I am having a confusion regarding conditional expectation. Let $X\sim \operatorname{Uniform}(0,2)$ and $Y$ be a random variable such that $Y=\max\{X,2-X\}$.
So basically $Y=X$ if $X>2-X$ i.e. $X>1$ and $Y=2-X$ if $X<1$. I have to find $E(Y)$, which can be written as $E(Y\mid X>1)×P(X>1)+ E(Y\mid X<1)×P(X<1)$. I write
$E(Y\mid X>1)=E(X)=1$
Is this correct? If not, can you suggest me an alternative way to solve the problem. Thank you!
By law of total expectation, you are correct that
$$E[Y]=E[Y\mid X>1]P(X>1)+E[Y\mid X<1]P(X<1)\tag{1}$$
Now for the conditional expectations, you have $E[Y\mid X>1]=E[X\mid X>1]$ (you cannot drop the condition $X>1$ because $X$ is not independent of $\{X>1\}$) and similarly for $E[Y\mid X<1]$.
Since $E[Y\mid A]=\frac{E[Y1_A]}{P(A)}$ for any event $A$ with positive probability, $(1)$ is just
$$E[Y]=E[Y1_{X>1}]+E[Y1_{X<1}]$$
Here $1$ is an indicator variable, so you have
\begin{align} E[Y1_{X>1}]=E[X1_{X>1}]&=\int x1_{x>1}f_X(x)\,dx \\&=\frac12\int_1^2 x\,dx \end{align}
But you can forget all of this and simply find the expectation using this theorem:
$$E[Y]=\int \max(x,2-x)f_X(x)\,dx$$