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?