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Suppose $X$ is exponentially distributed with rate $\lambda$ what is the $E(X^2|X>a)$? I am not sure whether it is right. I think the answer is ${1\over \lambda^2}+{1\over\lambda}+a$ as the second moment is ${1\over \lambda^2}+{1\over\lambda}$. Is it correct

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No, it is not. For one thing, your answer doesn't make sense dimensionally. If we think of $X$ as a time measured in, say, seconds, the rate $\lambda$ is in $1/$seconds, $a$ is in seconds, and $E[X^2 | X > a]$ should be in seconds$^2$. But $1/\lambda + a$ is in seconds, not seconds$^2$.

By the "lack of memory" property of the exponential distribution, the conditional distribution of $X$ given $X > a$ is the same as the distribution of $a+X$. Thus you should have $E[X^2 | X > a] = E[(a+X)^2]$