Given that $X$ is a exponential random variable. I want to calculate $E[X^2|X>1]$. I think it would be $E[(X+1)^2]$, but do not have any convincing explanation.
Asked
Active
Viewed 71 times
1 Answers
3
Memorylessness. The conditional distribution of $X$, given that $X\gt 1$, is the same as the distribution of the random variable $1+Y$, where $Y$ has exponential distribution with the same parameter as $X$. Informally, the additional waiting time, given that one has waited for time $1$, has the same distribution as $X$.
André Nicolas
- 507,029
-
I get it now. Thanks @ Andre. – Sanwar Apr 05 '16 at 01:26
-
You are welcome. – André Nicolas Apr 05 '16 at 01:28