Single-Asset VaR

December 11, 2021

In mathematical finance, value at risk (VaR) is a statistic that quantifies the level of risk in a position or portfolio over a specific time frame. VaR is a measue of probable maximum loss (PML)—the maximum loss that the position or portfolio would be expected to incur. Specifically, we are interested in answering the question, What are the worst 1% of possible outcomes?

Suppose we have a position in a stock. How might we go about determining the worst 1% of possible outcomes for tomorrow? Well, if $P$ is a random variable that models tomorrow's closing price, then the 99% probable minimum closing price $p_0$ satisfies the integral equation

$\int_{p_0}^\infty f_P(p) \, dp = \text{Pr}[P \geq p_0] = 0.99,$

where $f_P$ is the probability density function (PDF) of $P$. To solve this equation for $p_0$, we need to know $f_P$—but how do we get it? To that end, let $\mathbf{p} = (p_1, \dots, p_n)$ be the sequence of daily closing prices of the stock for the past $n$ days. For example, $\mathbf{p}$ might be the sequence of daily closing prices of Tesla for the past year, plotted below.

If we assume that tomorrow's closing price will be like past closing prices, then we can use the frequency distribution of $\mathbf{p}$ to estimate $f_P$. The frequency of a price $p$ in $\mathbf{p}$ is just the number of occurrences of $p$ in $\mathbf{p}$, given by $\sum_{i=1}^n \delta_{p_ip}$, where $\delta_{ij}$ is the Kronecker delta:

$\delta_{ij} = \begin{cases} 0, & i \neq j, \\ 1, & i = j. \end{cases}$

The relative frequency is just the normalized frequency: $\frac{1}{n}\sum_{i=1}^n \delta_{p_ip}$. It tells us the probability that a randomly selected $p_i$ will equal $p$. We define the relative frequency function (RFF) of $\mathbf{p}$ to be a function $f_\mathbf{p}(p)$ that gives the relative frequency of $p$ in $\mathbf{p}$. Symbolically,

$f_\mathbf{p}(p) \equiv \frac{1}{n}\sum_{i=1}^n \delta_{p_ip}.$

$f_\mathbf{p}$ for our example is represented below as a histogram.

Under our assumptions, $f_P \approx f_\mathbf{p}$. But this plot illustrates a clear problem with our approach to estimating $f_P$. Based on our approach, TSLA would be more likely to close around \$700—a massive drop—than around the same price as yesterday! This seems wildly wrong. The problem lies in our assumption: tomorrow's closing price will not be like past closing prices—a certain closing price is not necessarily likely just because the stock has closed at that price many times in the past. Thus, we are forced to consider alternative approaches.

A reasonable next step might be to look at changes in price. Let $C$ be a random variable that models tomorrow's change in price, and define the sequence $\mathbf{c} = (c_1, \dots, c_{n-1})$ via

$c_i = p_{i+1} - p_i,$

so that $c_i$ is the change in price on day $i + 1$. $\mathbf{c}$ and $f_\mathbf{c}$ for our example are plotted below.

This looks much more promising. Our RFF looks fairly normal, so suppose that $C \sim \mathcal{N(\mu, \sigma^2)}$ for some mean $\mu$ and variance $\sigma^2$. The problem with using changes in price arises when we try to estimate $\mu$ and $\sigma$ by applying maximum likelihood estimation (MLE) to this distribution. If you have studied linear regression, the idea here is similar: we want to optimize the parameters of our function to best fit our data. The difference is that we are working with a Gaussian (or "bell curve") as opposed to a linear function.

To see the problem, consider $\mathbf{c}$ to be a sample for the sequence of random variables $C_1, \dots, C_{n-1}$ that model the change in price on days $2, \dots, n$, respectively, and suppose that each $C_i$ is normally distributed, which is reasonable since $C$ is assumed to be normally distributed and in this context there is nothing special about tomorrow as opposed to, say, last Thursday. The problem here is that we should not assume that changes in price are identically distributed—each $C_i$ requires its own parameters: $C_i \sim \mathcal{N}(\mu_i, \sigma_i^2)$. The reason identical distribution would be a bad assumption here is that, as anyone who has traded stocks can attest, changes in price tend to scale with the price itself. (This is why stock prices often look exponential over long periods of time: the defining characteristic of an exponential function $f(x)$ is that it satisfies the differential equation

$\frac{df}{dx} = af(x),$

where $a$ is a constant. In this case, the derivative corresponds to the change in price and the function itself corresponds to the price. From a psychological perspective, this can be attributed to what is known as the Weber-Fechner law, which says that the perceived effect of a change depends on the starting point.) Because of this, estimating the $\mu_i$s and $\sigma_i^2$s would not get us any closer to estimating $\mu$ and $\sigma^2$. We can account for this scaling behavior, however—by converting to returns.

Let $R$ be a random variable that models tomorrow's return, and define the sequence $\mathbf{r} = (r_1, \dots, r_{n-1})$ via

$r_i = \frac{c_{i+1}}{p_{i}} = \frac{p_{i+1} - p_i}{p_i},$

so that $r_i$ is the return on day $i + 1$. $\mathbf{r}$ and $f_\mathbf{r}$ for our example are plotted below.

As before, our RFF appears normal, so suppose that $R \sim \mathcal{N}(\mu, \sigma^2)$ for some $\mu$ and $\sigma^2$ and consider $\mathbf{r}$ to be a sample for the sequence of random variables $\mathbf{R} = (R_1, \dots, R_{n-1})$ that model the return on days $2, \dots, n$, respectively. This time around, since we are dealing with percentage changes, we can assume identical distribution: each $R_i \sim \mathcal{N}(\mu, \sigma^2)$. Then the PDF of each $R_i$ is

$f_{R_i}(r \mid \mu, \sigma^2) = f_R(r \mid \mu, \sigma^2) = \frac{1}{\sqrt{2\pi\sigma^2}}e^{-(r - \mu)^2/2\sigma^2}.$

Furthermore, because we do not expect the return on any given day to depend too much on past returns, we assume that the $R_i$s are mutually independent. We can then apply the multiplication rule to obtain their joint cumulative distribution function (JCDF):

\begin{aligned} F_\mathbf{R}(\mathbf{r_0} \mid \mu, \sigma^2) &\equiv \text{Pr}[R_1 \leq r_{0,1}, \dots, R_{n-1} \leq r_{0,n-1} \mid \mu, \sigma^2] \\ &= \prod_{i=1}^{n-1} \text{Pr}[R_i \leq r_{0,i} \mid \mu, \sigma^2] \\ &= \prod_{i=1}^{n-1} \int_{-\infty}^{r_{0,i}} f_{R_i}(r \mid \mu, \sigma^2) \, dr, \\ &= \prod_{i=1}^{n-1} \int_{-\infty}^{r_{0,i}} f_R(r \mid \mu, \sigma^2) \, dr, \end{aligned}

where $\mathbf{r_0} = (r_{0,1}, \dots, r_{0,n-1})$ is an arbitrary sample for $\mathbf{R}$. Their joint probability density function is then, using the fundamental theorem of calculus,

\begin{aligned} f_\mathbf{R}(\mathbf{r_0} \mid \mathbf{\mu}, \mathbf{\sigma^2}) &\equiv \frac{\partial^{n-1}F_\mathbf{R}}{{\partial}r_{0,1}\cdots{\partial}r_{0,n-1}} \\ &= \frac{\partial^{n-1}}{{\partial}r_{0,1}\cdots{\partial}r_{0,n-1}}\prod_{i=1}^{n-1} \int_{-\infty}^{r_{0,i}} f_R(r \mid \mu, \sigma^2) \, dr \\ &= \prod_{i=1}^{n-1} \frac{\partial}{{\partial}r_{0,i}}\int_{-\infty}^{r_{0,i}} f_R(r \mid \mu, \sigma^2) \, dr \\ &= \prod_{i=1}^{n-1} f_R(r_{0,i} \mid \mu, \sigma^2) \\ \end{aligned}

For given parameters $\mu$ and $\sigma^2$, the relative likelihood of observing $\mathbf{r}$ is given by the likelihood function:

\begin{aligned} \mathcal{L}(\mu, \sigma^2) &\equiv f_\mathbf{R}(\mathbf{r} \mid \mu, \sigma^2) \\ &= \prod_{i=1}^{n-1} f_R(r_i \mid \mu, \sigma^2) \\ &= \prod_{i=1}^{n-1} \frac{1}{\sqrt{2\pi\sigma^2}}e^{-(r_i - \mu)^2/2\sigma^2} \\ &= \left(\prod_{i=1}^{n-1} \frac{1}{\sqrt{2\pi\sigma^2}}\right)\left(\prod_{i=1}^{n-1} e^{-(r_i - \mu)^2/2\sigma^2}\right) \\ &= \left(\frac{1}{2\pi\sigma^2}\right)^{(n-1) / 2}\exp\left(\sum_{i=1}^{n-1} -\frac{(r_i - \mu)^2}{2\sigma^2}\right) \\ &= \left(\frac{1}{2\pi\sigma^2}\right)^{(n-1) / 2}\exp\left(-\frac{1}{2\sigma^2}\sum_{i=1}^{n-1} (r_i - \mu)^2\right) \end{aligned}

We want to find values of $\mathbf{\mu}$ and $\mathbf{\sigma^2}$ that maximize $\mathcal{L}$. That is, we want to satisfy the equations

$\frac{\partial\mathcal{L}}{\partial\mathbf{\mu}} = 0 \quad \text{and} \quad \frac{\partial\mathcal{L}}{\partial\mathbf{\sigma^2}} = 0.$

Because of the exponential in $\mathcal{L}$, it is convenient to work with the log-likelihood function:

\begin{aligned} \ell(\mu, \sigma^2) &\equiv \ln \mathcal{L}(\mu, \sigma^2) \\ &= \ln\left[\left(\frac{1}{2\pi\sigma^2}\right)^{(n - 1) / 2}\exp\left(-\frac{1}{2\sigma^2}\sum_{i=1}^{n-1} (r_i - \mu)^2\right)\right] \\ &= \ln\left(\frac{1}{2\pi\sigma^2}\right)^{(n - 1) / 2} + \ln\exp\left(-\frac{1}{2\sigma^2}\sum_{i=1}^{n-1} (r_i - \mu)^2\right) \\ &= \frac{n - 1}{2}\ln \frac{1}{2\pi\sigma^2} - \frac{1}{2\sigma^2}\sum_{i=1}^{n-1} (r_i - \mu)^2. \end{aligned}

Since the natural logarithm function is monotonically increasing, $\ell$ is maximized by the same values of $\mu$ and $\sigma^2$ as $\mathcal{L}$. Thus, we can look at the equations

$\frac{\partial\ell}{\partial\mathbf{\mu}} = 0 \quad \text{and} \quad \frac{\partial\ell}{\partial\mathbf{\sigma^2}} = 0.$

For the $\mu$ equation, we have

\begin{aligned} & \begin{aligned} 0 & = \frac{\partial}{\partial\mu}\left(\frac{n - 1}{2}\ln \frac{1}{2\pi\sigma^2} - \frac{1}{2\sigma^2}\sum_{i=1}^{n-1} (r_i - \mu)^2\right) \\ & = -\frac{1}{2\sigma^2}\sum_{i=1}^{n-1} \frac{\partial}{\partial\mu}(r_i - \mu)^2 \\ & = \frac{1}{\sigma^2}\sum_{i=1}^{n-1}(r_i - \mu) \\ & = \frac{1}{\sigma^2}\left(\sum_{i=1}^{n-1} r_i - (n - 1)\mu\right) \end{aligned} \\ & \implies \sum_{i=1}^{n-1} r_i - (n - 1)\mu = 0 \\ & \implies (n - 1)\mu = \sum_{i=1}^{n-1} r_i \\ & \implies \mu = \frac{1}{n - 1}\sum_{i=1}^{n-1} r_i, \end{aligned}

which is just the arithmetic mean. For the $\sigma^2$ equation, we have

\begin{aligned} & \begin{aligned} 0 & = \frac{\partial}{\partial\sigma^2}\left(\frac{n - 1}{2}\ln \frac{1}{2\pi\sigma^2} - \frac{1}{2\sigma^2}\sum_{i=1}^{n-1} (r_i - \mu)^2\right) \\ & = -\frac{n - 1}{2\sigma^2} + \frac{1}{2\sigma^4}\sum_{i=1}^{n-1} (r_i - \mu)^2 \end{aligned} \\ & \implies -(n - 1)\sigma^2 + \sum_{i=1}^{n-1} (r_i - \mu)^2 = 0 \\ & \implies (n - 1)\sigma^2 = \sum_{i=1}^{n-1} (r_i - \mu)^2 \\ & \implies \sigma^2 = \frac{1}{n - 1}\sum_{i=1}^{n-1} (r_i - \mu)^2. \end{aligned}

With $\mu$ and $\sigma$ in hand, we are now in a position to consider the integral equation

$0.99 = \int_{r_0}^\infty f_R(r) \, dr = \frac{1}{\sqrt{2\pi\sigma^2}}\int_{r_0}^\infty e^{-(r - \mu)^2/2\sigma^2} \, dr.$

As it happens, Gaussians are not antidifferentiable, so we cannot compute this integral analytically, but we have, in principle, arrived at the answer to our question. Of course, principle does not help you assess real-world risk, so it is important to know that this integral can be calculated numerically. In fact, because Gaussians are so ubiquitous, they have been well-studied, and numeric approximations of their integrals can be readily found in tables. In this case, the solution is

$r_0 = -2.33\sigma.$

Therefore, to estimate the 99% probable minimum loss for tomorrow, you can simply multiply the historical standard deviation of the stock by $2.33$.