Conditional expectation of the square of a random variable Let the joint PDF of X and Y, $f(x, y) = \frac{1}{2}e^{-x}$ if $x \geq 0$, $|y| < x$, $f(x, y) = 0$ everywhere else. Calculate $\mathbb{V}(Y|X = x)$.
By definition, $\mathbb{V}(Y|X = x) = \mathbb{E}\Big\{\big[Y - \mathbb{E}(Y|X = x)\big]^2|X = x\Big\}$. Earlier, I calculated that $\mathbb{E}(Y|X = x) = 0$, so $\mathbb{V}(Y|X = x) = \mathbb{E}(Y^2|X = x)$.
Could I please have a hint for calculating the conditional expectation of the square of a random variable?

You are not calculating the (conditional) variance of the square of a random variable, but the (conditional) expectation of the square of a random variable.  The left hand side of $$\mathbb{V}(Y|X = x) = \mathbb{E}(Y^2|X = x)$$ is the conditional variance of $Y$, not the conditional variance of $Y^2$, and the right side is the conditional expectation of $Y^2$, and that's what you are looking for.
Hint: Given the value of $X$ is some $x \geq 0$, first show that the conditional distribution of $Y$ given $X=x$ is $\mathcal U(-x,x)$. Then the calculation of the conditional variance of $Y$ given that $X=x$ is straightforward. Whether you choose to formally write down the proof of what this expectation is, or use a canned formula such as $\dfrac{(b-a)^2}{12} = \dfrac{x^2}{3}$ is up to you.
