Why is $y\sim\mathcal{N}(Xβ,\sigma^{2} I_n)$? (i.e. $\varepsilon\sim \mathcal{N}(0, \sigma^{2} I_n)$)
How can I obtain $y\sim\mathcal{N}(Xβ,\sigma^{2} I_n)$?
$\begingroup$
$\endgroup$
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
8
Adding a constant $X\beta$ to a normally distributed random variable is again normal.
The mean of the $y = X\beta + \epsilon$ is: $$E(y) = E(X\beta + \epsilon) = E(X\beta) + E(\epsilon) = X\beta + 0 = X\beta$$ and the variance is $$\operatorname{var} (y) = E(y -X\beta)(y -X\beta)^T = E(\epsilon \epsilon^T) = \operatorname{var} (\epsilon) = \sigma^2 I$$ We conclude that $$y \sim N(X \beta, \sigma^2 I)$$
-
$\begingroup$ Hi, why is the variance of y equal to (y−Xβ)? should't it be Xβ+ϵ? $\endgroup$ Commented Oct 1, 2018 at 5:08
-
$\begingroup$ so by definition: $$\operatorname{var} (y) = E(y - E(y))(y-E(y))^T$$ $\endgroup$ Commented Oct 1, 2018 at 5:10
-
$\begingroup$ Hi again, I am a follow up question. How do you know var(y) is a vector rather than a variable? Since the definition you provided is for random vector. Thanks ! @Ahmad Bazzi $\endgroup$ Commented Oct 1, 2018 at 22:35
-
$\begingroup$ Hi @MathAvengers, assuming it to be a vector if it were a scalar is okay, because: $$\operatorname{var} (y) = E(y - E(y))(y-E(y))^T=E( y-E(y))^2$$ for scalar $y$. $\endgroup$ Commented Oct 1, 2018 at 22:42
-
$\begingroup$ @Admad Bazzi , what about assuming a vector as a scalar, is that okay? Thank you so much !!! $\endgroup$ Commented Oct 1, 2018 at 22:44
$\begingroup$
$\endgroup$
The variance will not change if you add a constant, i.e $Var(X+c)=Var(X)$, here $X\beta$ is a constant
The formula is :
$Y=X\beta+\varepsilon$ where $\varepsilon\sim \mathcal{N}(0, \sigma^{2} I_n)$
answered Oct 1, 2018 at 4:16