# Find distribution of $U=(Y-X_1\hat{\beta_1})^T(Y-X_1\hat{\beta_1})$.

I need to find distribution of $U=(Y-X_1\hat{\beta_1})^T(Y-X_1\hat{\beta_1})$.

Suppose a linear model $Y= X\beta +\varepsilon$ where $\varepsilon \sim N(0,\sigma ^2I)$. Write $X=[X_1;X_2]$ where $X_1$ are the first $p_1$ columns of $X$ nad $X_2$ are the last $p_2$ columns. Similary split $\beta ^T =(\beta_1^T; \beta_2^T)$.

I computed:

$\operatorname E(Y - X_1\hat{\beta_1}) = X\beta - X_1\beta _1$

$\operatorname{var}(Y - X_1\hat{\beta_1}) = \sigma ^2(I-H_1)$

where $(I-H_1)$ is idemotent matrix. Then I computed $(I-H_1)(X\beta - X_1\beta _1) =0$

Here I stopped and I do not know what I need to do next.

Thank you for any help.

• It depends on how the error terms are distributed, but I suspect you'll find the model's assumptions are such that $Y-X_1\hat{\beta}_1$ has a mean-zero multivariate Gaussian distribution, in which case $U$ has this distribution: en.wikipedia.org/wiki/Wishart_distribution – J.G. Mar 8 '18 at 10:17

Let $\beta_1 \in \mathbb{R}^{p_1}$ and $Y\in \mathbb{R}^n$, and recall that if $X_1,...,X_2 \sim \mathcal{N}(\mu, \sigma^2)$ then $\sum_{i=1}^n \sigma^{-2}(X_i - \mu)^2 \sim \chi^2_{(n)}$. So, $$U = (Y - X_1 \hat{\beta}_1)'( Y - X_1 \hat{\beta}_1) =e_1'e_1 = ||e||_2^2 \sim \sigma^2\chi^2_{(n-p_1)}.$$