We have that $X,Y$ are independent exponential random variables with rates $\mu,\lambda$ respectively. We're to find $E(\mathrm{max}\{X,Y\})$. I'm getting the wrong answer and I don't know why.
I have seen alternate proofs arriving at the correct answer---my question is not how to solve this problem, my question is what is wrong with the below reasoning.
\begin{align} E(\mathrm{max}\{X,Y\})&=E(X|Y<X)P(Y<X) + E(Y|X<Y)P(X<Y)\\ &= E(X)P(Y<X) + E(Y)P(X<Y)\\ &= \frac1\mu P(Y<X) + \frac1\lambda P(X<Y)\\ &= \frac 1\mu \cdot \frac \lambda{\lambda+\mu} + \frac1\lambda\cdot \frac \mu{\lambda+\mu}\\ &=\frac{\mu^2+\lambda^2}{\mu\lambda(\mu+\lambda)} \end{align}
A mistake is that you do not have $$ \mathbb{E}[X \mid Y < X] = \mathbb{E}[X]. $$ First, note that a priori, the left-hand side is a conditional expectation, i.e. a random variable, while the right-hand side is a real value (a "number"). So the "type" of the objects is not straightforwardly even compatible.
You would have equality if $X$ were independent of $\mathbb{1}_{\{X>Y\}}$, but this is not obvious this is the case (and actually it's not). $X$ is independent of $Y$, sure; but not of $\mathbb{1}_{\{X>Y\}}$.