关于计算随机变量$e^T\Sigma^{-1}(e\theta+u)$方差的技术问询

Hey there! Let’s walk through how to calculate the variance of this random variable step by step. I’ll use basic variance properties and matrix algebra to break this down clearly.

First, let’s simplify the expression for the random variable using matrix distribution laws:
$$
e^T\Sigma{-1}(e\theta+u) = \theta \cdot (e^T\Sigma{-1}e) + e^T\Sigma{-1}u
$$
Let’s define two constants to make this easier:

• Let $c = e^T\Sigma{-1}e$: this is a scalar constant (since $e$ is a $k \times 1$ vector and $\Sigma^{-1}$ is a $k \times k$ matrix, their product simplifies to a single number).
• Let $d^T = e^T\Sigma{-1}$: this is a $1 \times k$ constant row vector.

With these definitions, our random variable becomes $c\theta + d^Tu$.

Step 1: Use variance properties for independent variables #

Since $\theta$ and $u$ are independent, the terms $c\theta$ and $d^Tu$ are also independent. For independent random variables, the variance of their sum is the sum of their variances:
$$
\text{Var}(c\theta + d^Tu) = \text{Var}(c\theta) + \text{Var}(d^Tu)
$$

Step 2: Calculate $\text{Var}(c\theta)$ #

For any constant $c$ and random variable $\theta$, $\text{Var}(c\theta) = c^2\text{Var}(\theta)$. We know $\text{Var}(\theta) = \sigma_0^2$, so:
$$
\text{Var}(c\theta) = (e^T\Sigma{-1}e)^2 \sigma_0^2
$$

Step 3: Calculate $\text{Var}(d^Tu)$ #

For a random vector $u$ with covariance matrix $\Sigma$, the variance of a linear combination $d^Tu$ is given by $\text{Var}(d^Tu) = d^T\text{Var}(u)d$. Substituting $d = \Sigma^{-1}e$ and $\text{Var}(u) = \Sigma$:
$$
\text{Var}(d^Tu) = (e^T\Sigma{-1}) \Sigma (\Sigma^{-1}e)
$$
Simplify using matrix inverses ($\Sigma\Sigma^{-1} = I$, the identity matrix):
$$
\text{Var}(d^Tu) = e^T\Sigma{-1}Ie = e^T\Sigma{-1}e
$$

Step 4: Combine the results #

Add the two variance terms together to get the final result:
$$
\text{Var}\left(e^T\Sigma{-1}(e\theta+u)\right) = (e^T\Sigma{-1}e)^2 \sigma_0^2 + e^T\Sigma{-1}e
$$
We can also factor out the common term for a cleaner form:
$$
\text{Var}\left(e^T\Sigma{-1}(e\theta+u)\right) = (e^T\Sigma{-1}e)\left( (e^T\Sigma{-1}e)\sigma_0^2 + 1 \right)
$$

Just to confirm: since $\theta$ and $u$ are independent, the covariance between $c\theta$ and $d^Tu$ is zero, so we don’t need to account for any cross terms in the variance calculation—this simplifies things a lot!

备注：内容来源于stack exchange，提问作者Ypbor