# Is there a mistake in the expression of this variance?

I'm busy reading through an econometrics textbook (page 147), and I don't understand the step

$$\mathrm {Var}\left(n^{\frac 12}\left(\hat\beta - \beta\right)\right) = \boldsymbol{A^{-1}}\sigma^2\boldsymbol{AA^{-1}} = \sigma^2\boldsymbol{A^{-1}}$$

The relevant part of the text is:

and so on.



Did we get the two $$\boldsymbol{A^{-1}}$$'s from the rule that $$\mathrm{Var}(a\boldsymbol X) = a^2\mathrm{Var}\boldsymbol X$$ ? And should the two $$\boldsymbol{A^{-1}}$$'s be on the right-hand side of the expression? Why should they be on the right, when $$\left(\frac 1n\boldsymbol{X'X}\right)^{-1}$$ is on the left?

Secondly, I don't understand what the "$$d$$ " means in $$\overset d\to$$.

$$\beta-\hat{\beta}=\beta-(X'X)^{-1}X'Y=\beta-(X'X)^{-1}X'(X\beta+\epsilon)$$ $$\beta-\hat{\beta}=\beta-(X'X)^{-1}(X'X)\beta+(X'X)^{-1}X'\epsilon$$ $$\beta-\hat{\beta}=(X'X)^{-1}X'\epsilon$$ $$\beta-\hat{\beta}=(n^{-1}X'X)^{-1}(n^{-1}X'\epsilon)$$

Now the $$\overset d\to$$ means "convergence in distribution", by the central limit theorem we have that if the error term is homocedastic $$\sqrt{n}(n^{-1}X'\epsilon)\overset d\to N(0,\sigma^2 A)$$ We know that $$(n^{-1}X'X)\overset p\to A^{-1}$$ and hence by Slutzky's theorem we have that $$\sqrt{n}(\beta-\hat{\beta})\overset d\to A^{-1}N(0,\sigma^2 A)=N(0,(A^{-1})\sigma^2 A (A^{-1})')$$

Since $$(A^{-1})'=A^{-1}$$ we end up with $$\sqrt{n}(\beta-\hat{\beta})\overset d\to N(0,\sigma^2 A^{-1})$$

Notice that in order to obtain this you need ergodic stationarity of the variables $$X$$ and $$Y$$ (to ensure the convergence to the population value) and homoskedasticity of the error term in order to express the variance/covariance matrix as the product of an scalar and a matrix (if the matrix $$A$$ has constant elements in its diagonal of course)

• I'm not sure why left-multiplying a distribution by a matrix results in the variance being $M\Sigma M'$, where $M$ is the matrix and $\Sigma$ is the covariance matrix. – ahorn Mar 11 at 9:38
• You mentioned the fact that $Var(aX)=a^2Var(X)$. Think that "multiplying" the ditribution is like saying that you are multiplying the random variable itself. So for instance if you have that $X\sim N(0,\sigma^2)$ then $aX\sim N(0,a^2\sigma^2)$ right? Well with matrices is the same thing, the difference is that you cannot just write $M^2$ (see quadratic forms). Instead you write $M\Sigma M'$. – RScrlli Mar 11 at 9:52
• My question was more to do with this multiplication of the variance than with explaining the theorem. – ahorn Mar 11 at 10:05
• The variance result you quoted is for univariate random variables. The one for vector-random variables is $\text{Var}(AX) = A\text{Var}(X) A'$ where here $\text{Var}$ is the variance-covariance matrix of its argument (e.g. see Wikipedia here). Even if you're not familiar with this, it's easy to confirm the correctness of the result element-by-element. Note that $A$ needn't be square, as long as its column-dimension matches the length of $X$ – Glen_b Mar 11 at 10:51

$$\sigma^2$$ is commutative: $$\boldsymbol {A^{-1}}\sigma^2\boldsymbol{AA^{-1}} = \sigma^2\boldsymbol{A^{-1}A A^{-1}} = \sigma^2\boldsymbol{A^{-1}}.$$

Even if the $$\boldsymbol{A^{-1}}$$'s were at the left-hand side of the expression, the commutativity of $$\sigma^2$$ will ensure that the result $$\left(\sigma^2\boldsymbol{A^{-1}}\right)$$ is true.

$$\overset d\to$$ means "the distribution converges to".