Memoryless Property For Exponentials

If $X\sim \text{Exponential}(\lambda )$, then:

$$\begin{array}{r}P(X>s+|X>s)=P(X>t),\forall s,t\ge 0\\ P(X>s+t\mid X>s)=P(X>t),\forall s,t\ge 0\end{array}$$

The probability that the process continues for at least another $t$ units of time, given it has already lasted $s$, is the same as the original probability of lasting $t$.

For PDF:

$$f_X(x)=\{0,x<0f_X(x)=\{\begin{array}{ll}\lambda e^{-\lambda x},& x\ge 0\\ 0,& x<0\end{array}$$

For CDF:

$P(X>x)=1-FX(x)=e-\lambda xP(X>x)=1-F_X(x)=e^{-\lambda x}$

Proof:

$$\begin{array}{rl}P(X>s+t\mid X>s)& =P(X>s+t)P(X>s)\\ & =e-\lambda (s+t)e-\lambda s=e-\lambda t=P(X>t)\\ & \\ P(X>s+t\mid X>s)& =\frac{P(X>s+t)}{P(X>s)}\\ & =\frac{e^{-\lambda (s+t)}}{e^{-\lambda s}}\\ & =e^{-\lambda t}=P(X>t)\end{array}$$

The exponential distribution is memoryless — it “forgets” how much time has passed. It’s the only continuous distribution with this property.

This property is ideal for Poisson Problems:

• Time between radioactive decays
• Time between customer arrivals
• Time until hardware failure