question emphasize $x, y > 0$ and $\lambda > 0$? @Kabo Murphy $\endgroup$ – TrueWarrior09 Dec 6 '19 at 6:51
  • $\begingroup$ Since $X$ is a non-negative random variable $P(X>x)=1$ for all $x \leq 0$. So only $x>0$ is relevant. @TrueWarrior09 $\endgroup$ – Kavi Rama Murthy Dec 6 '19 at 7:16
  • $\begingroup$ @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 $x\to\infty$. $\endgroup$ – Shubham Johri Dec 6 '19 at 7:31
  • $\begingroup$ @TrueWarrior09 I have provided more details now. $\endgroup$ – 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 $x_1<x_2$ then $\Pr(X>x_1) = \Pr(x_1<X\le x)2) + \Pr(X>x_2).$

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 $\lambda = -\log_e a.$


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