Breeder's equation and equivalent expressions for narrow-sense heritability

Question

I am trying to model the phenotype of a trait as $X = G + E$, where $G$ and $E$ are the genetic and environmental effects. (I'll ignore the distinction between broad-sense and narrow-sense heritability in this question.) Wikipedia tells me that $$h^2 = \frac{\mathrm{Var}(G)}{\mathrm{Var}(X)}$$ The same page also gives the breeder's equation $R = h^2S$ but also says "Note that heritability in the [breeder's equation] is equal to the ratio [${\mathrm {Var}}(G)/{\mathrm {Var}}(X)$] only if the genotype and the environmental noise follow Gaussian distributions." So let $G \sim N(0, \sigma_G^2)$ and $E \sim N(0, \sigma_E^2)$ be distributed according to normal distributions. Then $X \sim N(0, \sigma_G^2 + \sigma_E^2)$ so $$h^2 = \frac{\mathrm{Var}(G)}{\mathrm{Var}(X)} = \frac{\sigma_G^2}{\sigma_G^2 + \sigma_E^2}$$

Now I want to perform a selection, choosing a parent with $X=a$. We can expect the child to have the trait $\mathbb E[G \mid X=a] + E'$ where $E' \sim N(0, \sigma_E^2)$ is some other environmental effect. (I don't want to bother with recombination and multiple parents, so we can just think of it as cloning the parent: given that $X=a$, we expect the parent's genetic part to have been $\mathbb E[G \mid X=a]$.)

To find $\mathbb E[G \mid X=a]$, note that $\mathbb E[X \mid X=a] = a$ and by linearity of expectation, $\mathbb E[X \mid X=a] = \mathbb E[G \mid X=a] + \mathbb E[E \mid X=a]$. To get the terms to cancel, we want to write $E$ as $\alpha E$ where $\alpha E$ has the same distribution as $G$. Since $\alpha E \sim N(0, \alpha^2 \sigma_E^2)$, we want $\alpha^2 \sigma_E^2 = \sigma_G^2$, so $\alpha = \sigma_G/\sigma_E$. Thus we have $$\begin{align}a &= \mathbb E[G \mid X=a] + \frac1\alpha\mathbb E[\alpha E \mid X=a] \\ &= \mathbb E[G \mid X=a] + \frac1\alpha\mathbb E[G \mid X=a] \\ &= \left(1+\frac1\alpha\right)\mathbb E[G \mid X=a]\end{align}$$ Hence $$\mathbb E[G \mid X=a] = \frac{a}{1+\frac1\alpha} = \frac{a}{1+\frac{\sigma_E}{\sigma_G}}$$

Since the parent was selected at $X=a$ and the child has mean $\frac{a}{1+\frac{\sigma_E}{\sigma_G}}$, we have $$h^2=\frac{R}{S} = \frac{1}{1+\frac{\sigma_E}{\sigma_G}} = \frac{\sigma_G}{\sigma_G+\sigma_E}$$

Now my problem is that the two expressions for $h^2$ don't match. They look pretty similar, but one of them has all the terms squared while the other one doesn't. What's going on/where did I go wrong?

ETA: To give a concrete example, suppose 80% of variance in the trait is explained by genes. Then the breeder's equation predicts the child to have $0.8a$. But computing $\mathbb E[G \mid X=a]$ gives $a/(1 + \sqrt{0.8^{-1} - 1}) = (2/3)a$. So the two approaches predict different things (although if it was 50% they would give the same prediction).

Answer 1

I found my mistake, which is where I said "To get the terms to cancel, we want to write $E$ as $\alpha E$ where $\alpha E$ has the same distribution as $G$." What we actually want is to find a constant $\alpha$ such that $\mathbb E[\alpha E\mid X=a] = \mathbb E[G\mid X=a]$. In my question, I assumed that this was the same thing as finding $\alpha$ such that $\mathbb E[\alpha E] = \mathbb E[G]$; hence I was trying to find $\alpha$ to make the distributions of $\alpha E$ and $G$ the same (the means are both 0, so I just needed to get the variances to equal each other). But actually, conditioning on $X=a$ shifts the distributions so the $\alpha$ I found in my question doesn't work.

Since my previous strategy doesn't work, I ended up skipping the step of finding $\alpha$ entirely. I realized that the random vector $(G, X)=(G,G+E)$ is a bivariate normal, because we can write $$\begin{pmatrix}G \\ G+E\end{pmatrix} = \begin{pmatrix}\sigma_G & 0\\ \sigma_G & \sigma_E\end{pmatrix}\begin{pmatrix}Z_1 \\ Z_2\end{pmatrix}$$ where $Z_1,Z_2$ are standard normal random variables.

Now Wikipedia gives the conditional distribution for bivariate normals, which in our case is $G\mid X=a$. We only need the mean, which is $\mu_G + \frac{\sigma_G}{\sigma_X}\rho (a - \mu_X)$. Both $\mu_G$ and $\mu_X$ are zero, and $\rho$ is the correlation which is $$\rho = \rho_{G,X} = \frac{\mathrm{cov}(G,X)}{\sigma_G \sigma_X} = \frac{\mathrm{cov}(G,G) + \mathrm{cov}(G,E)}{\sigma_G \sigma_X} = \frac{\sigma_G^2}{\sigma_G \sigma_X} = \frac{\sigma_G}{\sigma_X}$$ Thus we end up with $$\mu_G + \frac{\sigma_G}{\sigma_X}\rho (a - \mu_X) = \frac{\sigma_G^2}{\sigma_X^2}a$$ which is exactly what we want.