Why is $y\sim\mathcal{N}(Xβ,\sigma^{2} I_n)$?

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 $$y\sim\mathcal{N}(Xβ,\sigma^{2} I_n)$$?

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 $$y \sim N(X \beta, \sigma^2 I)$$

• Hi, why is the variance of y equal to (y−Xβ)? should't it be Xβ+ϵ? Oct 1 '18 at 5:08
• so by definition: $$\operatorname{var} (y) = E(y - E(y))(y-E(y))^T$$ Oct 1 '18 at 5:10
• 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 Oct 1 '18 at 22:35
• 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 $y$. Oct 1 '18 at 22:42
• @Admad Bazzi , what about assuming a vector as a scalar, is that okay? Thank you so much !!! Oct 1 '18 at 22:44

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 $$\varepsilon\sim \mathcal{N}(0, \sigma^{2} I_n)$$