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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?
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    $\begingroup$ 1). He wants to get exponent, and as long as $F_{exp} = 1 - e^{-\lambda x}$, he does this to get power function in the end, otherwise it should be more complicated. 2). Every real number is a limit of a sequence of rational numbers ($\mathbb{Q}$ is everywhere dense in $\mathbb{R}$), you need to re-read the course of calculus 3). The power function is easily transformed into exponent: $\text{something}^x = e^{ln(\text{something})*x}$. Recall how derivatives like $\frac{d}{dx}x^x$ are taken. $\endgroup$ – Slowpoke May 27 '16 at 7:00
  • $\begingroup$ 1) It is his aim to show that $G=1-F$ is an exponential function, so has all reason to focus on $G$ (and not on $1-G=F$). $\endgroup$ – drhab May 27 '16 at 7:38
  • $\begingroup$ math.stackexchange.com/questions/107075/… $\endgroup$ – StubbornAtom Dec 6 '19 at 9:52
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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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  • $\begingroup$ I think there is a typo: $\mathsf P(X>s+t\mid X>t)=\mathsf P(X>s)$ $\endgroup$ – Rajat Feb 11 '20 at 4:13
  • $\begingroup$ $\mathsf P(X>s+t \cap X > t) = \mathsf P(X>s+t) = \mathsf P(X>s)\mathsf P(X>t)$ $\endgroup$ – Rajat Feb 11 '20 at 4:20
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In the proposed order:

  1. You can use $G(x)$ instead of $(1-F(x))$ just to make you demonstration less clumsy by not carrying that difference around.
    For instance, the first line would be more readable:

    Try $s=t$, this gives us $G(2t)=G(t)^2$, $G(3t)=G(t)^3$, ..., $G(kt)=G(t)^k$
    Try $s=t$, this gives us $1-F(2t)=(1-F(t))^2$, $1-F(3t)=(1-F(t))^3$, ..., $1-F(kt)=(1-F(t))^k$

  2. Consider the set $E$ of exponents $x$ for which $G(xt)=G(t)^x$.
    Until that point of the proof, all they proved was that $\mathbb{Q}^+\subset E$.
    They did not prove, in particular, that $\mathbb{R}^+\subset E$.
    Now, $\mathbb{R}^+$ is the closure of $\mathbb{Q}^+$ (i.e., the set of all possible limits $a$ for all possibile sequences $(a_i)$ in $\mathbb{Q}^+$) and that is why they mentioned "taking limit of $m/n$".
    Once $\mathbb{R}^+\subset E$, $G(xt)=G(t)^x$ would be valid for all $x\in\mathbb{R}^+$.

  3. Well, $G(1)$ is, indeed, equal to some constant $\kappa\in(0,1)$.
    So saying that $G(x)=G(1)^x$ is the same as saying $G(x)=\kappa^x$, which is an exponential function.
    Since $\kappa = \ln(e^\kappa) = \ln(\exp(\kappa)) = e^{\ln(\kappa)} = \exp(\ln(x))$ (for $\ln$ and $\exp$ are inverse function of one another), then $$G(x) = G(1)^x = \exp(\ln(G(1)))^x = \left(e^{\ln(G(1))}\right)^x = e^{x\ln(G(1))} = e^{x\ln(\kappa)} = e^{x(-\lambda)} = e^{-\lambda x}$$ for some $\lambda>0$, as $\kappa\in(0,1) \implies \ln(\kappa)<0$.

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