probability: nonlinear best predictor $\hat{Y} = g(X)$

Question

Consider $X\sim\mathcal{U}(-1,1)$ and $Y = X^2$. The nonlinear predictor is defined as $$ \hat{Y} = g(X) = E_{Y|X}[Y|x_i] $$ Now $E_{Y|X}[Y|x_i] = \int_{-\infty}^{\infty}y\frac{f_{X, Y}(x, y)}{f_X(x)}dy$. How do I find $f_{X, Y}(x, y)$? \begin{align} f_Y(y) &= \frac{1}{2\sqrt{y}}(f_X(\sqrt{y}) + f_X(-\sqrt{y}))\\ &= \begin{cases} 1/\sqrt{y}, & 0<y<1\\ 0, & \text{otherwise} \end{cases} \end{align} Should it actually be $\frac{1}{2\sqrt{y}}$ instead?

Answer 1

$\forall x_i\in [-1,1]$   When given $X=x_i$ then $Y$ is certainly equal to $x_i^2$, so we have a probability mass with a support of one point:

$$\begin{align} \mathsf P[Y=y\mid X=x_i] & = \begin{cases} 1 & : y=x_i^2 \\ 0 & : y\neq x_i^2\end{cases} \\[4ex] \mathsf E_{Y\mid X}(Y\mid x_i) & = \sum_{y=x_i^2} y\, \mathsf P(Y=y\mid X=x_i) \\ & = x_i^2 \end{align}$$

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Remark   It is usually considered that clearly: $\mathsf E[h(X)\mid X] = h(X)$ , so you needn't bother showing this work.

$$\mathsf E\left(X^2\,\middle|\, X=x_i\right) = x_i^2$$