January 26, 2025
Portfolio managers adjust their positions by trading to tailor risk and return. In general, lower volatility and higher returns are preferred. They also have to consider multi-period risk management, transaction costs, counterparty risk, tax considerations, and other messy details salient to being successful in their career. We have nothing to say about these important matters and only consider the simplest case: a one-period model that permits rigorous mathematical results.
Our model allows for any return distribution, not just multivariate normal. However, we do not address the difficult and crucial problem of determining the distribution of returns. See this for recent literature on this topic.
We define the one-period model, portfolio, return, and realized return. In general, there is a unique portfolio having minimum realized return. If that has zero variance it is a zero coupon bond. This can be used to tailor the standard deviation and return of a portfolio.
An efficient portfolio has the minimum variance for a given return. The Capital Asset Pricing Model shows every efficient portfolio is a linear combination of any two independent efficient portfolios. A zero coupon bond is clearly efficient. Classical literature assumes the “market” portfolio is efficient, usually accompanied with hand waving arguments involving “equilibrium.”
The key insight of Markowitz was to move from Graham and Todd’s detailed Security Analysis of individual stocks to leveraging market wisdom. There are many people who get up every day and try to beat the market. The prices they pay for instruments indicate what people with skin in the game believe. How to make use of that information is an ongoing topic of research.
Recall if A and B are sets, then the set exponential {B^A = \{f\colon A\to B\}} is the set of all functions from A to B.
Exercise. If A and B are finite, then \#(B^A) = \#B^{\#A}.
Hint. Where \# counts the number of elements in a set.
If S is a set then \mathbf{R}^S is a vector space over the real numbers \mathbf{R} with scalar multiplication and vector addition defined pointwise: (ax)(s) = a(x(s)) and (x + y)(s) = x(s) + y(s), a\in\mathbf{R}, x,y\in\mathbf{R}^S. The dimension of \mathbf{R}^S is the cardinality of S. For x\in\mathbf{R}^S, we write x(s) as x_s and {x = (x_s)_{s\in S}}.
A subspace of a vector space is a subset closed under scalar multiplication and vector addition.
Exercise. Show if U and V are subspaces then {U + V = \{u + v\mid u\in U, v\in V\}} is a subspace.
Hint: Show a(u + v) \in U + V, a\in\mathbf{R}, and u + v\in U + V.
A cone is a subset closed under positive scalar multiplication and vector addition.
Exercise. Show if U and V are cones then {U + V = \{u + v\mid u\in U, v\in V\}} is a cone.
Hint: Same as the previous hint using a > 0.
If S is finite define the inner/dot product to be the scalar {x^*y = \sum_{s\in S} x_s y_s} and the outer product to be the matrix xy^* = [x_s y_t]_{s,t\in S} for x,y\in\mathbf{R}^S.
A probability space is a set \Omega, the sample space, and a probability measure P on \Omega. The elements of the sample space are the possible outcomes. A measure is a function from events, subsets of \Omega, to the real numbers \mathbf{R} satisfying P(E\cup F) = P(E) + P(F) - P(E\cap F), E, F\subseteq\Omega; measures don’t count things twice. Measures must also satisfy P(\emptyset) = 0; the measure of nothing is zero. A measure is a probability measure if P(E)\ge0 for E\subseteq\Omega and P(\Omega) = 1.
Exercise. If \Omega is finite then p_j = P(\{\omega\}), \omega_j\in\Omega determine the probability measure P.
Hint: Show P(E) = \sum_{\omega_j\in E} p_j for E\subseteq\Omega.
The one-period model consists of a probability space (\Omega, P), a set of instruments I, their prices x\in\mathbf{R}^I at the beginning of the period, and their random prices {X\colon\Omega\to\mathbf{R}^I} at the end of the period. If outcome {\omega\in\Omega} occurs, the prices at the end of the period are {X(\omega)\in\mathbf{R}^I}.
Holding an instrument may entail cash flows. For example, bonds have coupons and stocks have dividends. We implicitly assume these are added to the prices at the end of the period.
Estimating the probability measure P is a difficult and important problem having no canonical solution. Often it is assumed to be multivariate normal. The mean and covariance can be estimated from historical returns assuming independent increments and stationarity. Even then, there are still questions. How far back in history? Should the estimates use exponential weighting? What parameter should be used in the exponential weighting?
We will assume the one-period model is arbitrage-free. Arbitrage exists (for a model) if there is a portfolio \xi\in\mathbf{R}^I with negative cost \xi^*x < 0 and non-negative return \xi^*X\ge0. You make money acquiring the portfolio and never lose money when it is liquidated.
The Fundamental Theorem of Asset Pricing states there is no arbitrage if and only if x belongs to the smallest closed cone containing the range of X. This is equivalent to the existence of a positive measure D on \Omega with x = \int_\Omega X,dD. In general, D is not unique. If we let P = D/D(\Omega) then P is a positive measure with weight 1 and x = E[X]D(\Omega).
Exercise: Show if there exists \zeta\in\mathbf{R}^I with \zeta^*X = 1 then D(\Omega) = \zeta^*x.
Hint: \zeta^*x = \int_\Omega \zeta^*X dD.
We call this portfolio a zero coupon bond. Its price \zeta^*x is the discount over the period.
The binomial model is when \Omega has only two possibilities.
Exercise: If X_i is not constant for some i\in I show every X_j is a linear function of X_i.
Hint: Evaluating X_j = a + b X_i at each point gives two linear equations in two unknowns.
This limits the usefulness of one-period binomial models.
A two instrument model of a bond and a stock is \Omega = \{L,H\}, x = (1, s), and X(\omega) = (R,\omega). The bond has realized return R and the stock can go to either a high or low price.
Exercise. Show every payoff Y\colon\Omega\to\mathbf{R} is a linear combination of the bond and stock.
Hint: Find a,b\in\mathbf{R} with Y(L) = aR + bL and Y(H) = aR + bH.
The interval model is a two instrument model where the stock can go to any price between a given low and high price. Its sample space is the interval \Omega = [L, H] with x = (1, s), and X(\omega) = (R,\omega).
The jointly normal model has sample space {\Omega = \mathbf{R}^I}, any vector of terminal prices is possible, and P is a jointly normal distribution on \mathbf{R}^I. Every jointly normal distribution is uniquely determined by the expected prices {E[X] = (E[X_i])_{i\in I}} and the covariance matrix \operatorname{Var}(X) = \operatorname{Cov}(X,X) = {[\operatorname{Cov}(X_i, X_j)]_{i,j\in I}}.
Exercise Show \operatorname{Var}(X) = E[(X - E[X])(X - E[X])^*] = E[XX^*] - E[X]E[X]^*.
Hint: XX^*(\omega) = X(\omega)X(\omega)^*.
If I has n instruments, then the jointly normal model is specified by n + n(n-1)/2 numbers since the covariance matrix is symmetric.
A portfolio is a vector \xi\in\mathbf{R}^I representing the number of shares held in each instrument. We assume perfect liquidity so the cost of acquiring \xi is \xi^*x. The return on \xi at the end of the period is the random variable \xi^*X\colon\Omega\to\mathbf{R}. In the one-period model it is implicitly assumed the portfolio is liquidated at the end of the period.
The market is the set of all possible portfolio returns {\mathcal{M}= \mathcal{M}_X = \{\xi^*X\mid\xi\in\mathbf{R}^I\} \subseteq \mathbf{R}^\Omega}. The model is complete if \mathcal{M}= \mathbf{R}^\Omega; every possible return is attainable from some portfolio.
If \xi^*X = 0 for some \xi\not=0 then X_i = -\sum_{j\not= i}\xi_j/\xi_i X_j where \xi_i is a non-zero component of \xi. In this case we can remove instrument i\in I and still have the same market.
Assuming we have done this for all instruments, the market is a vector space of dimension \#I. The one period model is not complete if \#\Omega is larger this. The number of possible outcomes are vastly larger than the number of instruments in any realistic model.
If x_i\not=0 we can replace x_i by 1 and X_i by X_i/x_i without affecting the market.
A portfolio \xi\in\mathbf{R}^I is feasible if its cost at the beginning of the period is one unit, i.e., \xi^*x = 1.
Exercise: Show if \xi^*x is not zero then \xi/\xi^*x is feasible.
The realized return of a portfolio \xi\in\mathbf{R}^I is {R_\xi = \xi^*X/\xi^*x} when {\xi^*x\not=0}. Note R_{w\xi} = R_\xi for any non-zero w\in\mathbf{R}.
Exercise: Show the return of a feasible portfolio equals its realized return.
Hint: If \xi is feasible then \xi^*x = 1.
Assuming a portfolio is feasible simplifies formulas.
Exercise. If \xi_0 and \xi_1 are feasible portfolios, then \xi_w = (1 - w)\xi_0 + w\xi_1 is a feasible portfolio for any w\in\mathbf{R}.
Hint: (1 - w) + w = 1.
Exercise. Show \xi_w = \xi_0 if w = 0 and \xi_w = \xi_1 if w = 1.
We begin by considering how to use two portfolios to tailor return and standard deviation over a single time period. Portfolio managers do not buy or sell a fraction of every instrument in a portfolio.
The (realized) return of R_w = \xi_w^*X is R_w = (1 - w)R_0 + w R_1 = R_0 + w(R_1 - R_0) and has variance \operatorname{Var}(R_w) = \operatorname{Var}(R_0) + 2w\operatorname{Cov}(R_0, R_1 - R_0) + w^2\operatorname{Var}(R_1 - R_0)
If E[R_0] = E[R_1], then E[R_w] is constant. We assume E[R_0] \not= E[R_1] in what follows.
Given a target return R we can find a weight w\in\mathbf{R} with R = E[R_w].
Exercise. Show the expected return of R_w is R if w = (R - E[R_0])/(E[R_1] - E[R_0]).
Hint: Solve R = E[R_0] + w(E[R_1] - E[R_0]) for w.
If R_0 and R_1 have correlation 1 then \operatorname{Var}(R_0 - R_1) = 0. If R_0 and R_1 have correlation -1 then \operatorname{Var}(R_0 + R_1) = 0. In either case we have a redundant securities. We assume R_0 and R_1 are not perfectly correlated in what follows.
Given a target standard deviation \sigma it might not be possible to find a weight w with \sigma = \sigma_w, where \sigma_w^2 is the variance of R_w. Variance is quadratic in w so there is a minimum.
Exercise: Show A - 2Bw + Cw^2 is minimized when {w = B/C} and the minimum is {A - B^2/C}.
Hint: Show A - 2Bw + Cw^2 = A + (B/\sqrt{C} - w\sqrt{C})^2 - B^2/C.
Exercise. Show \sigma_w is minimized when {w = (\operatorname{Var}(R_0) - \operatorname{Cov}{R_0,R_1})/(\operatorname{Var}(R_0) - 2\operatorname{Cov}(R_0, R_1) + \operatorname{Var}(R_1))} and the minimum is {\operatorname{Var}(R_0) - (\operatorname{Var}(R_0) - \operatorname{Cov}(R_0, R_1))^2/(\operatorname{Var}(R_0) - 2\operatorname{Cov}(R_0, R_1) + \operatorname{Var}(R_1))}.
For any values of \sigma greater than the minimum there will be two weights giving the same standard deviation.
Exercise: If \xi_0 is risk-free show \operatorname{Var}(R_w) = w^2\operatorname{Var}(R_1).
This shows if a risk-free instrument is available, we can achieve any non-negative target standard deviation.
Exercise. Show R_0 + (R_1 - R_0)\sigma/\sigma_1 has standard deviation \sigma.
This is called the capital allocation line in the literature. We call it matching a standard deviation target.
Exercise. Show (E[R_{\sigma/\sigma_1}] - R_0)/\sigma = (E[R_1] - R_0)/\sigma_1 for any \sigma > 0.
The Sharpe ratio of a realized return R is S(R) = (E[R] - R_0)/\sigma where R_0 is the risk-free realized return. The exercise shows we cannot use a risk-free instrument to change the Sharpe ratio.
A portfolio is efficient if its variance is less than or equal to all other portfolios having the same expected realized return. If \operatorname{Var}(R_\zeta) = 0 for some \zeta\in\mathbf{R}^I then \zeta is clearly efficient and we say \zeta is a risk-free zero coupon bond.
Recall \operatorname{Var}(\xi^*X) = \xi^*V\xi where V = \operatorname{Var}(X). If there is no risk-free security then V is invertible.
Minimizing \operatorname{Var}(R_\xi) given \xi^*x = 1 and E[R_\xi] = \rho yields \xi = \lambda V^{-1}x + \mu V^{-1} E[X], where V = \operatorname{Var}(X), \lambda = (C - \rho B)/D, \mu = (-B + \rho A)/D with \begin{aligned} A &= x^* V^{-1}x \\ B &= x^* V^{-1}E[X] = E[X]^*V^{-1}x \\ C &= E[X]^* V^{-1}E[X] \\ \end{aligned} and D = AC - B^2. Note that A, B, C, and D depend only on x, E[X], and E[XX^*]. This shows every efficient portfolio is a linear combination of V^{-1}x and V^{-1} E[X].
The minimum variance solution is \xi = \frac{C - \rho B}{D} V^{-1}x + \frac{-B + \rho A}{D} V^{-1}E[X]. with variance \operatorname{Var}(R_\xi) = \xi^* V\xi = (C - 2B\rho + A\rho^2)/D.
(A + vw^*)^{-1} = A^{-1} - A^{-1}vw^*A^{-1}/(1 + w^*A^{-1}v).
((1 - \rho)I + \rho 11^*)^{-1} = \frac{1}{1 - \rho} I - \frac{1}{(1 - \rho)(1 - \rho + n)} 11^*
x = 1, X = R + Z where R\in\mathbf{R}^I and E[Z] = 0.
\operatorname{Var}(X) = \operatorname{Var}(Z) = E[Z^*Z] = \delta(\sigma)((1 - \rho)I + \rho 11^*)\delta(\sigma).
V^{-1} = 1/(1 - \rho)\delta(1/\sigma)(I - 11^*/(1 - \rho + n))\delta(1/\sigma)
V^{-1}1 = 1/(1 - \rho)\delta(1/\sigma)(1 - n1)/(1 - \rho + n))\delta(1/\sigma)
1^*V^{-1}1 = 1/(1 - \rho)\delta(1/\sigma)(n - n^2)/(1 - \rho + n))\delta(1/\sigma)
x = (1, 1), X = R + Z, E[Z] = 0
V = \begin{bmatrix} \operatorname{Var}(R_0) & \sigma_0\sigma_1\rho \\ \sigma_0\sigma_1\rho & \operatorname{Var}(R_1) \\ \end{bmatrix} V^{-1} = \frac{1}{\operatorname{Var}(R_0)\operatorname{Var}(R_1)(1 - \rho^2)} \begin{bmatrix} \operatorname{Var}(R_1) & -\sigma_0\sigma_1\rho \\ -\sigma_0\sigma_1\rho & \operatorname{Var}(R_0) \\ \end{bmatrix}
V^{-1}x = \frac{1}{\operatorname{Var}(R_0)\operatorname{Var}(R_1)(1 - \rho^2)} \begin{bmatrix} \operatorname{Var}(R_1) - \sigma_0\sigma_1\rho \\ -\sigma_0\sigma_1\rho + \operatorname{Var}(R_0) \\ \end{bmatrix}
x^*V^{-1}x = \frac{1}{\operatorname{Var}(R_0)\operatorname{Var}(R_1)(1 - \rho^2)} (\operatorname{Var}(R_1) - 2\sigma_0\sigma_1\rho + \operatorname{Var}(R_0))