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

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).

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.