Proof involving Cauchy Equation

Question

In Probability, a probability distribution is said to have the Memoryless Property (https://en.wikipedia.org/wiki/Memorylessness) if the following condition is true:

\begin{equation} \text{Memoryless Property:} \quad P(X > s + t | X > s) = P(X > t) \quad \text{for all $s, t \geq 0$} \quad \end{equation}

In general, the above condition only holds for the Exponential Probability Distribution:

\begin{equation} \text{Exponential PDF:} \quad f(x; \lambda) = \lambda e^{-\lambda x} \end{equation}

My Question: I am trying to understand why the Memoryless Property only holds for the Exponential Distribution.

• In a previous question I posted (https://math.stackexchange.com/posts/4806515), I was told (in the comments section) that the Memoryless Property satisfies the Cauchy Functional Equation $G(t+s)=G(t)+G(s)$: but I don't know why this is true. Why does the Memoryless Property mean that the Cauchy Functional Equation is satisfied?

• While I don't fully understand why, it was so explained to me that the only monotonic (I don't know why the emphasis on "monotonic" is relevant here - I think its because Probability Distribution Functions are strictly required to be monotonic?) function that satisfies the Cauchy Functional Equation is $G(t)=A$ for some constant $A$.

• Some additional comments were provided, such that the above implies $FX(t)=1−e−\lambda t$ for some $\lambda >0$ - but I don't understand why this is implied.

• A last comment is provided where the Cauchy Functional Equation is written as $H(x+y)=H(x)H(y)$, but on the Wikipedia page (https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation) its written in the form of $H(x+y)=H(x) + H(y)$

All in all, I am quite confused on how the Cauchy Functional Equation can be used to prove that the Memoryless Property only holds for the Exponential Distribution. Can someone please help me understand this proof?

Thanks!

References:

• https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation

Note: Here is my own proof that shows that the Exponential Distribution satisfies the Memoryless Property:

\begin{equation} \text{Exponential PDF:} \quad f(x; \lambda) = \lambda e^{-\lambda x} \end{equation}

\begin{equation} \text{Exponential CDF:} \quad P(X \leq x) = F(x; \lambda) = 1 - e^{-\lambda x} \end{equation}

\begin{equation} \text{Survival Function of the Exponential Distribution: 1 - Exponential CDF:} \quad 1 - P(X \leq x) = P(X > x) = e^{-\lambda x} \end{equation}

Using Law of Conditional Probability:

\begin{equation} P(X > s + t | X > s) = \frac{P(X > s + t \text{ and } X > s)}{P(X > s)} = \frac{P(X > s + t)}{P(X > s)} \end{equation}

\begin{equation} P(X > s + t | X > s) = \frac{e^{-\lambda (s + t)}}{e^{-\lambda s}} = e^{-\lambda t} = P(X > t) \end{equation}

Answer 1

OP here: I tried to approach this proof using concepts from Survival Analysis (https://en.wikipedia.org/wiki/Survival_analysis)

Part 1: Typically, we say that the Survival Function is given by $S(t) = P(x>t) = 1 - F(t)$.

In the case of the Exponential Distribution, the Survival Function is given by : $S(t) = P(x>t) = 1 - F(t) = e^{-\lambda t}$

The law of conditional probability tells us that $P(A | B) = \frac{P(A \cap B)}{P(B)}$ . If $A$ and $B$ are independent, then $P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A) *P(B)}{P(B)}$ . We can apply this law for $P(x>s+t | x>s)$ :

$$ P(X>s+t | x>s) = \frac{P(x>s+t \cap x>s)}{P(x>s)} = \frac{P(x>s+t) }{P(x>s)} = \frac{S(s+t) }{S(s)} $$

Note that $P( x>s)$ is implicitly contained within $P(x>s+t)$. This is because the probability of surviving time $s+t$ implies that a minimum of time $s$ has already been survived.

Part 2:

(A) If the Memoryless Property were to hold, then:

$$P(X>s+t | X>s) = P(X>t)$$

and $$S(s+t | s) = S(t)$$

(B) But in general (whether the Memoryless Property is true or not true, does not matter):

$$P(X>s+t | X>s) = \frac{P(x>s+t \cap x>s)}{P(x>s)} = \frac{P(x>s+t )}{P(x>s)} $$

and

$$S(s+t | s) = \frac{S(s+t \cap s)}{S(s)} = \frac{S(s+t )}{S(s)} $$

Now, equating (A) and (B) :

$$ S(t) = \frac{S(s+t )}{S(s)} $$ $$S(s+t) = S(s)*S(t)$$

Part 3: Using the relationship $S(s+t) = S(s)*S(t)$, we can take the natural logarithm on both sides:

$$ln[S(s+t)] = ln[S(s)*S(t)] = ln[S(s)] + ln[S(t)] $$

If we now take the derivative of both sides with respect to $s$:

$$ \text{LHS:} \quad \frac{d}{ds} \ln[S(s+t)] = \frac{S'(s+t)}{S(s+t)} $$ $$ \text{RHS:} \quad \frac{d}{ds} \ln[S(s)] + \frac{d}{ds} \ln[S(t)] = \frac{S'(s)}{S(s)} $$

Since $ \text{LHS = RHS}$, then:

$$\frac{S'(s+t)}{S(s+t)} = \frac{S'(s)}{S(s)} = c \quad \text{where c is some constant, e.g.,} \quad c = - \lambda$$

Now, we integrate the above equation with respect to $s$:

$$\int \frac{S'(s)}{S(s)} \, ds = \int -\lambda \, ds$$

$$ ln[S(s)] = - \lambda * S$$

$$ S(s) = e^{ -\lambda * s} $$

And $S(s) = e^{ -\lambda * s}$ is the survival function of the Exponential Distribution. This means that only a distribution with a Survival Function of $S(s) = e^{ -\lambda * s}$ would satisfy the Memoryless Property. And I think that only the Exponential Distribution will have a Survival Function of $S(s) = e^{ -\lambda * s}$.

Ideally, this would have concluded the proof.

However, am not sure if this is a valid proof - for example, perhaps some other probability distribution $g(x)$ might somehow share the same survival function ... but I am not sure how to proceed (note: apparently this is not possible - survival functions are unique: Can we Prove that a Function of a Function is Unique?).

Note: If the memoryless property is true in general, then:

$$P(X>t+s | x>s) = P(x>t)$$ $$\frac{P(x>t+s)}{P(x>s)} = S(t)$$ $$\frac{S(t+s)}{S(s)} = S(t)$$ $$S(t+s) = S(t)*S(s)$$