Memoryless

Think of this as the “continuous” version of “if I flip a coin 100 times and get 50 heads, then the next one must be tails, because we’re due for one.” It is incorrect.

Example

Support $X$ is the waiting time from today until the next earthquake, and $X\sim \text{Expo}(17)$. Suppose the world doesn’t have an earthquake for 3 months.

This does not make us due for an earthquake.

If we’ve waited $\frac{1}{4}$ of a year, then the additional waiting time left is given by $X-\frac{1}{4}$. However, we know that $X\ge \frac{1}{4}$, as we’ve already passed 3 months. Then, the distribution is given by

$$P\left(X-\frac{1}{4}\mid X\ge \frac{1}{4}\right)=P(X\ge t)$$

Here, $X\ge \frac{1}{4}$ is the event that we’re still waiting for the next earthquake, 3 months down the line. The distribution is the same as the original distribution of $X$.

It doesn’t matter how long we’ve “waited.” An additional wait length $t$ is just as likely now as it was when we first started.

• The Exponential distribution is the only continuous, memoryless distribution.
• Realistic for some processes (radioactive decay), less so for others (human life).
• Connected to Poisson processes and the “finite” version of the Geometric distribution, which counts the “finite” waiting time until the first success.

Formal definition

Formally, we say a distribution is memoryless if a r.v. $X$ from that distribution satisfies

$$P(X\ge s+t\mid X\ge s)=P(X\ge t)$$

for all $s,t>0$. Furthermore, these two distributions are Independent of one another.