VAR(1)过程方差推导疑问：C、A转置项由来求解

Hey there! Let's work through why those transpose terms ($C'$ and $A'$) show up in your variance calculations—this hinges on a key property of variance for vector-valued random variables, which is easy to forget if your linear algebra is a bit rusty.

Key Linear Algebra Property for Vector Variance #

First, let's recap the critical rule we'll use: For any random vector $\mathbf{Y}$ and constant matrix $\mathbf{M}$, the variance of the linear transformation $\mathbf{M}\mathbf{Y}$ is:
$$\text{Var}(\mathbf{M}\mathbf{Y}) = \mathbf{M} \cdot \text{Var}(\mathbf{Y}) \cdot \mathbf{M}'$$
This is the vector version of the scalar rule $\text{Var}(aY) = a^2\text{Var}(Y)$—the transpose accounts for the fact we're dealing with matrices (not scalars) and ensures the result is a symmetric, valid variance matrix.

Step 1: 1-Period Prediction Error Variance #

Let's start with the 1-period case you already understand:

$x_{t+1} - E_{t}(x_{t+1}) = Cw_{t+1}$

We know $\text{Var}(w_{t+1}) = I$ (the identity matrix). Applying our variance rule:
$$\text{Var}t(x{t+1}) = \text{Var}(Cw_{t+1}) = C \cdot \text{Var}(w_{t+1}) \cdot C' = C \cdot I \cdot C' = CC'$$
The $C'$ here isn't just a random add-on—it's required to make the matrix multiplication work (matching dimensions) and to produce a symmetric variance matrix (which all variance matrices must be).

Step 2: 2-Period Prediction Error Variance #

Now for the 2-period case:

$x_{t+2} - E_{t}(x_{t+2}) = Cw_{t+2} + ACw_{t+1}$

First, remember that $w_{t+2}$ and $w_{t+1}$ are independent (they're distinct white noise terms, so $E(w_{t+2}w_{t+1}') = 0$). For independent random vectors, the variance of their sum is the sum of their variances. Let's break down each term:

1. Variance of $Cw_{t+2}$: This is identical to the 1-period case, so $\text{Var}(Cw_{t+2}) = CC'$.
2. Variance of $ACw_{t+1}$: Here, we treat $AC$ as a single constant matrix. Apply our variance rule again:
   $$\text{Var}(ACw_{t+1}) = (AC) \cdot \text{Var}(w_{t+1}) \cdot (AC)'$$
   Now use the matrix transpose property $(AB)' = B'A'$—so $(AC)' = C'A'$. Substitute that in:
   $$(AC) \cdot I \cdot (C'A') = ACC'A'$$

Adding the two variances together gives the total 2-period prediction error variance:
$$\text{Var}t(x{t+2}) = CC' + ACC'A'$$

Why the Transposes Are Non-Negotiable #

Beyond the variance rule, transposes are necessary for dimension matching:

• Suppose $x_t$ is an $n \times 1$ vector, $w_t$ is a $k \times 1$ white noise vector, $C$ is $n \times k$, and $A$ is $n \times n$.
• $CC'$ is $n \times k$ multiplied by $k \times n$, resulting in an $n \times n$ symmetric matrix (the correct size for the variance of an $n \times 1$ vector).
• Without the transpose, $CC$ would be $n \times k$ multiplied by $n \times k$—which is only valid if $k=n$, and wouldn't produce a symmetric matrix.

That's the core of it! The transposes come directly from how variance works for vector-valued random variables, paired with basic matrix transpose properties.

内容的提问来源于stack exchange，提问作者BenBernke