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I am currently reading a book "A first course in Linear model Theory". A linear regression model is defined as $$ y = X\beta+\epsilon$$ where $y = (Y_i, ... , Y_N )$ is an N-dimensional vector of observed responses,$\beta = ( \beta_0,\beta_1,...\beta_n)$'is a(k+1)-dimensional vector of unknown parameters,X is an $N*(k+1)$ matrix of rank r of known predictors, and $\epsilon= (\epsilon_1,\epsilon_2,...,\epsilon_n)$ is N-dimensional random vector of unobserved errors. Suppose,

$E(\epsilon)=0$ and $cov(\epsilon)=\delta ^2 I_N$ (1)

The book then writes :

Using (1) and the properties of the expectation and covariance operators:

$E(y)=E(X\beta+\epsilon)=X\beta+E(\epsilon)=X\beta$ (2)

$Cov(y)=Cov(X\beta+\epsilon)=Cov(\epsilon)=\delta ^2I_N$ (3)

I would appreciate so much a proof on how (1) and properties of expectation and covariance lead to equation (2) and (3) conclusions above

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1 Answer 1

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When $X$ and $\beta$ are assumed to be fixed, which is probably the case mentioned in the book, $X\beta$ term will be a constant inside the expressions. So, $$\operatorname{var}(\epsilon+c)=\operatorname{\epsilon}, \ \ \ E[\epsilon+c]=E[\epsilon]+c$$ Here, $c=X\beta$.

In a fully bayesian model, $X$ and $\beta$ could have been random as well. In that case, these expressions would have been $\operatorname{var}(y|X,\beta)$ and $E[y|X,\beta]$.

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  • $\begingroup$ ${ x_1,...x_n}$ are random variables. can this be considered fixed. what disturbs me is $E(X\beta )=X\beta$ $\endgroup$ Commented Feb 17, 2021 at 12:24
  • $\begingroup$ No, if they're assumed to be RVs in this analysis, the correct expression would be $E[X\beta|X]=X\beta$. Either they're assumed fixed in this particular analysis (not referring to another section of the book where $X$ can be assumed as random), or the expectation and variance notation is a bit slack. $\endgroup$
    – gunes
    Commented Feb 17, 2021 at 12:28

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