22
Jun 20

Solution to Question 2 from UoL exam 2019, zone B

Solution to Question 2 from UoL exam 2019, zone B

Suppose the parameters in a GARCH (1,1) model

\sigma _{t + 1}^2 = \omega + \beta \sigma _t^2 + \alpha \varepsilon _t^2   (1)

are \omega = 0.000004,\ \alpha = 0.06,\ \beta = 0.93, the index t refers to days and {\varepsilon _t} is zero-mean white noise with conditional variance \sigma _t^2.

(a) What are the requirements for this process to be covariance stationary, and are they satisfied here? [20 marks]

If the coefficients satisfy the condition for positivity, \omega>0,\ \alpha,\beta\ge0, then the condition for covariance-stationarity is \alpha + \beta < 1. They are barely satisfied.

(b) What is the long-run average volatility? [20 marks]

We use the facts that {\sigma ^2} = E\sigma _{t + 1}^2 = E\left[ {E(\varepsilon _{t + 1}^2|{F_t})} \right] for all t. Applying the unconditional mean to regression (1) and using the LIE we get

{\sigma ^2}=E\sigma _{t+1}^2=E\left[{\omega+\beta\sigma _t^2+\alpha\varepsilon _t^2}\right]=\omega+\beta{\sigma^2}+\alpha{\sigma^2}

and

{\sigma^2}=\frac{\omega }{{1-\alpha-\beta}}=\frac{{0.000004}}{{1-0.06-0.93}}=0.0004.

(c) If the current volatility is 2.5% per day, what is your estimate of the volatility in 20, 40, and 60 days? [20 marks]

On p.107 of the Guide there is the derivation of the equation

\sigma _{t + h,t}^2 = \sigma _y^2 + {(\alpha + \beta )^{h - 1}}(\sigma _{t + 1,t}^2 - \sigma _y^2),\,\,h \ge 1.    (2)

I gave you a slightly easier derivation in my class, please use that one. If we interpret "current" as t+1 and "in twenty days" as t+1+20, then

\sigma _{t+21}^2=\sigma^2+(\alpha + \beta )^{20}(\sigma _{t+1}^2-\sigma^2)

= 0.0004+\exp\left[ 20\ln(0.06+0.93)\right](0.025-0.0004) = 0.020521.

For h=41,61 use the same formula to get 0.016692, 0.013725, resp. I did it in Excel and don't envy you if you have to do it during an exam.

(d) Suppose that there is an event that decreases the current volatility by 1.5%to 1% per day. Estimate the effect on the volatility in 20, 40, and 60 days. [20 marks]

Calculations are the same, just replace 0.025 by 0.01. Alternatively, one can see that the previous values will go down by \exp[(h-1)\ln(0.06+0.93)]0.015, which results in volatility values 0.012146, 0.009934 and 0.008125.

(e) Explain what volatility should be used to price 20-, 40-and 60-day options, and explain how you would calculate the values. [20 marks]

The only unobservable input to the Black-Scholes option pricing formula is the stock price volatility. In the derivation of the formula the volatility is assumed to be constant. The value of the constant should depend on the forecast horizon. If we, say, forecast 20 days ahead, we should use a constant value for all 20 days. This constant can be obtained as an average of daily forecasts obtained from the GARCH model.

If the GARCH is not used, a simpler approach is applied. If the average daily volatility is {\sigma _d}, then assuming independent returns, over a period of n days volatility is {\sigma _{nd}} = \sqrt n {\sigma _d}.

In practice, traders go back from option prices to volatility. That is, they use observed option prices to solve the Black-Scholes formula for volatility (find the root of an equation with the price given). The resulting value is called implied volatility. If it is plugged back into the Black-Scholes formula, the observed option price will result.

4
Mar 18

Interest rate - the puppetmaster behind option prices

Interest rate - the puppetmaster behind option prices

Call as a function of interest rate

Figure 1. Call as a function of interest rate

The interest rate is the last variable we need to discuss. The dependence of the call price on the interest rate that emerges from the Black-Scholes formula is depicted in Figure 1. The dependence is positive, right? Not so fast. This is the case when common sense should be used instead of mathematical models. One economic factor can influence another through many channels, often leading to contradicting results.

John Hull offers two explanations.

  1. As interest rates in the economy increase, the expected return required by investors from the stock tends to increase. This suggests a positive dependence of the call price on the interest rate.
  2. On the other hand, when the interest rate rises, the present value of any future cash flow received by the long call holder decreases. In particular, this reduces the payoff if at expiration the option is in the money.

The combined impact of these two effects embedded in the Black-Scholes formula is to increase the value of the call options.

However, experience tells the opposite

The Fed changes the interest rate at discrete times, not continually. Two moments matter: when the rumor about the upcoming interest rate change hits the market and when the actual change takes place. The market reaction to the rumor depends on the investors' mood - bullish or bearish. A bullish market tends to shrug off most bad news. In a bearish market, even a slight threat may have drastic consequences. By the time the actual change occurs, it is usually priced in.

In 2017, the Fed raised the rate three times: on March 15 (no reaction, judging by S&P 500 SPDR SPY), June 14 (again no reaction) and December 13 (a slight fall). The huge fall in the beginning of February 2018 was not caused by any actual change. It was an accumulated result of cautiousness ("This market has been bullish for too long!") and fears that the Fed would increase the rates in 2018 by more than had been anticipated. Many investors started selling stocks and buying bonds and other less risky assets. The total value of US bonds is about $20 trillion, while the total market capitalization of US stocks is around $30 trillion. The two markets are comparable in size, which means there is enough room to move from one to another and the total portfolio reshuffling can be considerable. Thus far, the mere expectation that the interest rate will increase has been able to substantially reduce stock prices and, consequently, call prices.

All this I summarized to my students as follows. When interest rates rise, bonds become more attractive. This is a substitution effect: investors switch from one asset to another all the time. Therefore stock prices and call prices fall. Thus the dependence of call prices on interest rates is negative.

The first explanation suggested by Hull neglects the substitution effect. The second explanation is not credible either, for the following reason. As I explained, stock volatility has a very strong influence on options. Options themselves have an even higher volatility. A change in interest rates by a couple percent is nothing in comparison with this volatility. Most investors would not care about the resulting reduction in the present value of future cash flows.

25
Feb 18

Volatility - king among option price determinants

Volatility - king among option price determinants

Here we apply the simple probabilistic model developed previously to see how a call price depends on volatility. Assumption 1 and the method to find ranges of initial stock prices that can lead to ITM areas give Figure 1.

Dependence on volatility

Figure 1. Dependence on volatility

As before, we are interested in the event \{K<S(T)\} that the call with the strike price K is in the money at expiration. In Figure 1, S_1 corresponds to volatility \sigma_1 and S_2 corresponds to volatility \sigma_2, where \sigma_1<\sigma_2S_1 is the highest stock price such that its area of influence (the base of the corresponding rectangle) does not contribute to the ITM area. Hence, only stock prices higher than S_1 do (these prices are indicated by the green arrow). The same goes for S_2, the range of prices that contribute to the ITM area being shown by the blue arrow. We see that the area corresponding to the larger volatility is wider, and therefore the corresponding call is more expensive.

Call as a function of volatility

Figure 2. Call as a function of volatility

In Figure 1, the independent variable (on the horizontal axis) is time. Figure 2 from Mathematica illustrates the same dependence with volatility as the independent variable.

The Black-Scholes formula doesn't tell the full story

In the Black-Scholes formula, the volatility parameter is fixed. In practice, it changes all the time. The dependence on volatility, when the latter is very large, can dominate all other determinants. This is best illustrated with an example.

AVGO stock

Figure 3. AVGO stock

Figure 3 is the chart of AVGO (symbol of Broadcom Ltd) in the period of high volatility 10/30/2017-1/8/2018. In general, it's difficult to pick exactly tops and bottoms. But suppose you were able to do so buying the stock at $248 and selling at $286. Usually stocks are traded in batches of 100. In that case, your profit would be (286-248)*100=3800 and the return would be 3800/24800=15.3%. Let's compare this return with the one you could have trading calls.

AVGO 260 call

Figure 4. AVGO 260 call

In Figure 4 you can see the behavior of 260 call option that expired on January 19, 2018. The period is the same and both charts are hourly charts. One call option commands 100 shares of stock. Again, assuming that you were able to pick the bottom and top exactly, your profit would be (29-10)*100=1900 and the return would be 1900/1000=190%. You would have a much higher return with a much smaller investment!

Note also that in the subsequent period, when the volatility and stock price both fell, the stock price did not fall below the previous low of $248, while the option price broke the previous low of $10 and hit $5 at some point.

Conclusion. The key to trading options is to catch a moment when volatility is low and is about to jump or when it is high and is going to fall.

19
Feb 18

Call options and probabilistic intuition - dependence on strike

Call options and probabilistic intuition - dependence on strike

Awhile ago I gave a course in options following the excellent book by Hull. Frankly, at that time I didn't understand a fraction of what a trader needs to understand about options. I plan to make several posts to explain the intuition behind the Black-Scholes formula for options valuation. The geniality of this formula consists in the fact that it correctly reflects the likelihood of various events. More precisely, the more likely a certain event, the higher the call option price. The level of explanation will be midway between the theoretical view of Hull and practical view of option traders. We start with a series of definitions.

Why do we want options?

A stock price moves up and down, and this movement is pretty erratic. You don’t want to buy the stock, because it’s an expensive investment and the price may fall right after you buy the stock. However, in case the price goes up, you want to gain from that movement, without the financial commitment required by the stock purchase. You are willing to pay a relatively small amount for the guarantee that if the stock goes up, you will be able to buy it at a fixed (presumably, lower) price. At the same time, you don’t want to buy the stock if it goes down. This is the idea behind the call option.

Basic definitions

A call option gives its buyer the right, but not the obligation, to buy the stock at the price fixed in the option contract. This price is called a strike price and denoted K. The contract is valid until a certain date called an expiration date and denoted T. More generally, T may denote any time before expiration and S(T) will then denote the stock price at that time (it is random and depends on time, while the strike price is fixed for the life of the contract and is deterministic).

There are two parties in the transaction: the call buyer owns a long call and the call seller owns a short call. We discuss only long positions; for short positions most of the time the opposite is true.

At any point in time during the life of the option, the strike price K may be higher than the stock price. In this case people say that it is out of the money (OTM). If K=S, the option is at the money (ATM). If K<S, the option is in the money (ITM). The call buyer wants the call to be ITM at expiration.

If at expiration the call option is ITM, the long call holder buys the stock at price K and sells at S(T), profiting S(T)-K-c, where c is the call price paid at the outset. If at expiration the call option is ATM or OTM, the holder doesn't gain from buying the stock and gives up the right to buy it, because buying is not an obligation. The option expires worthless and the call holder’s loss is c.

The market price c of a call option depends on 5 variables: the strike price K, stock price S, time to expiration T (measured in years), volatility of the stock price \sigma and risk-free interest rate r (annualized). Our purpose is to understand the option price dynamics when one of the arguments of the call option changes, while others are kept fixed.

The call price changes during the life of the option, depending on market conditions. The call holder can sell it at any time prior to expiration. The call price at that time depends on the event \{S(T)>K\} that the option is in the money. In all our explanations we will use this event. The higher its likelihood, the higher the call price.

So, how does the call price depend on the strike price?

In the money ranges for two strikes

Figure 1. In the money ranges for two strikes

Consider two strike prices K_1<K_2. Since \{K_2<S(T)\}\subset\{K_1<S(T)\} (see Figure 1), the K_1 call should be more expensive than the K_2 call because the market values higher events with higher probabilities. As the strike increases, the call price falls. This is illustrated in Figure 2.

Call as a function of strike price

Figure 2. Call as a function of strike price

Figure 2 was produced in Mathematica using the Black-Scholes formula, which is valid for European options that can be exercised only at expiration. Our simple probabilistic argument is true also for American options that can be exercised any time prior to expiration. In Figure 2 we assume that the stock price is $50 (the other assumptions are Expiration=0.5, Interest Rate=0.1, Volatility=0.5, Dividend=0.05). The red part of the curve corresponds to in the money strikes, and the out of the money strikes are shown by the blue part of the curve.

Note also that in Mathematica it's better to use the command FinancialDerivative than to program the Black-Scholes formula directly. The latter for some reason sometimes doesn't work.