How can I show that E[Y|X]= X^2

Question

I have X distributed as a standard normal and $$Y = X^2$$. How can I show that $$E[Y|X]= X^2$$

Answer 1

$Y=X^2 = X \cdot X$ is measurable as product of $X$ in the $\sigma$-Algebra generated by $X$, therefore $$E[Y\mid X] = Y = X^2$$

Adding a proof for what @MPW has mentioned in the comments, let $X$ and $Z$ be random variables (with discrete values), then for any (Borel-measurable) transformation $g(X)$ it follows that:

$$E[Z\cdot g(X) | X ] = g(X) \cdot E[Z|X]$$

We have to compute the conditional expectation given $X=x$, whereas the value of $Z$ varies over the entire range (here denoted with $z$), therefore: $$\begin{align*}E[Z\cdot g(X) | X=x ] &= \sum_{z} zg(x) P(Z=z|X=x)\\ &= g(x) \sum_{z} zP(Z=z|X=x) \\ &= g(x) E[Z|X=x]\end{align*}$$

It follows that $E[Z\cdot g(X)|X] = g(X)E[Z|X]$ with a similar argument for the continous case (see: For $X,Y $ random variables, $h $ a function, show that $E (Xh(Y)|Y)=h (Y)E (X|Y) $ almost surely) and for your exercise $Z = 1$ and the transformation $g$ is squaring.