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 $P(X>5|X>1) = P(X>5)$?
If
- $P(X>5|X>1) = P(X>4)$
- $P(X>5|X>2) = P(X>3)$
- $P(X>5|X>3) = P(X>2)$
...
past definitely has an effect on present.
Can anyone kindly explain it to me?
Based on $P(X>5\mid X>3)=P(X>2)$ you conclude that the past has effect on the present. That is not true in this context.
Maybe a bell will ring if you rewrite this as:$$P(X-3>2\mid X-3>0)=P(X>2)$$
Yes, we have been waiting for the train now for $3$ minutes (condition $X>3)$ but apparantly under that condition the time that we still have to wait is equipped with the same distribution as it was at the moment we arrived at the station. The $3$ minutes of waiting apparantly did not change that distribution: the past did not have an effect on the present.