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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.

Let y be a k x 1 random vector with E[y] = μ and Var[y] = V. Let A be a k x k matrix of real numbers. Show that E[y' A y] = tr(AV) + μ' A μ.

Explore a step-by-step solution to proving the expectation of a quadratic form involving a random vector. Learn about key concepts like linear algebra, the trace property, and multivariate statistics.

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