Case study: portfolio risk and conditional expectation

The fund's variance, from the rules

The data are case study 3.2.5's, invented but typical, in annual decimals. The expected returns are $\boldsymbol{\mu} = (0.06, 0.08, 0.11)$ and the volatilities 0.12, 0.18 and 0.25. The correlations are $\rho_{AB} = 0.3$, $\rho_{AC} = 0.1$ and $\rho_{BC} = 0.5$, and the weights $\mathbf{w} = (0.5, 0.3, 0.2)$.

Before the year starts, its returns are random. Write them as a random vector $\mathbf{R} = (R_A, R_B, R_C)$, in capitals; case study 3.1.4 wrote one year's values as $\mathbf{r}$.

Its mean is $\boldsymbol{\mu}$, and its covariance matrix $\Sigma$ has entries $\Sigma_{ij} = \operatorname{Cov}(R_i, R_j) = \rho_{ij}\sigma_i\sigma_j$ (lesson 5.5.2). It is the matrix case study 3.2.5 built from the volatilities and correlations.

The fund's return is $R_p = \mathbf{w}^\top\mathbf{R} = w_AR_A + w_BR_B + w_CR_C$ (case study 3.1.4). Its mean follows from linearity (lesson 5.3.2): $\mathbb{E}[R_p] = \mathbf{w}^\top\boldsymbol{\mu} = 0.076$. Its variance takes three steps.

1. 1

   $\operatorname{Var}(R_p) = \operatorname{Cov}(R_p,\ R_p)$

   Why?
2. 2

   $\operatorname{Cov}\big(\sum_i w_iR_i,\ \sum_j w_jR_j\big) = \sum_i\sum_j w_iw_j\operatorname{Cov}(R_i, R_j)$

   Why?
3. 3

   $\sum_i\sum_j w_iw_j\,\Sigma_{ij} = \mathbf{w}^\top\Sigma\,\mathbf{w}$

   Why?

The double sum has nine terms. The three on the diagonal, each asset with itself, add up to 0.009016; the six off it, each pair twice, add 0.005244. The total is 0.01426, a volatility of 11.94%: case study 3.2.5's numbers, now derived.

Two assets: case study 1.2.4. With weight $w$ in A and $1 - w$ in B, lesson 5.5.2's rule for $\operatorname{Var}(aX + bY)$ gives

with $\operatorname{Cov}(R_A, R_B) = \rho\sigma_A\sigma_B$. That is the formula case study 1.2.4 took as given.

Its factor 2 counts $\operatorname{Cov}(R_A, R_B)$ and $\operatorname{Cov}(R_B, R_A)$. With that case study's $\sigma_A = 0.2$, $\sigma_B = 0.3$ and $\rho = 0.2$, the minimum-variance weight 0.7358 still gives a variance of 0.0326.

What the formula needs, and what it does not

The steps used only expectations and covariances. The formula needs no normal distribution and no independence: any returns with this $\Sigma$ give a variance of 0.01426.

A simulation shows it. Draw 200,000 years as $\boldsymbol{\mu} + L\mathbf{U}$, with $L$ the Cholesky factor of $\Sigma$ (case study 3.7.4) and $\mathbf{U}$ normal.

Then do it again with $\mathbf{U}$ uniform on $[-\sqrt{3}, \sqrt{3}]$, which also has mean 0 and variance 1. By lesson 5.5.3's rule both have covariance matrix $LL^\top = \Sigma$.

In this seeded run both give a sample variance close to 0.01426. Their chances of a loss beyond 10% differ, though: 7.03% with normal draws and 7.55% with uniform ones. The variance comes from $\Sigma$ alone; a probability needs the whole distribution.

With jointly normal returns, $R_p \sim \mathcal{N}(0.076,\ 0.01426)$ (lesson 5.5.3). So $P(R_p < -0.10) = N\big(\frac{-0.10 - 0.076}{0.119415}\big) = N(-1.4738) = 0.0703$, matching the normal draws.

From months to a year: when √12 works

Log returns add across time: a year's log return is the sum of its twelve monthly ones (case study 1.4.7). Lesson 5.5.2's rule for a sum of many variables then gives

with $r_t$ the log return of month $t$. If the months are uncorrelated, every covariance is 0 and only the twelve variances are left. Independence is enough for that, since independent variables are uncorrelated (lesson 5.5.2).

This is what case studies 3.2.5 and 3.6.5 assumed.

• Months to a year. The fund's monthly volatility is 3.4472%, from $\Sigma/12$; times $\sqrt{12}$ it is the annual 11.94% (case study 3.2.5).
• Days to a year. The level component of case study 3.6.5's yield changes has a daily volatility of 10.52 bp. Yield changes add up exactly over days, so with uncorrelated days it is $10.52\sqrt{252} = 167.0$ bp a year.

When neighbouring months are correlated

Suppose each month's return has correlation $\rho_1$ with the month before, and none with months further apart: a lag-one correlation. Twelve months have eleven neighbouring pairs, each counted twice, so the variance becomes $\sigma^2(12 + 22\rho_1)$.

With $\rho_1 = 0.2$, the $\sqrt{12}$ rule reports 11.94% when the risk is 13.96%: 14.5% too low. Positive lag-one correlation is typical of smoothed or stale prices, such as appraisal values of property and thinly traded assets. A negative one, as in a series that tends to reverse its moves, makes $\sqrt{12}$ overstate the risk.

What B's return says about C's

Now take two assets, B and C, and treat their returns as jointly normal (lesson 5.5.3). Their means are 8% and 11%, their volatilities 18% and 25%, and $\rho = 0.5$. Build them from independent standard normals $Z_1$ and $Z_2$ with the Cholesky factor, B first, as case study 4.7.5 built A and B:

Learning that $R_B = b$ fixes $Z_1 = \frac{b - \mu_B}{\sigma_B}$. It says nothing about $Z_2$, which is independent of $Z_1$, so $Z_2$ is still standard normal (lesson 5.5.1). Given $R_B = b$, then, $R_C$ is a known number plus $\sigma_C\sqrt{1 - \rho^2}\,Z_2$.

Here the slope is $0.5 \times \frac{0.25}{0.18} = 0.6944$, and the conditional volatility is $0.25\sqrt{0.75} = 21.65\%$, against 25% before anything is known.

At $b = 8\%$, B's mean, the line gives C's mean, 11%. Each point that B's return lies above or below 8% moves the expected C by 0.6944 points the same way.

Probabilities shift too. Unconditionally, C loses money in $N\big(\frac{-0.11}{0.25}\big) = 33.0\%$ of years. Given that B lost 10%, the chance is $N\big(\frac{0.015}{0.2165}\big) = 52.76\%$.

The hedge ratio again

The slope $\frac{\operatorname{Cov}(R_B, R_C)}{\operatorname{Var}(R_B)} = 0.6944$ is case study 3.5.5's hedge ratio for C hedged with B alone. That case study found it as the coefficient of a projection, with $R^2 = \rho^2 = 0.25$.

Probability adds a reading. Selling 0.6944 of B per dollar of C leaves $R_C - 0.6944R_B$, which is $\sigma_C\sqrt{1 - \rho^2}\,Z_2$ plus a constant.

That depends only on $Z_2$, while $R_B$ depends only on $Z_1$, so the two are independent (lesson 5.5.1). What the hedge leaves, 21.65% a year, is a risk that B's moves cannot explain at all.

The same rule works for any pair. A's return given B's has slope $\rho_{AB}\frac{\sigma_A}{\sigma_B} = 0.3 \times \frac{0.12}{0.18} = 0.2$.

The line does not reverse. Given C's return instead, $\mathbb{E}[R_B \mid R_C = c]$ has slope $\rho\frac{\sigma_B}{\sigma_C} = 0.36$, not $\frac{1}{0.6944} = 1.44$. Both lines pass through the two means, but they differ unless $\rho = \pm 1$.

Estimated from simulated years

In the explorer, $X$ and $Y$ stand for B's and C's returns, each measured from its mean in units of B's volatility, 18%. Set the standard deviation of $Y$ to 1.4, close to $\frac{\sigma_C}{\sigma_B} = 1.39$. The dashed line of conditional means then has slope 0.7.

Turn on the fitted line and cut the number of draws to 20. Press "Draw again" a few times: a slope estimated from a short sample wanders.

With many draws the estimate is close. In 200,000 simulated years, the years with $R_B$ within 2 points of each $b$ in the table have average $R_C$ close to the line. The sample slope, $\frac{s_{BC}}{s_B^2}$ (lesson 5.5.2), is 0.696.

Ten years of monthly returns are another matter. Case study 3.2.5's 120 months give a slope of 0.5109. Across 1,000 such ten-year samples, the middle 90% of the estimates runs from 0.5169 to 0.881.

When the line is not straight

The straight line is a property of the normal distribution. For other joint distributions, $\mathbb{E}[Y \mid X = x]$ can bend: lesson 5.5.1's density $x + y$ had $\mathbb{E}[Y \mid X = x] = \frac{x/2 + 1/3}{x + 1/2}$, a curve.

The slope $\frac{\operatorname{Cov}}{\operatorname{Var}}$ then still gives the best straight line through the cloud, the least-squares line of lesson 3.5.3. But it is only an approximation to the conditional mean. Module 5.10 studies such lines as regressions.

On the desk

import math

import numpy as np

MU = np.array([0.06, 0.08, 0.11])
VOLS = np.array([0.12, 0.18, 0.25])
CORR = np.array([[1.0, 0.3, 0.1], [0.3, 1.0, 0.5], [0.1, 0.5, 1.0]])
SIGMA = np.diag(VOLS) @ CORR @ np.diag(VOLS)


def _raises(f, *args, what=""):
    try:
        f(*args)
    except ValueError:
        return
    raise AssertionError(f"expected ValueError: {what}")


def test_hook():
    """If B returns −10%, C's conditional mean is −1.5% with standard deviation about 21.65%"""
    expect_close(conditional_normal(MU, SIGMA, 2, 1, -0.10), (-0.015, 0.25 * math.sqrt(0.75)), what="conditional_normal(C given B = -0.10)")


def test_at_the_mean():
    """Given B at its mean of 8%, C's conditional mean is its own mean, 11%"""
    expect_close(conditional_normal(MU, SIGMA, 2, 1, 0.08)[0], 0.11, what="mean of C given B = 0.08")


def test_slope():
    """The conditional mean of C moves 0.6944 points per point of B"""
    up = conditional_normal(MU, SIGMA, 2, 1, 0.20)[0]
    down = conditional_normal(MU, SIGMA, 2, 1, 0.10)[0]
    expect_close((up - down) / 0.10, 0.5 * 0.25 / 0.18, what="slope of C on B")


def test_reverse():
    """B given C has slope 0.36, not 1 / 0.6944, and standard deviation 18% × √0.75"""
    up = conditional_normal(MU, SIGMA, 1, 2, 0.21)
    down = conditional_normal(MU, SIGMA, 1, 2, 0.11)
    expect_close((up[0] - down[0]) / 0.10, 0.36, what="slope of B on C")
    expect_close(up[1], 0.18 * math.sqrt(0.75), what="standard deviation of B given C")


def test_a_given_b():
    """A given B = 20% has mean 8.4%: slope 0.2"""
    expect_close(conditional_normal(MU, SIGMA, 0, 1, 0.20)[0], 0.084, what="mean of A given B = 0.20")


def test_bad_cov():
    """A non-symmetric covariance matrix, a bad index or target equal to given raises ValueError"""
    bad = SIGMA.copy()
    bad[0, 1] = 0.0648
    _raises(conditional_normal, MU, bad, 0, 1, 0.1, what="a non-symmetric covariance matrix")
    _raises(conditional_normal, MU, SIGMA, 3, 1, 0.1, what="index 3 with three assets")
    _raises(conditional_normal, MU, SIGMA, 1, 1, 0.1, what="target equal to given")
    _raises(conditional_normal, MU, SIGMA[:2, :2], 0, 1, 0.1, what="a 2 x 2 matrix with three means")


def test_months():
    """The fund's monthly volatility 3.4472%, over 12 uncorrelated months, is 11.94% a year"""
    monthly = math.sqrt(np.array([0.5, 0.3, 0.2]) @ (SIGMA / 12) @ np.array([0.5, 0.3, 0.2]))
    expect_close(annual_vol(monthly, 12), 0.11941524190239804, what="annual_vol(monthly, 12)")


def test_days():
    """10.52 bp a day over 252 uncorrelated days is about 167.0 bp"""
    expect_close(annual_vol(10.52, 252), 10.52 * math.sqrt(252), what="annual_vol(10.52, 252)")


def test_lag_one():
    """A lag-one correlation of 0.2 raises 12 months of 3.4472% to √16.4 times as much; −0.2 lowers it to √7.6 times"""
    expect_close(annual_vol(0.034472, 12, 0.2), 0.034472 * math.sqrt(16.4), what="annual_vol(0.034472, 12, 0.2)")
    expect_close(annual_vol(0.034472, 12, -0.2), 0.034472 * math.sqrt(7.6), what="annual_vol(0.034472, 12, -0.2)")


def test_one_period():
    """One period is the period's own volatility, whatever lag1 is"""
    expect_close(annual_vol(0.02, 1, 0.5), 0.02, what="annual_vol(0.02, 1, 0.5)")


def test_bad_inputs():
    """No periods, a negative volatility, a correlation above 1 or a negative variance raises ValueError"""
    _raises(annual_vol, 0.01, 0, what="periods = 0")
    _raises(annual_vol, -0.01, 12, what="a negative volatility")
    _raises(annual_vol, 0.01, 12, 1.5, what="lag1 = 1.5")
    _raises(annual_vol, 0.01, 12, -0.6, what="lag1 = -0.6, which gives 12 - 13.2 < 0")