To prove the given equation, let's analyze the expectation of the quadratic form involving the random vector $\mathbf{y}$ and matrix $\mathbf{A}$. We are given:

• $\mathbf{y}$ is a $k \times 1$ random vector.
• $\mathbb{E}[\mathbf{y}] = \boldsymbol{\mu}$ and $\text{Var}[\mathbf{y}] = \mathbf{V}$.
• $\mathbf{A}$ is a $k \times k$ matrix.

We need to show that: $\mathbb{E}[\mathbf{y}' \mathbf{A} \mathbf{y}] = \text{tr}(\mathbf{A} \mathbf{V}) + \boldsymbol{\mu}' \mathbf{A} \boldsymbol{\mu}.$

Step-by-Step Proof

1. Express $\mathbf{y}$ in terms of its mean and deviation: Let $\mathbf{y} = \boldsymbol{\mu} + \mathbf{z}$, where $\mathbf{z}$ is the deviation vector such that $\mathbb{E}[\mathbf{z}] = \mathbf{0}$ and $\text{Var}[\mathbf{y}] = \mathbf{V} = \mathbb{E}[\mathbf{z} \mathbf{z}']$.

2. Expand the quadratic form: Substituting $\mathbf{y} = \boldsymbol{\mu} + \mathbf{z}$ into $\mathbb{E}[\mathbf{y}' \mathbf{A} \mathbf{y}]$: $\mathbb{E}[(\boldsymbol{\mu} + \mathbf{z})' \mathbf{A} (\boldsymbol{\mu} + \mathbf{z})].$

   Expanding this, we get: $\mathbb{E}[\boldsymbol{\mu}' \mathbf{A} \boldsymbol{\mu} + \boldsymbol{\mu}' \mathbf{A} \mathbf{z} + \mathbf{z}' \mathbf{A} \boldsymbol{\mu} + \mathbf{z}' \mathbf{A} \mathbf{z}].$

3. Simplify each term:

   • $\boldsymbol{\mu}' \mathbf{A} \boldsymbol{\mu}$ is deterministic, so: $\mathbb{E}[\boldsymbol{\mu}' \mathbf{A} \boldsymbol{\mu}] = \boldsymbol{\mu}' \mathbf{A} \boldsymbol{\mu}.$

   • For $\mathbb{E}[\boldsymbol{\mu}' \mathbf{A} \mathbf{z}]$: $\boldsymbol{\mu}' \mathbf{A}$ is a constant vector, and $\mathbb{E}[\mathbf{z}] = \mathbf{0}$, so: [ \mathbb{E}[\boldsymbol{\mu}' \mathbf{A} \mathbf{z}] = \boldsymbol{\mu}' \mathbf{A} \mathbb{E}[\mathbf{z}] = \boldsymbol{\mu}' \mathbf{A} \mathbf{0} = 0. ]

   • Similarly, $\mathbb{E}[\mathbf{z}' \mathbf{A} \boldsymbol{\mu}] = \mathbb{E}[\mathbf{z}' \mathbf{A}] \boldsymbol{\mu} = 0$.

   • For the term $\mathbb{E}[\mathbf{z}' \mathbf{A} \mathbf{z}]$, use the trace property: [ \mathbb{E}[\mathbf{z}' \mathbf{A} \mathbf{z}] = \text{tr}(\mathbf{A} \mathbb{E}[\mathbf{z} \mathbf{z}']) = \text{tr}(\mathbf{A} \mathbf{V}). ]

4. Combine the results: Adding up the non-zero terms, we get: $\mathbb{E}[\mathbf{y}' \mathbf{A} \mathbf{y}] = \boldsymbol{\mu}' \mathbf{A} \boldsymbol{\mu} + \text{tr}(\mathbf{A} \mathbf{V}).$

Thus, we have shown that: $\mathbb{E}[\mathbf{y}' \mathbf{A} \mathbf{y}] = \text{tr}(\mathbf{A} \mathbf{V}) + \boldsymbol{\mu}' \mathbf{A} \boldsymbol{\mu}.$

Do you have any questions or need further details?

Here are 5 related questions for further exploration:

1. How does the trace property help in simplifying the expectation of quadratic forms?
2. What happens if $\mathbf{A}$ is not symmetric in this derivation?
3. How is this result used in multivariate normal distributions?
4. What would the expression look like if $\mathbf{A}$ was a diagonal matrix?
5. How does the variance matrix $\mathbf{V}$ affect the outcome of $\mathbb{E}[\mathbf{y}' \mathbf{A} \mathbf{y}]$?

Tip: When working with expectations of quadratic forms, breaking down the random variable into its mean and deviation can simplify the calculations.