Let $X$ be an exponential random variable with rate $\lambda$
Use the identity below to solve for $E[X|X < c]$
$$E[X] = E[X|X < c]*P(X < c) + E[X|X > c]*P(X > c)$$
So right off the bat, I know I can rearrange this to find what I'm looking for as follows
$$E[X|X<c] = \frac{E[X] – E[X|X > c]*P(X > c)}{P(X<c)}$$
Of course $$E[X] = \frac{1}{\lambda}$$
$$P(X<c) = F_X(c) = 1-e^{-\lambda t}$$
$$P(X>c) = 1-F_X(c) = 1-(1-e^{-\lambda t}) = e^{-\lambda t}$$
I can plug these in and solve easilly. However, I'm not entirely sure how to solve for $E[X|X>c]$. I could use the definition of conditional expectation I suppose, but I feel like that's defeating the purpose of the problem; is there another way to compute this value so I can complete the equation?
Best Answer
I would exploit the memoryless property of the exponential distribution: that is, if $X \sim {\rm Exponential}(\lambda)$, then $\Pr[X > x + c \mid X > c] = \Pr[X > x]$. From this, it easily follows that $${\rm E}[X \mid X > c] = {\rm E}[X] + c = \frac{1}{\lambda} + c.$$