 ### Volatility

After completing this reading you should be able to:

• Define and distinguish between volatility, variance rate, and implied volatility.
• Describe the power law.
• Explain how various weighting schemes can be used in estimating volatility.
• Apply the exponentially weighted moving average (EWMA) model to estimate volatility.
• Describe the generalized autoregressive conditional heteroskedasticity (GARCH($$p,q$$)) model for estimating volatility and its properties.
• Calculate volatility using the GARCH(1,1) model.
• Explain mean reversion and how it is captured in the GARCH(1,1) model.
• Explain the weights in the EWMA and GARCH(1,1) models.
• Explain how GARCH models perform in volatility forecasting.
• Describe the volatility term structure and the impact of volatility changes.

## What is Volatility?

The volatility of a variable denoted as $$\sigma$$, is the standard deviation of return which the variable per unit of time gives when continuous compounding expresses return.

Assuming that at the end of day $$i$$, a variable is valued at $${ S }_{ i }$$. Then, on day $$i$$ the variable has the following continuously compounded return per day:

$$ln\frac { { S }_{ i } }{ { S }_{ i-1 } }$$

Which is almost equivalent to:

$$\frac { { S }_{ i }-{ S }_{ i-1 } }{ { S }_{ i-1 } }$$

This implies that the standard deviation of the proportional change in the variable in the day defines daily volatility.

Supposing that each day the returns are independent having a constant variance, then, over $$T$$ days the returns will have a variance that is $$T$$ times that of a single day’s return. Therefore, over $$T$$ days the return’s standard deviation will be $$\sqrt { T }$$ times that of a single day.

### Variance Rate

The square of the volatility is simply the variance rate. Variance rate per day, in this case, happens to be the one day return’s variance. As time elapses, variance experiences a linear increase. Ignoring weekends and holidays, a year is considered to have 252 days.

Supposing that we are provided with independent returns on successive days, with a constant standard deviation, therefore:

$${ \sigma }_{ yr }={ \sigma }_{ day }\sqrt { 252 }$$

$$\Rightarrow { \sigma }_{ day }=\frac { { \sigma }_{ yr } }{ \sqrt { 252 } }$$

From the above, we can deduce that the daily volatility is slightly over 6% the annual volatility.

### Implied Volatility

When substituted into the pricing model, this volatility gives the option’s market price. Indices of implied volatility are published by the CBOE with the most popular being the VIX index.

## The Power Law

According to the law, for most real-life variables, when $$x$$ is large, the value of the variable, $$v$$:

$$prob\left( v>x \right) =k{ x }^{ -\alpha }$$

In the equation, $$k$$ and $$\alpha$$ are constants.

From the above equation:

$$lnprob\left( v>x \right) =ln k\quad -\alpha \quad ln x$$

To test whether the above equation holds, a graph of $$lnprob\left( v>x \right)$$ plotted against $$ln⁡x$$ is necessary. In this case, $$v$$ should be defined as the number of standard deviations which an average rate changes in a day.

### Monitoring Daily Volatility

Assuming that on day $$n$$, a market variable has a volatility per day of $${ \sigma }_{ n }$$, as estimated at the end of day $$n – 1$$. Therefore, the variance rate is, $${ \sigma }_{ n }^{ 2 }$$. It is assumed that at the end of day $$i$$, the market is valued at $${ S }_{ i }$$.

If $${ U }_{ i }$$ is the continuously compounded return during day $$i$$, then:

$${ U }_{ i }=ln\frac { { S }_{ i } }{ { S }_{ i-1 } }$$

To estimate $${ \sigma }_{ n }$$, we equate it to the standard deviations of the $${ U }_{ i }$$’s. The latest $$m$$ observations are applied on the $${ U }_{ i }$$ together with the usual standard deviation formula.

This results in:

$${ \sigma }_{ n }^{ 2 }=\frac { 1 }{ m-1 } { \Sigma }_{ i=1 }^{ m }{ \left( { u }_{ n-1 }-\bar { u } \right) }^{ 2 }\quad \quad \quad \quad equation\quad I$$

The $${ U }_{ i }$$’s have a mean of $$\bar { u }$$, such that:

$$\bar { u } =\frac { 1 }{ m } \sum _{ i=1 }^{ m }{ { u }_{ n-1 } }$$

Risk management may require us to vary equation $$I$$ in several ways:

1. Between the closure of day $$i$$ and the closure of day $$i – 1$$, the market variable changes by a percentage $${ u }_{ i }$$, such that:

$${ u }_{ i }=\frac { { S }_{ i }-{ S }_{ i-1 } }{ { S }_{ i-1 } }$$

2. Since there is a very small expected change in a variable within a day – in comparison to the change in standard deviation, then we assume $$\bar { u }$$ to be zero.
3. Replace $$m – 1$$ with $$m$$. We, therefore, shift from an unbiased estimate of the volatility to an estimate of the maximum likelihood.

With these changes, the variance rate formula can be simplified as:

$${ \sigma }_{ n }^{ 2 }=\frac { 1 }{ m } { \Sigma }_{ i=1 }^{ m }{ u }_{ n-i }^{ 2 }\quad \quad \quad \quad equation\quad II$$

## Weighting Schemes

Equal weight has been given to each of $${ u }_{ n-1 }^{ 2 }, { u }_{ n-2 }^{ 2 },\dots { u }_{ n-m }^{ 2 }$$ in equation $$II$$. To estimate the volatility on day $$n$$, $${ \sigma }_{ n }$$, more weight should be accorded to recent data. To achieve this, we apply the following model:

$${ \sigma }_{ n }^{ 2 }={ \Sigma }_{ i=1 }^{ m }{ \alpha }_{ i }{ u }_{ n-i }^{ 2 }\quad \quad \quad \quad equation\quad III$$

The $$i$$ days ago observations’ weight is denoted by the variable $${ \alpha }_{ i }$$, with the $$\alpha$$’s being positive. If $${ \alpha }_{ i }<{ \alpha }_{ j }$$ and $$i>j$$, older observations are accorded less weight. The weights are such that:

$$\sum _{ i=1 }^{ m }{ { \alpha }_{ i }=1 }$$

The idea in equation $$II$$ can be extended to consider the existence of a long-run average variance rate that should be accorded some weight. The resulting model will take the form:

$${ \sigma }_{ n }^{ 2 }={ \gamma { V }_{ L }+\Sigma }_{ i=1 }^{ m }{ \alpha }_{ i }{ u }_{ n-i }^{ 2 }\quad \quad \quad \quad equation\quad IV$$

The long-run variance rate is denoted as $${ V }_{ L }$$ and is assigned a weight of $$\gamma$$. Since the sum of the weights must be equal to one, then:

$$\gamma + \sum _{ i=1 }^{ m }{ { \alpha }_{ i }=1 }$$

This is the ARCH($$m$$) model, with variance being estimated on the basis of a long-run average variance and $$m$$ observations.

Let:

$$\omega =\gamma { V }_{ L }$$

Equation $$IV$$ can be written as:

$${ \sigma }_{ n }^{ 2 }=\omega +\sum _{ i=1 }^{ m }{ { \alpha }_{ i }{ u }_{ n-i }^{ 2 } }$$

## The Exponentially Weighted Moving Average (EWMA) Model

The model is a special case of the model in equation $$III$$. As we move back in time, we will notice an exponential decrease in the weights, $${ \alpha }_{ i }$$. Let $$\lambda$$ be a constant between 0 and 1, then:

$${ \alpha }_{ i+1 }=\lambda { \alpha }_{ i }$$

With this weighting scheme, we can obtain the following volatility estimates updating formula:

$${ \sigma }_{ n }^{ 2 }=\lambda { \sigma }_{ n-1 }^{ 2 }+\left( 1-\lambda \right) { u }_{ n-i }^{ 2 }\quad \quad \quad \quad \quad equation\quad V$$

$${ \sigma }_{ n }$$ is the day $$n$$ volatility estimate, and is computed from $${ \sigma }_{ n-1 }$$ and $${ u }_{ n-1 }$$. To see how the above equation corresponds to exponentially decreasing weights, let’s substitute for $${ \sigma }_{ n-1 }^{ 2 }$$, such that:

$${ \sigma }_{ n }^{ 2 }=\lambda \left[ { \sigma }_{ n-2 }^{ 2 }+\left( 1-\lambda \right) { u }_{ n-2 }^{ 2 } \right] +\left( 1-\lambda \right) { u }_{ n-1 }^{ 2 }$$

Alternatively:

$${ \sigma }_{ n }^{ 2 }=\left( 1-\lambda \right) \left( { u }_{ n-1 }^{ 2 }+\lambda { u }_{ n-2 }^{ 2 } \right) +{ \lambda }^{ 2 }{ \sigma }_{ n-2 }^{ 2 }$$

A similar substitution for $${ \sigma }_{ n-2 }^{ 2 }$$ yields in:

$${ \sigma }_{ n }^{ 2 }=\left( 1-\lambda \right) \left( { u }_{ n-1 }^{ 2 }+\lambda { u }_{ n-2 }^{ 2 }+{ \lambda }^{ 2 }{ u }_{ n-3 }^{ 2 } \right) +{ \lambda }^{ 3 }{ \sigma }_{ n-3 }^{ 2 }$$

Therefore:

$${ \sigma }_{ n }^{ 2 }=\left( 1-\lambda \right) \sum _{ i=1 }^{ m }{ { \lambda }^{ i-1 }{ u }_{ n-i }^{ 2 }+ } { \lambda }^{ m }{ \sigma }_{ n-m }^{ 2 }$$

When $$m$$ is large, we obtain a sufficiently small $${ \lambda }^{ m }{ \sigma }_{ n-m }^{ 2 }$$ and equation $$V$$ will be similar to equation $$III$$. Then:

$${ \alpha }_{ i }=\left( 1-\lambda \right) { \lambda }^{ i-1 }$$

We notice a decline in the weights for $${ u }_{ i }$$ by a rate of $$\lambda$$ as we move back in time.

## GARCH (1, 1) Model

The analogous difference between equation $$III$$ and equation $$IV$$, is the only difference between the EWMA model and the GARCH (1, 1) model. We will apply the long-run average variance rate, $${ V }_{ L }$$ to compute $${ \sigma }_{ n }^{ 2 }$$, from $${ \sigma }_{ n-1 }$$ and $${ u }_{ n-1 }$$, in this model.

Therefore, the GARC (1, 1) equation is as follows:

$${ \sigma }_{ n }^{ 2 }=\gamma { V }_{ L }+\alpha { u }_{ n-1 }^{ 2 }+\beta { \sigma }_{ n-1 }^{ 2 }$$

$${ V }_{ L }$$ has been assigned a weight of $$\gamma$$, $${ \sigma }_{ n-1 }^{ 2 }$$ has been assigned a weight of $$\alpha$$, and $${ \sigma }_{ n-1 }^{ 2 }$$ assigned a weight of $$\beta$$. The sum of the weights must be one:

$$\gamma +\alpha +\beta =1$$

We notice that the EWMA model is a special case of the GARC (1, 1) model with:

$$\gamma=0$$,

$$\alpha=1-\lambda$$, and

$$\beta=\lambda$$

In this model, (1, 1) is to inform us that the basis of $${ \sigma }_{ n }^{ 2 }$$ is the latest observation of the variance rate and $${ u }^{ 2 }$$. In general, GARCH ($$p, q$$) model is a computation of $${ \sigma }_{ n }^{ 2 }$$ from the latest $$p$$ observations on $${ u }^{ 2 }$$, and the variance rate’s latest $$q$$ estimates.

Recall that:

$$\omega =\gamma { V }_{ L }$$

Therefore, the model can be written as:

$${ \sigma }_{ n }^{ 2 }=\omega +\alpha { u }_{ n-1 }^{ 2 }+\beta { \sigma }_{ n-1 }^{ 2 }\quad \quad \quad equation\quad VI$$

In parameter estimation, we often apply this form of the model. After estimating for $$\omega$$,$$\alpha$$,and $$\beta$$, then:

$$\gamma=1- \alpha – \beta$$

And:

$${ V }_{ L }=\frac { \omega }{ \gamma }$$

For stability of the model, it is necessary that:

$$\alpha + \beta < 1$$

If the condition is not satisfied, we will have a negative weight applied to the long term variance.

## The Weights

Let us substitute for $${ \sigma }_{ n-1 }^{ 2 }$$ in equation $$VI$$, then:

$${ \sigma }_{ n }^{ 2 }=\omega +\alpha { u }_{ n-1 }^{ 2 }+\beta \left( \omega +\alpha { u }_{ n-2 }^{ 2 }+\beta { \sigma }_{ n-2 }^{ 2 }\quad \right)$$

Implying that:

$${ \sigma }_{ n }^{ 2 }=\omega +\beta \omega +\alpha { u }_{ n-1 }^{ 2 }+\alpha \beta { u }_{ n-2 }^{ 2 }+{ \beta }^{ 2 }{ \sigma }_{ n-2 }^{ 2 }$$

Let us now substitute for $${ \sigma }_{ n-2 }^{ 2 }$$, therefore:

$${ \sigma }_{ n }^{ 2 }=\omega +\beta \omega +\alpha { u }_{ n-1 }^{ 2 }+\alpha \beta { u }_{ n-2 }^{ 2 }+\alpha { \beta }^{ 2 }{ u }_{ n-3 }^{ 2 }+{ \beta }^{ 3 }{ \sigma }_{ n-3 }^{ 2 }$$

Therefore, if the procedure is continued, it is observed that $$\alpha { \beta }^{ i-1 }$$ is the weight applied to $${ u }_{ n-i }^{ 2 }$$. Where $$\beta$$ is the rate of decay and is similar to $$\lambda$$ in the EWMA model.

### Maximum Likelihood Methods

We apply the maximum likelihood method to determine the estimation of the above models from historical data. We consider the parameters that maximize the probability of the occurrence of the data.

### Constant Variance Estimation

Supposing that we have been provided with observations $${ u }_{ 1 },{ u }_{ 2 },\dots,{ u }_{ m }$$, and the underlying normal distribution has a mean of zero. We will denote variance as $$v$$. Then, the PDF for $$X$$ with $$X={ u }_{ 1 }$$ is the probability of observing $${ u }_{ i }$$:

$$\frac { 1 }{ \sqrt { 2\pi v } } exp\left( \frac { -{ u }_{ i }^{ 2 } }{ 2v } \right)$$

The following is an expression of the probability of the occurrence of $$m$$ observations in the order they have been observed:

$${ \Pi }_{ i=1 }^{ m }\left[ \frac { 1 }{ \sqrt { 2\pi v } } exp\left( \frac { -{ u }_{ i }^{ 2 } }{ 2v } \right) \right] \quad \quad \quad equation\quad VII$$

The value that can maximize this expression happens to be the best $$v$$ estimate, applying the maximum likelihood method. If we consider the logarithms of the expression in the equation $$VII$$, while ignoring the multiplicative factors, then it will be apparent that our wish is to maximize:

$$\sum _{ i=1 }^{ m }{ \left[ -ln\left( v \right) -\left( \frac { { u }_{ i }^{ 2 } }{ v } \right) \right] }$$

Equivalently:

$$-m\quad ln\left( v \right) -\sum _{ i=1 }^{ m }{ \frac { { u }_{ i }^{ 2 } }{ v } }$$

If the above expression is differentiated with respect to $$v$$ and the result equation sold to zero, then $$v$$ has the following maximum likelihood estimator:

$$\frac { 1 }{ m } \sum _{ i=1 }^{ m }{ { u }_{ i }^{ 2 } }$$

## Estimating EWMA or GARCH (1, 1) Model

In this section, we study the application of the maximum likelihood method in parameter estimation when using the EWMA, GARCH (1, 1) or any other procedure for updating volatility. Suppose that the estimated variance for day $$i$$ is $${ v }_{ i }={ \sigma }_{ i }^{ 2 }$$, and a normal probability distribution of $${ u }_{ i }$$ unconditional on the variance. Therefore, when parameters maximize the following expression then they are considered as the best:

$$\prod _{ j=1 }^{ m }{ \left[ \frac { 1 }{ \sqrt { 2\pi { v }_{ j } } } exp\left( \frac { -{ u }_{ j }^{ 2 } }{ 2{ v }_{ j } } \right) \right] }$$

And with logarithms included, we are maximizing an expression of the form:

$$\sum _{ i=1 }^{ m }{ \left[ -ln\left( { v }_{ i } \right) -\left( \frac { { u }_{ i }^{ 2 } }{{ v }_{ i }}\right) \right] }$$

### How good is the Model?

With a perfectly functioning GARCH (1, 1) model, autocorrelation should be done away with. We consider the variables’ autocorrelation structure $${ { u }_{ i }^{ 2 } }/{ { \sigma }_{ i }^{ 2 } }$$ to evaluate whether the model has removed the autocorrelation. The Ljung-Box statistic is considered to offer a more scientific test.

Supposing that there are $$m$$ observations in a series, then the Ljung-Box statistic is expressed as:

$$m\sum _{ k=1 }^{ K }{ { w }_{ k }{ c }_{ k }^{ 2 } }$$

A lag of $$k$$ has an autocorrelation denoted as $${ c }_{ k }$$. The equation consider $$K$$ lags, and:

$${ w }_{ k }=\frac { m+2 }{ m-k }$$

## Application of the GARCH (1, 1) Model in Forecasting Future Volatility

If we apply the GARCH (1, 1) model, at the closure of day $$n – 1$$ we will have the following estimate of the variance rate for day $$n$$:

$${ \sigma }_{ n }^{ 2 }=\left( 1-\alpha -\beta \right) { V }_{ L }+\alpha { u }_{ n-1 }^{ 2 }+\beta { \sigma }_{ n-1 }^{ 2 }$$

Therefore:

$${ \sigma }_{ n }^{ 2 }-{ V }_{ L }=\alpha \left( { u }_{ n-1 }^{ 2 }-{ V }_{ L } \right) +\beta \left( { \sigma }_{ n-1 }^{ 2 }-{ V }_{ L } \right)$$

On a future day $$n + t$$, we have:

$${ \sigma }_{ n+t }^{ 2 }-{ V }_{ L }=\alpha \left( { u }_{ n+t-1 }^{ 2 }-{ V }_{ L } \right) +\beta \left( { \sigma }_{ n+t-1 }^{ 2 }-{ V }_{ L } \right)$$

$${ \sigma }_{ n+t-1 }^{ 2 }$$ is the expected value of $${ u }_{ n+t-1 }^{ 2 }$$. Therefore:

$$E\left[ { \sigma }_{ n+t }^{ 2 }-{ V }_{ L } \right] =\left( \alpha +\beta \right) E\left[ { \sigma }_{ n+t-1 }^{ 2 }-{ V }_{ L } \right]$$

Continuously repeating this equation results in:

$$E\left[ { \sigma }_{ n+t }^{ 2 }-{ V }_{ L } \right] ={ \left( \alpha +\beta \right) }^{ t }\left( { \sigma }_{ n }^{ 2 }-{ V }_{ L } \right)$$

Implying that:

$$E\left[ { \sigma }_{ n+t }^{ 2 } \right] ={ V }_{ L }+{ \left( \alpha +\beta \right) }^{ t }\left( { \sigma }_{ n }^{ 2 }-{ V }_{ L } \right) \quad \quad \quad equation\quad VIII$$

With this equation, we can apply the information provided at the closure of day $$n – 1$$ to forecast the volatility on day $$n + t$$.

For the EWMA model, $$\alpha + \beta = 1$$. Therefore, according to equation $$VIII$$, the current variance rate and the expected future variance are equal.

## Volatility Term Structures

Let us assume that we are on day $$n$$. then we define:

$$V\left( t \right) =E\left[ { \sigma }_{ n+t }^{ 2 } \right]$$

And:

$$a=ln\frac { 1 }{ \alpha +\beta }$$

Therefore, equation $$VIII$$ changes to:

$$V\left( t \right) ={ V }_{ L }+{ e }^{ -at }\left[ V\left( 0 \right) -{ V }_{ L } \right]$$

The instantaneous variance rate in $$t$$ days is estimated by $$V\left( t \right)$$. Between today and time $$T$$, the average variance per day is expressed as follows:

$$\frac { 1 }{ T } \int _{ 0 }^{ T }{ V\left( t \right) dt } ={ V }_{ L }+\frac { 1-{ e }^{ -at } }{ aT } \left[ V\left( 0 \right) -{ V }_{ L } \right]$$

This will tend to $${ V }_{ L }$$ as $$T$$ increases.

Let $$\sigma \left( T \right)$$ be the volatility per annum applied in the pricing of a $$T$$-day option under GARC (1, 1) model. Recall that we assumed a year has 252 days.

Then:

$$\sigma { \left( T \right) }^{ 2 }=252\left\{ { V }_{ L }+\frac { 1-{ e }^{ -at } }{ aT } \left[ V\left( 0 \right) -{ V }_{ L } \right] \right\}$$

The above equation can also be written as:

$$\sigma { \left( T \right) }^{ 2 }=252\left\{ { V }_{ L }+\frac { 1-{ e }^{ -at } }{ aT } \left( \frac { \sigma { \left( 0 \right) }^{ 2 } }{ 252 } -{ V }_{ L } \right) \right\}$$

A variation in $$\sigma { \left( 0 \right) }$$ by $$\Delta \sigma { \left( 0 \right) }$$ causes $$\sigma { \left( T \right) }$$ to vary by the following factor:

$$\frac { 1-{ e }^{ -at } }{ aT } \frac { \sigma { \left( 0 \right) } }{ \sigma { \left( T \right) } } \Delta \sigma { \left( 0 \right) }$$

## Question 1

Suppose that we know from experience that α = 3 for a particular financial variable and we observe that the probability that v > 10 is 0.04.

Determine the probability that v is greater than 20.

1. 0.0125%
2. 0.5%
3. 4%
4. 0.1%

The correct answer is B.

From the given probability, we can get the value of constant k as follows:

prob(v>x) = kx(-α)

0.04 = k(10)(-3)

k = 40

Thus,

P(v > 20) = 40(20)(-3)  = 0.005 or 0.5%

Note: The power law provides an alternative to assuming normal distributions.

## Question 2

The current estimate of daily volatility is 2%. The closing price of an asset yesterday was $50.00. The closing price of the asset today is$50.50.

Using the EWMA model with a decay factor of 0.94, what is the updated estimate of volatility?

1. 2.125%
2. 0.0425%
3. 1.95%
4. 0.0382%

The correct answer is C.

We know that the updated squared volatility formula is:

$${ \sigma }_{ n }^{ 2 }=\lambda { \sigma }_{ n-1 }^{ 2 }+\left( 1-\lambda \right) { u }_{ n-i }^{ 2 }$$

Remember: Return has to be compounded continuously, that is:

$$Return = ln\frac { { S }_{ i } }{ { S }_{ i-1 } } = ln\frac { 50.5 }{ 50 } = 0.995\%$$

$$Updated \quad squared \quad volatility = { \sigma }_{ n }^{ 2 }=0.94{ 0.02 }^{ 2 }+ 0.06 { (0.009950 }^{ 2 } = 0.000382$$

$$Updated \quad volatility = { \sigma }_{ n } = \sqrt{0.000382} = 0.0195 \quad or \quad 1.95\%$$