Why is $E(u^2)=Var(y)$? (Binary Response Model)

Question

I'm trying to show some results in binary response models, and a couple of the proofs use the "fact" that $E(u^2)=Var(y)$, but I can't see why this is.

The setup is that $y$ takes on the value $0$ or $1$, and we could write \begin{equation} P(y=1|\mathbf{x})=g(\mathbf{x\beta}) \end{equation} where $g:\mathbb{R} \to (0,1)$ ($g$ could represent the Normal CDF in the case of a Probit for example). Or, you could take the Poisson case where \begin{equation} P(y=1|\mathbf{x})=\exp\left[-E\left(y\middle\vert\mathbf{x}\right)\right]\frac{E\left(y\middle\vert\mathbf{x}\right)^y}{y!} \end{equation}

In showing some results use, specifically, $E(u^2|\mathbf{x})=Var(y|\mathbf{x})$ (where note that we define $u=y-g(\mathbf{x\beta}$). I'm trying to figure out why this would be true, but can't see why. Anyone know why this is?

Thanks!

Answer 1

Under the model, $E(y|x)=g(\mathbf{x} \beta)$. Therefore $E(u|x)=0$, making $Var(u|x) = E(u^2|x) - E(u|x)^2 = E(u^2|x)$. The last step is to note that $Var(y|x) = Var(u|x)$ since $y=u+g(\mathbf{x}\beta)$ and the last term is constant wrt $\mathbf{x}$.