Conditional expectation of exponential random variable

Question

Let $X$ be an exponential random variable with parameter $\lambda$. For any fixed $s>0$, I would like to compute $E(X\vert X>s)$. I have read in multiple places that this conditional expectation is equal to $1/\lambda +s$, however, the definition of a conditional expectation implies that $E(E(X\vert X>s))=E(X)=1/\lambda\ne E(1/\lambda +s)=1/\lambda +s$. (For any $\sigma$-field $\mathcal{F}$, it is a fact that $E(E(X\vert\mathcal{F}))=E(X)$, so here we just have $\mathcal{F}=\sigma(\{X>s\})=\{\emptyset, \Omega, \{X>s\}, \{X\le s\}\}.)$

Why is there a discrepancy here? There shouldn't be...I'm probably missing something, but I don't know what.

EDIT: I suppose the discrepancy could be if $E[\cdot \vert X>s]$ was meant to be the expectation of $X$ with respect to the probability measure $P(\cdot \vert X>s)$. However, does one know what the conditional expectation $E[X\vert X>s]$ should be as a RV?

Answer 1

Recall that the exponential distribution is the unique continuous probability distribution with the memoryless property. That is, if $X\sim\mathrm{Exp}(\lambda)$ then for $s,t>0$, $$ \mathbb P(X>t+s\mid X>s) = \mathbb P(X>t). $$ (This follows immediately from $\mathbb P(X>t)=e^{-\lambda t}$ and the definition of conditional probability.) Therefore we may compute \begin{align} \mathbb E[X\mid X>s] &= \int_0^\infty tf_{X\mid X>s}(t)\ \mathsf dt\\ &= \int_0^\infty tf_X(t-s)\ \mathsf dt\\ &= \int_0^\infty (s+t)f_X(t)\ \mathsf dt\\ &= s\int_0^\infty \lambda e^{-\lambda t}\ \mathsf dt + \int_0^\infty t\lambda e^{-\lambda t}\ \mathsf dt\\ &= s + \frac1\lambda. \end{align}