# Derivation for marginal effect sizes' distribution in linear model

Model: $$Y_{N\times 1} = X_{N\times M} \beta_{M\times 1} + \epsilon_{N\times 1}$$ the design matrix $$X_{N\times M}$$ is known, each column of the design matrix has been standardized to have mean 0 and variance 1. $$Y$$ is also known and has been standardized to mean 0 and variance 1. Also assume $$\beta \sim \mathcal{N}(0, \sigma^2 I_{M\times M})$$ and $$\epsilon \sim \mathcal{N}(0, \sigma_\epsilon^2 I_{N\times N})$$. In addition $$M \sigma^2 + \sigma_\epsilon^2 = 1$$

The marginal effect sizes are $$\hat{\beta} = \frac{1}{N}X^\top Y$$. The distribution the marginal effect sizes $$\hat{\beta}$$ can be determined as follows: \begin{align*} \hat{\beta} & = \frac{X^\top Y}{N} \\ & = \frac{X^\top ( X \beta + \epsilon) }{N} \\ & = \frac{XX^\top}{N} \beta + \frac{1}{N}X^\top \epsilon \end{align*} $$E[\hat{\beta} | X, \beta] = \frac{XX^\top}{N} \beta$$ $$\text{Var}[\hat{\beta} | X,\beta] = \frac{1}{N^2}X^\top E[\epsilon \epsilon^\top]X=\frac{1}{N^2}\sigma_\epsilon^2X^\top X$$ Thus $$\hat{\beta} \sim \mathcal{N}(\frac{XX^\top}{N} \beta, \frac{1}{N^2}\sigma_\epsilon^2X^\top X)$$ My question is that if we consider a small chunk of variable of size b, where the corresponding design matrix is of shape $$X_{N \times b}$$, the corresponding marginal effect sizes is $$\hat{\beta}_b$$: $$\hat{\beta}_b = \frac{X^\top_b Y}{N}$$ How is the following two derived: $$E[\hat{\beta}_b | \beta_b, X_b] = \frac{X_b^\top X_b}{N} \beta_b$$ $$\text{Var}[\hat{\beta}_b | \beta_b, X_b] = \frac{1}{N}( 1 - b \sigma^2) \frac{X_b^\top X_b}{N}$$

One thing I am confused about is the $$X_b$$ being conditioned in $$E[\hat{\beta}_b | \beta_b, X_b]$$. Does this mean that we don't assume anything on the rest of the design matrix? For more context, this is from the methods section in this paper.