Why can't I write $P(X>5|X>1) = P(X>5)$? I have a confusion with the memorylessness property of exponential distribution.
If exponential distribution is memoryless (i.e. the past has no bearing on its future behavior), why can't I write $P(X>5|X>1) = P(X>5)$? Why  $P(X>5|X>1) = P(X>4)$?
Can anyone kindly explain it to me?
 A: This property emerges from the fact that the exponential distribution has a constant hazard rate.  If you look at the exponential random variable as representing a time-to-failure, this means that conditioning on past survival does not alter the distribution of the remaining time-to-failure.  The property can be verified from the fact that $X \sim \text{Exp}(\lambda)$ leads to $\mathbb{P}(X>x) = \exp(-\lambda x)$.  Using this form for the survival function, for all $t \geqslant 0$ you have:
$$\begin{equation} \begin{aligned}
\mathbb{P}(X>x+t | X>x) 
&= \frac{\mathbb{P}(X>x+t, X>x)}{\mathbb{P}(X>x)} \\[6pt]
&= \frac{\mathbb{P}(X>x+t)}{\mathbb{P}(X>x)} \\[6pt]
&= \frac{\exp(- \lambda(x+t))}{\exp(- \lambda x)} \\[6pt]
&= \frac{\exp(- \lambda x) \exp(- \lambda t)}{\exp(- \lambda x)} \\[6pt]
&= \exp(- \lambda t) \\[6pt]
&= \mathbb{P}(X>t). \\[6pt]
\end{aligned} \end{equation}$$
In the case where $t<0$ we have:
$$\begin{equation} \begin{aligned}
\mathbb{P}(X>x+t | X>x) 
&= \frac{\mathbb{P}(X>x+t,X>x)}{\mathbb{P}(X>x)} \\[6pt]
&= \frac{\mathbb{P}(X>x)}{\mathbb{P}(X>x)} = 1. \\[6pt]
\end{aligned} \end{equation}$$
