Probability Theory – Proof That Positive Continuous Random Variables with Memoryless Property are Exponentially Distributed

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The theorem to prove is: $X$ is a positive continuous random variable with the memoryless property, then $X \sim Expo(\lambda)$ for some $\lambda$. The proof is explained in this video, but I will type it out here as well. I would like to get some clarification on certain parts of this proof.

Proof

Let $F$ be the CDF of $X$, and let $G(x)=P(X>x)=1-F(x)$. The memoryless property says $G(s+t)=G(s)G(t)$, we want to show that only the exponential will satisfy this.

Try $s=t$, this gives us $G(2t)=G(t)^2,G(3t)=G(t)^3,…,G(kt)=G(t)^k$.

Similarly, from the above we see that $G(\frac{t}{2})=G(t)^\frac{t}{2},…,G(\frac{t}{k})=G(t)^{\frac{1}{k}}$.

Combining the two, we get $G(\frac{m}{n}t)=G(t)^\frac{m}{n}$ where $\frac{m}{n}$ is a rational number.

Now, if we take the limit of rational numbers, we get real numbers. Thus, $G(xt)=G(t)^x$ for all real $x>0$.

If we let $t=1$, we see that $G(x)=G(1)^x$ and this looks like the exponential. Thus, $G(1)^x=e^{xlnG(1)}$, and since $0 <G(1) \leq 1$, we can let $lnG(1)=-\lambda$.

Therefore $e^{xlnG(1)}=e^{-\lambda x}$ and only exponential can be memoryless.

So there are several parts that I am confused about:

  1. Why do we use $G(x)=1-F(x)$ instead of just $F(x)$?
  2. What does the professor mean when he says that you can get real numbers by taking the limit of rational numbers. That is, how did he get from the rational numbers $\frac{m}{n}$ to the real numbers $x$?
  3. In the video, he just says that $G(x)=G(1)^x$ looks like an exponential and thus, $G(x)=G(1)^x=e^{xlnG(1)}$. How did he know that this is an exponential?

Best Answer

Why do we use $G(x)=1−F(x)$ instead of just $F(x)$?

Because the memoryless property is that: $\mathsf P(X>t+s\mid X>s)=\mathsf P(X>t)$

So we can use this to state: $$\begin{align}\mathsf P(X>s+t) =&~\mathsf P(X>s)\mathsf P(X>s+t\mid X>s)\\=&~\mathsf P(X>s)\mathsf P(X>t)\\[2ex] 1-F_X(s+t)=&~ (1-F_X(s))(1-F_X(t))\\[1ex] F_X(s+t)=&~ F_X(s)+F_X(t)-F_X(s)F_X(t)\\[3ex] G_X(s+t) =&~ G_X(s)G_X(t) & G_X(z):=1-F_X(z)\end{align}$$

Using $G$ gives a more useful result.

What does the professor mean when he says that you can get real numbers by taking the limit of rational numbers.

Every real number is the limits of some sequence of rational numbers. $$\forall r\in\Bbb R :\lim_{n\in\Bbb N, n\to\infty}\frac{\lfloor{rn}\rfloor}{n}=r$$

In the video, he just says that $G(x)=G(1)^x$ looks like an exponential and thus, $G(x)=G(1)^x=e^{x\ln G(1)}$.   How did he know that this is an exponential?

Because it looks like it.   It is an easily recognised pattern.

By definition of the $\ln()$ function, for any $a$ (except zero), $a^x = e^{x\ln a}$.

Thus $G(1)^x=e^{x\ln G(1)}$

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