Of course. Here is the step-by-step solution to the Markowitz portfolio problem.

Let the portfolio weights be denoted by the vector $w = [w_1, w_2, w_3]^T$. The constraints are $w_1+w_2+w_3=1$, which

Question

Of course. Here is the step-by-step solution to the Markowitz portfolio problem.

Let the portfolio weights be denoted by the vector $w = [w_1, w_2, w_3]^T$. The constraints are $w_1+w_2+w_3=1$, which can be written as $w^T 1 = 1$, where $1 = [1, 1, 1]^T$.

The given covariance matrix and expected returns vector are:
$$\Sigma = \begin{bmatrix} 1 & 1 & 0 \ 1 & 2 & 1 \ 0 & 1 & 2 \end{bmatrix}, \quad \bar{r} = \begin{bmatrix} 0.4 \ 0.8 \ 0.8 \end{bmatrix}$$

### (a) Find the minimum-variance portfolio. What is $\mu$ in this case?

To find the minimum-variance portfolio, we need to solve the following optimization problem:
Minimize: $\sigma_p^2 = w^T \Sigma w$
Subject to: $w^T 1 = 1$

The Lagrangian for this problem is $L = \frac{1}{2} w^T \Sigma w - \lambda(w^T 1 - 1)$. The first-order condition is $
abla_w L = 0$, which gives:
$$\Sigma w - \lambda 1 = 0 \implies \Sigma w = \lambda 1$$
This yields the following system of linear equations:
1.  $w_1 + w_2 = \lambda$
2.  $w_1 + 2w_2 + w_3 = \lambda$
3.  $w_2 + 2w_3 = \lambda$
And the constraint:
4.  $w_1 + w_2 + w_3 = 1$

From (1) and (2), we get $(w_1 + 2w_2 + w_3) - (w_1 + w_2) = \lambda - \lambda$, which simplifies to $w_2 + w_3

VIPSolutions · Accepted Answer

Certainly! Here is the step-by-step solution, rewritten as requested, with all the details and equations as in your original:

---

Let the portfolio weights be denoted by the vector
$$
w = \begin{bmatrix} w_1 \ w_2 \ w_3 \end{bmatrix}.
$$
The constraint is $ w_1 + w_2 + w_3 = 1 $, or $ w^T 1 = 1 $, where $ 1 = \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} $.

The given covariance matrix and expected returns vector are:
$$
\Sigma = \begin{bmatrix} 1 & 1 & 0 \ 1 & 2 & 1 \ 0 & 1 & 2 \end{bmatrix}, \qquad \bar{r} = \begin{bmatrix} 0.4 \ 0.8 \ 0.8 \end{bmatrix}
$$

---

## (a) **Find the minimum-variance portfolio. What is $\mu$ in this case?**

To find the minimum-variance portfolio, solve:
$$
\min \ w^T \Sigma w \quad \text{subject to} \quad w^T 1 = 1
$$

The Lagrangian is
$$
L = \frac{1}{2} w^T \Sigma w - \lambda(w^T 1 - 1)
$$
The first-order condition ($\nabla_w L = 0$):
$$
\Sigma w - \lambda 1 = 0 \implies \Sigma w = \lambda 1
$$
This yields:
$$
\begin{cases}
w_1 + w_2 = \lambda \
w_1 + 2w_2 + w_3 = \lambda \
w_2 + 2w_3 = \lambda \
w_1 + w_2 + w_3 = 1
\end{cases}
$$
From the first two equations: $ (w_1 + 2w_2 + w_3) - (w_1 + w_2) = w_2 + w_3 = 0 \implies w_3 = -w_2 $.

Substitute $ w_3 = -w_2 $ into the third equation:
$$
w_2 + 2(-w_2) = \lambda \implies -w_2 = \lambda \implies w_2 = -\lambda
$$
So $ w_3 = \lambda $.

From the first equation:
$$
w_1 + w_2 = \lambda \implies w_1 = \lambda - w_2 = \lambda - (-\lambda) = 2\lambda
$$

Now use the sum constraint:
$$
w_1 + w_2 + w_3 = 2\lambda - \lambda + \lambda = 2\lambda = 1 \implies \lambda = 0.5
$$

Thus,
$$
w_1 = 2\lambda = 1,\quad w_2 = -\lambda = -0.5, \quad w_3 = \lambda = 0.5
$$
Therefore,
$$
w_{mvp} = \begin{bmatrix} 1 \ -0.5 \ 0.5 \end{bmatrix}
$$

The expected return is:
$$
\mu = w_{mvp}^T \bar{r} = [1, -0.5, 0.5] \cdot [0.4, 0.8, 0.8]^T = 1 \times 0.4 - 0.5 \times 0.8 + 0.5 \times 0.8 = 0.4 - 0.4 + 0.4 = 0.4
$$

**Answer for (a):**  
The minimum-variance portfolio is $ w_{mvp} = [1, -0.5, 0.5]^T $.  
The expected return is $ \mu = 0.4 $.

---

## (b) **Find the optimal portfolio with $\bar{r}_p = 0.7$ directly.**

We want to minimize variance subject to $ w^T 1 = 1 $ and $ w^T \bar{r} = 0.7 $.

Lagrangian:
$$
L = \frac{1}{2} w^T \Sigma w - \lambda_1 (w^T \bar{r} - 0.7) - \lambda_2 (w^T 1 - 1)
$$
First-order condition:
$$
\Sigma w = \lambda_1 \bar{r} + \lambda_2 1
$$
Write out:
$$
\begin{cases}
w_1 + w_2 = 0.4 \lambda_1 + \lambda_2 \
w_1 + 2 w_2 + w_3 = 0.8 \lambda_1 + \lambda_2 \
w_2 + 2 w_3 = 0.8 \lambda_1 + \lambda_2
\end{cases}
$$
From equations 2 and 3:
$$
w_1 + 2w_2 + w_3 = w_2 + 2w_3 \implies w_1 + w_2 = w_3
$$
Let’s use this and the constraints:
$$
\begin{align*}
w_3 &= w_1 + w_2 \
w_1 + w_2 + w_3 &= 1 \
0.4 w_1 + 0.8 w_2 + 0.8 w_3 &= 0.7
\end{align*}
$$
From the first two: $ (w_1 + w_2) + (w_1 + w_2) = 1 \implies 2(w_1 + w_2) = 1 \implies w_1 + w_2 = 0.5 \implies w_3 = 0.5 $.

Let $ w_2 = x $, then $ w_1 = 0.5 - x $, $ w_3 = 0.5 $.
Plug into expected return constraint:
$$
0.4(0.5 - x) + 0.8x + 0.8 \times 0.5 = 0.7 \
0.2 - 0.4x + 0.8x + 0.4 = 0.7 \
0.6 + 0.4x = 0.7 \implies 0.4x = 0.1 \implies x = 0.25
$$
So,
$$
w_2 = 0.25, \quad w_1 = 0.25, \quad w_3 = 0.5
$$

**Answer for (b):**  
The optimal portfolio is $ w = [0.25, 0.25, 0.5]^T $.

---

## (c) **Find another efficient portfolio by setting $\lambda = 1, \mu = 0$, and thus the optimal portfolio with expected return 1 by the Two Fund Theorem.**

The FOC is $ \Sigma w = \lambda \bar{r} + \mu 1 $. Setting $\lambda = 1$, $\mu = 0$:
$$
\Sigma w = \bar{r} \implies w = \Sigma^{-1} \bar{r}
$$
Given
$$
\Sigma^{-1} = \begin{bmatrix} 3 & -2 & 1 \ -2 & 2 & -1 \ 1 & -1 & 1 \end{bmatrix}
$$
So,
$$
w_A = \Sigma^{-1} \bar{r} = \begin{bmatrix} 3 & -2 & 1 \ -2 & 2 & -1 \ 1 & -1 & 1 \end{bmatrix} \begin{bmatrix} 0.4 \ 0.8 \ 0.8 \end{bmatrix} = \begin{bmatrix} 1.2 - 1.6 + 0.8 \ -0.8 + 1.6 - 0.8 \ 0.4 - 0.8 + 0.8 \end{bmatrix} = \begin{bmatrix} 0.4 \ 0 \ 0.4 \end{bmatrix}
$$
Note $ w_A^T 1 = 0.8 $, not 1.

Now, let $ w_B = w_{mvp} = [1, -0.5, 0.5]^T $.

Any efficient portfolio is a linear combination: $ w_p = a w_A + b w_B $.

We want $ w_p^T 1 = 1 $ and $ w_p^T \bar{r} = 1 $.
Compute:
$$
\begin{align*}
w_A^T 1 &= 0.4 + 0 + 0.4 = 0.8 \
w_B^T 1 &= 1 - 0.5 + 0.5 = 1 \
w_A^T \bar{r} &= 0.4 \times 0.4 + 0 \times 0.8 + 0.4 \times 0.8 = 0.16 + 0 + 0.32 = 0.48 \
w_B^T \bar{r} &= 0.4 \text{ (from part a)}
\end{align*}
$$
So,
$$
\begin{cases}
0.8a + 1b = 1 \
0.48a + 0.4b = 1
\end{cases}
$$
From the first: $ b = 1 - 0.8a $.

Substitute into the second:
$$
0.48a + 0.4(1 - 0.8a) = 1 \
0.48a + 0.4 - 0.32a = 1 \
0.16a = 0.6 \implies a = 3.75 \
b = 1 - 0.8(3.75) = 1 - 3 = -2
$$

So,
\[
w_p = 3.75 w_A - 2 w_B = 3.75 \begin{bmatrix} 0.4 \ 0 \ 0.4 \end{bmatrix} - 2 \begin{bmatrix} 1 \ -0.5 \ 0.5 \end{bmatrix} = \begin{bmatrix} 1.5 \ 0 \ 1.5 \end{bmatrix} - \begin{bmatrix} 2 \ -1 \ 1 \end{bmatrix} = \begin{bmatrix} -0.5 \ 1 \ 0.5 \en

VIPSolutions · Answer

The minimum-variance portfolio is found by minimizing the portfolio variance, $w^T \Sigma w$, subject to the constraint that the weights sum to one, $w^T 1 = 1$, using the method of Lagrange multipliers. Solving the resulting system of linear equations gives the optimal weights, which are then used to calculate the portfolio's expected return, $\mu = w^T \bar{r}$.

VIPSolutions · Answer

The optimal portfolio with $\bar{r}_p=0.7$ is found by solving the system formed from the Lagrangian conditions, leading to weights $w = [0.25, 0.25, 0.5]^T$. This ensures the portfolio meets the expected return constraint while minimizing variance.

VIPSolutions · Answer

A: The portfolio obtained by solving the Lagrangian conditions ensures the expected return constraint is satisfied while minimizing variance, resulting in weights $w = [0.25, 0.25, 0.5]^T$. The process involves solving the system derived from the first-order conditions, confirming the optimal weights.

VIPSolutions · Answer

A: The portfolio obtained by solving the Lagrangian conditions ensures the expected return constraint is satisfied while minimizing variance, resulting in weights $w = [0.25, 0.25, 0.5]^T$. The process involves solving the system derived from the first-order conditions, confirming the optimal weights.

Q: (d) If the risk-free rate is $\ce{r_f = 0.1 }$r_f = 0.1 , find the efficient portfolio of risky assets required by the One-Fund Theorem.

The tangency portfolio maximizing the Sharpe ratio is:

$\ce{w^* = \frac{\Sigma^{-1} \bar{r}_e}{1^T \Sigma^{-1} \bar{r}_e}}$w^* = \frac{\Sigma^{-1} \bar{r}_e}{1^T \Sigma^{-1} \bar{r}_e}

where $\ce{\bar{r}_e = \bar{r} - r_f 1 }$\bar{r}_e = \bar{r} - r_f 1 .

Compute:

$\ce{\bar{r}_e = \begin{bmatrix} 0.4 \ 0.8 \ 0.8 \end{bmatrix} - 0.1 \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} = \begin{bmatrix} 0.3 \ 0.7 \ 0.7 \end{bmatrix}}$\bar{r}_e = \begin{bmatrix} 0.4 \ 0.8 \ 0.8 \end{bmatrix} - 0.1 \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} = \begin{bmatrix} 0.3 \ 0.7 \ 0.7 \end{bmatrix}

Now,

$\ce{z = \Sigma^{-1} \bar{r}_e = \begin{bmatrix} 3 & -2 & 1 \ -2 & 2 & -1 \ 1 & -1 & 1 \end{bmatrix} \begin{bmatrix} 0.3 \ 0.7 \ 0.7 \end{bmatrix} = \begin{bmatrix} 0.9 - 1.4 + 0.7 \ -0.6 + 1.4 - 0.7 \ 0.3 - 0.7 + 0.7 \end{bmatrix} = \begin{bmatrix} 0.2 \ 0.1 \ 0.3 \end{bmatrix}}$z = \Sigma^{-1} \bar{r}_e = \begin{bmatrix} 3 & -2 & 1 \ -2 & 2 & -1 \ 1 & -1 & 1 \end{bmatrix} \begin{bmatrix} 0.3 \ 0.7 \ 0.7 \end{bmatrix} = \begin{bmatrix} 0.2 \ 0.1 \ 0.3 \end{bmatrix}

Sum:

$\ce{1^T z = 0.2 + 0.1 + 0.3 = 0.6}$1^T z = 0.2 + 0.1 + 0.3 = 0.6

So,

$\ce{w^* = \frac{1}{0.6} \begin{bmatrix} 0.2 \ 0.1 \ 0.3 \end{bmatrix} = \begin{bmatrix} 1/3 \ 1/6 \ 1/2 \end{bmatrix}}$w^* = \frac{1}{0.6} \begin{bmatrix} 0.2 \ 0.1 \ 0.3 \end{bmatrix} = \begin{bmatrix} 1/3 \ 1/6 \ 1/2 \end{bmatrix}

explanations in 2 lines
A: The portfolio of risky assets that maximizes the Sharpe ratio (the tangency portfolio) is found by calculating $\Sigma^{-1} \bar{r}_e$ and normalizing it by its sum, resulting in weights $\left[\frac{1}{3}, \frac{1}{6}, \frac{1}{2}\right]^T$. This portfolio forms the optimal risky asset component in the two-fund separation when combined with the risk-free asset.

VIPSolutions · Answer

A: The portfolio obtained by solving the Lagrangian conditions ensures the expected return constraint is satisfied while minimizing variance, resulting in weights $w = [0.25, 0.25, 0.5]^T$. The process involves solving the system derived from the first-order conditions, confirming the optimal weights.

Q: (d) If the risk-free rate is $\ce{r_f = 0.1 }$r_f = 0.1 , find the efficient portfolio of risky assets required by the One-Fund Theorem.

The tangency portfolio maximizing the Sharpe ratio is:

$\ce{w^* = \frac{\Sigma^{-1} \bar{r}_e}{1^T \Sigma^{-1} \bar{r}_e}}$w^* = \frac{\Sigma^{-1} \bar{r}_e}{1^T \Sigma^{-1} \bar{r}_e}

where $\ce{\bar{r}_e = \bar{r} - r_f 1 }$\bar{r}_e = \bar{r} - r_f 1 .

Compute:

$\ce{\bar{r}_e = \begin{bmatrix} 0.4 \ 0.8 \ 0.8 \end{bmatrix} - 0.1 \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} = \begin{bmatrix} 0.3 \ 0.7 \ 0.7 \end{bmatrix}}$\bar{r}_e = \begin{bmatrix} 0.4 \ 0.8 \ 0.8 \end{bmatrix} - 0.1 \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} = \begin{bmatrix} 0.3 \ 0.7 \ 0.7 \end{bmatrix}

Now,

$\ce{z = \Sigma^{-1} \bar{r}_e = \begin{bmatrix} 3 & -2 & 1 \ -2 & 2 & -1 \ 1 & -1 & 1 \end{bmatrix} \begin{bmatrix} 0.3 \ 0.7 \ 0.7 \end{bmatrix} = \begin{bmatrix} 0.9 - 1.4 + 0.7 \ -0.6 + 1.4 - 0.7 \ 0.3 - 0.7 + 0.7 \end{bmatrix} = \begin{bmatrix} 0.2 \ 0.1 \ 0.3 \end{bmatrix}}$z = \Sigma^{-1} \bar{r}_e = \begin{bmatrix} 3 & -2 & 1 \ -2 & 2 & -1 \ 1 & -1 & 1 \end{bmatrix} \begin{bmatrix} 0.3 \ 0.7 \ 0.7 \end{bmatrix} = \begin{bmatrix} 0.2 \ 0.1 \ 0.3 \end{bmatrix}

Sum:

$\ce{1^T z = 0.2 + 0.1 + 0.3 = 0.6}$1^T z = 0.2 + 0.1 + 0.3 = 0.6

So,

$\ce{w^* = \frac{1}{0.6} \begin{bmatrix} 0.2 \ 0.1 \ 0.3 \end{bmatrix} = \begin{bmatrix} 1/3 \ 1/6 \ 1/2 \end{bmatrix}}$w^* = \frac{1}{0.6} \begin{bmatrix} 0.2 \ 0.1 \ 0.3 \end{bmatrix} = \begin{bmatrix} 1/3 \ 1/6 \ 1/2 \end{bmatrix}

explanations in 2 lines
A: The portfolio of risky assets that maximizes the Sharpe ratio (the tangency portfolio) is found by calculating $\Sigma^{-1} \bar{r}_e$ and normalizing it by its sum, resulting in weights $\left[\frac{1}{3}, \frac{1}{6}, \frac{1}{2}\right]^T$. This portfolio forms the optimal risky asset component in the two-fund separation when combined with the risk-free asset.

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