After completing this reading you should be able to:

- Explain the lognormal property of stock prices, the distribution of rates of return, and the calculation of expected return.
- Compute the realized return and historical volatility of a stock.
- Describe the assumptions underlying the Black-Scholes-Merton option pricing model.
- Compute the value of a European option using the Black-Scholes-Merton model on a non-dividend-paying stock.
- Define implied volatilities and describe how to compute implied volatilities from market prices of options using the Black-Scholes-Merton model.
- Explain how dividends affect the decision to exercise early for American call and put options.
- Compute the value of a European option using the Black-Scholes-Merton model on a dividend-paying stock.
- Describe warrants, calculate the value of a warrant and calculate the dilution cost of the warrant to existing shareholders.

**Introduction**

Suppose we have a random variable X. This variable will have a lognormal distribution if its natural log (ln X) is normally distributed. In other words, when the natural logarithm of a random variable is normally distributed, then the variable itself will have a lognormal distribution.

The two most important characteristics of the lognormal distribution are as follows:

- It has a lower bound of zero, i.e., a lognormal variable cannot take on negative values
- The distribution is skewed to the right, i.e., it has a long right tail.

These characteristics are in direct contrast to those of the normal distribution which is symmetrical (zero skew) and can take on both negative and positive values. As a result, the normal distribution cannot be used to model stock prices because stock prices cannot fall below zero. The lognormal distribution is also used to value options.

## The Lognormal Property of Stock Prices

A crucial part of the BSM model is that it assumes stock prices are lognormally distributed. Precisely,

$$ ln{ S }_{ T }\sim N\left( ln{ S }_{ 0 }+\left( \mu -\frac { { \sigma }^{ 2 } }{ 2 } \right) T,\sigma^{ 2 } T \right) $$

Where:

\({ S }_{ T }\)=stock price at time \(T\)

\({ S }_{ 0 }\)=stock price at time 0

\(\mu\)=expected return on stock per year

\(\sigma\)=annual volatility of the stock price

Note: The above relationship holds because mathematically, if the natural logarithm of a random variable \(x\left( ln\quad x \right)\) is normally distributed, then \(x\) has a lognormal distribution. It’s also imperative to note that the BSM model assumes stock prices are lognormally distributed, with **stock returns being normally distributed**. Specifically, continuously compounded annual returns are normally distributed with:

a mean of \(\left[ \mu -\frac { { \sigma }^{ 2 } }{ 2 } \right] \) and a variance of \(\frac { { \sigma }^{ 2 } }{ T } \)

### Example:

\(ABC\) stock has an initial price of $60, an expected annual return of 10%, and an annual volatility of 15%. Calculate the mean and the standard deviation of the distribution of the stock price in six months.

$$ ln{ S }_{ T }\sim N\left( ln{ S }_{ 0 }+\left( \mu -\frac { { \sigma }^{ 2 } }{ 2 } \right) T,\sigma \sqrt { T } \right) =N\left[ ln60+\left( 0.10-\frac { { 0.15 }^{ 2 } }{ 2 } \right) 0.5,{ 0.15 }^{ 2 }\times 0.5 \right] $$

$$ ln{ S }_{ T }\sim N\left[ 4.139, 0.011 \right] $$

Sometimes the examiner may want to test your understanding of the lognormal concept by involving confidence intervals. Since \(ln{ S }_{ T }\) is lognormally distuributed, 95% of values will fall within 1.96 standard deviations of the mean. Similarly, 99% of the values will fall within 2.58 standard deviations of the mean. For example, to obtain the 99% confidence interval for stock prices using the above data, we will proceed as follows:

$$ ln{ S }_{ T }\sim N\left[ 4.139, 0.011 \right] $$

$$ ln{ S }_{ T }=\mu \pm { z }_{ \alpha }\times \sigma $$

*(In this case, we have *σ = √0.011 = 0.1049)

$$ 4.139-{ z }_{ \alpha }\times \sigma <ln{ S }_{ T }<4.139+{ z }_{ \alpha }\times \sigma $$

$$ { e }^{ 4.139-{ z }_{ \alpha }\times \sigma }<{ S }_{ T }<{ e }^{ 4.139+{ z }_{ \alpha }\times \sigma } $$

$$ { e }^{ 4.139-{ 2.58 }\times 0.1049 }<{ S }_{ T }<{ e }^{ 4.139+{ 2.58 }\times 0.1049 } $$

$$ 47.86 <{ S }_{ T }< 82.24 $$

$$ \left[ { e }^{ \wedge }\left( lnx \right) =x \right] $$

### Expected Stock Price

Using the properties of a **lognormal distribution***, *we can show that the expected value of \({ S }_{ T }\), \(E(T)\), is:

$$ E\left( { S }_{ T } \right) ={ S }_{ 0 }{ e }^{ \mu T } $$

\(\mu\)=expected rate of return

**Example**

A stock is currently priced at $40 with an expected annul return of 15%. What is the expected value of the stock in six months?

**Solution**

$$ Expected \quad stock \quad price = $40𝑒^{0.15×0.5}=$43.11 $$

### Realized Portfolio Return

$$ { R }_{ pr }={ \left[ { r }_{ 1 }\times { r }_{ 2 }\times { r }_{ 3 }\times \dots \times { r }_{ n } \right] }^{ \frac { 1 }{ n } }-1 $$

\({ r }_{ i }\)=portfolio return at time \(i\)

The **continuously compounded** return realized over a period of time of length T is given by:

$$ \frac { 1 }{ T } ln( \frac { S_T }{ S_0 } ) $$

**Example**

The realized return of a stock initially priced at $50 growing, with volatility,

to $87 over five periods would simply be:

$$ Realized \quad return = \frac { 1 }{ 5 } ln($87/$50) = 11.08\% $$

### Estimating Historical Volatility

Historical volatility can be computed from daily price data of a stock. We simply need to compute continuously compounded returns per day and then determine the standard deviation.

The continuously compounded return for day *i* is calculated as:

$$ ln( \frac { S_i }{ S_{i-1} } ) $$

The volatility of short time periods can be scaled to give the volatility of longer time periods.

For example,

$$ annual\quad volatility=daily\quad volatility\times \sqrt { \left( no.\quad of\quad trading\quad days\quad in\quad a\quad year \right) } $$

*Note that this formula is useful throughout the whole FRM part 1 and FRM part 2 exams in estimating volatility.*

Conversely,

$$ daily\quad volatility=\frac { annual\quad volatility }{ \sqrt { no.\quad of\quad trading\quad days\quad in\quad a\quad year } } $$

**Black-Scholes-Merton Model**

The Black-Scholes-Merton model is used to price European options and is undoubtedly the single most important tool for the analysis of derivatives. It is a product of Fischer Black, Myron Scholes, and Robert Merton.

The model takes into account the fact that the investor has the option of investing in an asset earning the risk-free interest rate. The overriding argument is that the option price is purely a function of the volatility of the stock’s price (option premium increases as volatility increases).

### Assumptions Underlying the Black-Scholes-Merton Option Pricing Model

- There is no arbitrage
- The price of the underlying asset follows a lognormal distribution
- The continuous risk-free rate of interest is constant and known with certainty
- The volatility of the underlying asset is constant and known
- The underlying asset has no cash flow, such as dividends, or interest payments
- Markets are frictionless – no transaction costs, taxes, or restrictions on short sales
- Options can only be exercised at maturity, i.e. they are European-style. The model cannot be used to accurately value American options.

### Determining the Value of Zero-dividend European Options Using the Black-Scholes-Merton Model

The value of a call option is given by:

$$ { C }_{ 0 }={ S }_{ 0 }\times N( { d }_{ 1 }) – K{ e }^{ -rT }\times N( { d }_{ 2 } ) $$

The value of a put option is given by:

$$ { P }_{ 0 }=K{ e }^{ -rT }\times N( { -d }_{ 2 } ) – { S }_{ 0 }\times N( { -d }_{ 1 }) $$.

Where

$$ { d }_{ 1 }=\frac { ln\frac { { S }_{ 0 } }{ K } +\left[ { r }+\left( { \sigma }^{ 2 }/2 \right) \right] T }{ \sigma \sqrt { T } } $$

$$ { d }_{ 2 }={ d }_{ 1 }-\left( \sigma \sqrt { T } \right) $$

\(T\)=time to maturity,assuming 365 days per year

\({ S }_{ 0 }\)=asset price

\(K\)=exercise price

\({ R }_{ f }^{ c }\)=continuously compounded riskfree rate

\(\sigma\)=volatility of continuously compounded returns on the stock

\(N\left( { d }_{ i } \right)\)=cumulative distribution function for a standardized normal distribution variable

#### Example of valuing a call option using the Black-Scholes-Merton model

Assume S_{0} = $100, *K* = $90, *T* = 6 months, *r* = 10%, and σ = 25%.

Calculate the value of a call option.

$$ { d }_{ 1 }=\frac { ln\frac { { S }_{ 0 } }{ K } +\left[ { r }+\left( { \sigma }^{ 2 }/2 \right) \right] T }{ \sigma \sqrt { T } } $$

$$ =\frac { ln\frac { 100 }{ 90 } +\left[ { 0.10 }+\left( { 0.25 }^{ 2 }/2 \right) \right] 0.5 }{0.25 \sqrt { 0.5 } } = (0.1053 + 0.0656)/0.1768 = 0.9672 $$

$$ { d }_{ 2 }={ d }_{ 1 }-\left( \sigma \sqrt { T } \right) $$

$$ ={ 0.9672 }-\left( 0.25 \sqrt { 0.5 } \right) = 0.7904 $$

From a standard normal probability table, look up *N*(0.97) = 0.8333 and *N*(0.79) = 0.7852.

$$ { C }_{ 0 }={ S }_{ 0 }\times N( { d }_{ 1 }) – K{ e }^{ -rT }\times N( { d }_{ 2 } ) $$

$$ ={ 100 }\times N( { 0.8340 }) – 90{ e }^{ -0.10\times0.50 }\times N( { 0.7852 } ) = $16.17 $$

Note that the *intrinsic* *value* of the option is $10—our answer must be at least that amount.

**Exam tips:**

**Tip 1:** Given one of either the put value or the call value, you can use the put-call parity to find the other. Precisely,

$$ { C }_{ 0 }=P_{ 0 }+{ S }_{ 0 }-\left( Ke^{ -{ R }_{ f }^{ c }T } \right) $$

$$ P_{ 0 }={ C }_{ 0 }-{ S }_{ 0 }+\left( Ke^{ -{ R }_{ f }^{ c }T } \right) $$

**Tip 2:** \(N\left(-{ d }_{ i } \right)=1-N\left( { d }_{ i } \right)\)

**Tip 3:** As \({ S }_{ 0 }\) becomes very large, calls (puts) are extremely in-the-money (out-of-the-money)

**Tip 4:** As \({ S }_{ 0 }\) becomes very small, calls (puts) are extremely out-of-the-money (in-the-money)

**Tip 5:** Although \(N\left( { d }_{ 1 } \right)\) and \(N\left( { d }_{ 2 } \right)\) can easily be identified from statistical tables, sometimes you’ll be asked to compute \({ d }_{ 1 }\) and \({ d }_{ 2 }\) without the formulas.

## The value of a European Option Using the Black-Scholes-Merton Model on a Dividend-paying Stock

Assume that we have a known dividend \(d\) distributed a time \({ T }_{ 1 }\),\({ T }_{ 1 }<T\), where \(T\) is the maturity date. To value calls and puts when there are such dividends, we modify the BSM model by replacing \({ S }_{ 0 }\) with \(S\), where:

$$ S={ S }_{ 0 }-D $$

\(D\) is the sum of the \(PV\) (discounted at \({ R }_{ f }^{ c }\)) of the dividend payments during the life of the option.

For example, with dividends \({ D }_{ 1 }\) and \({ D }_{ 2 }\) at times \(\Delta { t }_{ 1 }\) and \(\Delta { t }_{ 2 }\) ,

$$ S={ S }_{ 0 }-{ D }_{ 1 }{ e }^{ -\left( { R }_{ f }^{ c } \right) \frac { \Delta t_{ 1 } }{ m } }-{ D }_{ 2 }{ e }^{ -\left( { R }_{ f }^{ c } \right) \frac { \Delta t_{ 2 } }{ m } } $$

\(\Delta t_{ i }\) represent the amount of time until the ex-dividend date

\(m\) is a division factor to bring the \(\Delta t\) to a full year. e.g. \(\Delta t=2\) months, \(m=12\) months,so \(\frac { \Delta t }{ m } =\frac { 2 }{ 12 } =0.1667 \) years

After this, everything else in the computational formulas remains the same, i.e.,

The value of a call option is given by:

$$ { C }_{ 0 }=\left[ { S }_{ 0 }\times N\left( { d }_{ 1 } \right) \right] -|K\times { e }^{ { -R }_{ f }^{ c }\times T }\times N\left( { d }_{ 2 } \right) | $$

The value of a put option is given by:

$$ { P }_{ 0 }=\left[ K{ e }^{ { -R }_{ f }^{ c }\times T }\times \left( 1-N\left( { d }_{ 2 } \right) \right) \right] -\left[ S \times \left( 1-N\left( { d }_{ 1 } \right) \right) \right] $$

$$ { d }_{ 1 }=\frac { ln\frac { { S } }{ K } +\left[ { R }_{ f }^{ c }+\left( 0.5\times { \sigma }^{ 2 } \right) \right] T }{ \sigma \sqrt { T } } $$

$$ { d }_{ 2 }={ d }_{ 1 }-\left( \sigma \sqrt { T } \right) $$

\(S\) is simply \({ S }_{ 0 }\) adjusted to include dividends payable.

The underlying argument here is that on the ex-dividend dates, the stock prices are expected to reduce by the amounts of the dividend payments.

**Exam tip:** Sometimes the examiner will give you, not a dollar amount “d” of the dividend, but a dividend yield \(q\). For example, you may be told that the dividend yield is 2%, continuously compounded. In such a case, you’re still expected to replace \({ S }_{ 0 }\) with \(S\), where:

$$ S={ e }^{ -qT }\times { S }_{ 0 } $$

## How Dividends Affect the Early Decision for American Calls and Puts

Call option holders have the right but not the obligation to buy shares as per the terms of the contract, but they do not hold shares. As such, they cannot benefit from the rights of shareholders such as the right to receive dividends – as long as the call options have not been exercised.

When the underlying stock pays dividends, a call option holder will not receive it unless they exercise the contract before the dividend is paid. Whoever owns the stock as of the ex-dividend date receives the cash dividend, so an investor who owns in-the-money call options may exercise early to capture the dividend. In summary, a call option should only be exercised early to take advantage of dividends if:

- The option is in-the-money
- The time value of the option needs to be less than the value of the dividend.

It wouldn’t make sense to exercise an out-of-the-money call option and pay an above-market price just to receive a dividend.

Suitable conditions for early exercise of a put option include:

- The option must be deep in-the-money
- High-interest rates
- Sufficiently low volatility

Provided these conditions have been met, the holder of an American put option can exercise early, but only after the dividend has been paid. Clearly, it would make a whole lot more sense to exercise the put option the day after the dividend is paid in order to collect the dividend, instead of exercising the day before and missing out.

## Black’s Approximation in Calculating the Value of an American Call Option On a Dividend-paying Stock

Black’s approximation sets the value of an American call option as the **maximum of two European prices:**

- A European call with the same maturity as the American call being valued, but with the stock price
**reduced**by the present value of the dividend. This implies that \({ S }_{ 0 }\) is reduced by the present value of the dividends payable, but all other variables remain the same. For example, if we anticipate two dividends,$$ S={ S }_{ 0 }-PV $$Where

$$ PV={ D }_{ 1 }{ e }^{ -\left( r \right) \frac { \Delta { t }_{ 1 } }{ m } }+{ D }_{ 2 }{ e }^{ -\left( r \right) \frac { \Delta { t }_{ 2 } }{ m } } $$

\({ D }_{ 1,2 }\) are the dividends on the ex-dividend dates

\(r\) is the risk-free rate

\(\Delta { t }_{ i }\) represent the amount of time until the ex-dividend date

\(m\) is a division factor to bring the \(\Delta { t }\) to a full year. If \(\Delta { t }=2\) months,\(m=12\) months,so \(\frac { \Delta t }{ m } =\frac { 2 }{ 12 } =0.1667\) years

Note: All other variables (\({ d }_{ 1 },{ d }_{ 2 },{ C }_{ 0 },K\),etc) remain the same.

- A European option maturing
**just before**the final ex-dividend date of the A-option. This implies that time to maturity is trimmed down to just before the final dividend is paid. The \(PV\) of dividends other than the final one must be deducted from \({ S }_{ 0 }\).

The largest of the two values (\(I\)) and (\(II\)) above is the desired Black’s approximation for the American call.

**Exam tips:**

- An American call on a non-dividend-paying stock should never be exercised early
- An American call on a dividend-paying stock should only be exercised immediately prior to an ex-dividend date

## Complications Involving the Valuation of Warrants

Warrants are securities issued by a company, which give their owners the right to purchase shares in the company at a specific price at a future date. They are much like options, the only difference being that while options are traded on an exchange, warrants are issued by a company directly to investors in bonds, rights issues, preference shares, and other securities. They are basically used as **sweeteners** to make offers more attractive.

When warrants are exercised, they have the potential to rattle and create some turbulence in the market price of the share. This is because the amount paid for each share is, in all likelihood, less than the share’s market price in case of **call warrants**, and greater than the market price in case of **put warrants**. When substantial call warrants are exercised, the value of equity per share will fall (dilution).

For detachable warrants, their value can be estimated as the difference between the market price of bonds with the warrants and the market price of the bonds without the warrants.

The Black-Scholes-Merton Model can also be used to value warrants using the BSM call/put option formulas, i.e.

$$ { C }_{ 0 }=\left[ { S }_{ 0 }\times N\left( { d }_{ 1 } \right) \right] -|K\times { e }^{ { -R }_{ f }^{ c }\times T }\times N\left( { d }_{ 2 } \right) | $$

However, the following adjustments must be made:

- The stock price \({ S }_{ 0 }\) is replaced by an “adjusted” stock price. Suppose a company has N outstanding shares worth \({ S }_{ 0 }\). This means that the value of the company’s equity is \(N { S }_{ 0 }\). Further, assume that the company has decided to issue M number of warrants with each warrant giving the holder the right to buy one share for K. If the stock prices changes to \({ S }_{ T }\) at time T, the (adjusted) stock price which accounts for the dilution effect of the issued warrants, is: \({ S }_{ adjusted } =\frac { N{ S }_{ 0 } +MK}{N+M}\)
- The volatility input is calculated on equity (volatility of the value of the shares plus the warrants, not just the shares).
- A multiplier (haircut) that captures dilution, given by N/(N+M).

## Implied Volatility

Volatility of the stock price is the only unobservable parameter in the BSM pricing formula. The implied volatility of an option is the volatility for which the BSM option price equals the market price.

Implied volatility represents the expected volatility of a stock over the life of the option. It is influenced by market expectations of the share price as well as by supply and demand of the underlying options. As expectations rise, and the demand for options increases, the implied volatility increases. The opposite is true.

If we use the observable parameters in the BSM formula \(\left( { S }_{ 0 },K,r\quad and\quad T \right) \) and set the BSM formula equal to the market price, then it’s possible to solve for volatility that satisfies the equation. However, there is no closed-form solution for the volatility, and the only way to find it is through iteration.

## Questions

### Question 1

\(ABC\) stock is currently trading at $70 per share. A dividend of $1 is expected after three months and another one of $1 after six months. An American option on \(ABC\) stock has a strike price of $65 and 8 months to maturity. Given that the risk-free rate is 10% and the volatility is 32%, compute the price of the option:

- $9.85
- $12.5
- $10
- $10.94

The correct answer is **D**.

The current price of the share must be adjusted to take into account the expected dividends.

The present value of the dividends is

$$ { e }^{ -0.25\times 0.1 }+{ e }^{ -0.50\times 0.1 }=1.9265 $$

$$ { S }_{ 0 }=70-1.9265=68.0735 $$

Next, calculate the variables required,

\({ S }_{ 0 }=68.0735\)

\(K=65\)

\(\sigma=0.32\)

\(r=0.1\)

\(T=0.6667\)

$$ { d }_{ 1 }=\frac { ln\left( { 68.0735 }/{ 65 } \right) +\left( { 0.1+{ 0.32 }^{ 2 } }/{ 2 } \right) 0.6667 }{ 0.32\sqrt { 0.6667 } } =0.5626 $$

$$ { d }_{ 2 }={ d }_{ 1 }-0.32\sqrt { 0.6667 } =0.3013 $$

\(N\left( d_{ 1 } \right) =0.7131\)

\(N\left( d_{ 2 } \right) =0.6184\)

The call price is

$$ 68.0735\times 0.7131-65{ e }^{ -0.1\times 0.6667 }\times 0.6184=10.94 $$

### Question 2

A stock price is currently $100. Assume that the expected return from the stock is 35% per annum and its volatility is 20% per annum. Calculate the mean and standard deviation of the distribution, and determine the 95% confidence interval for the stock price.

$$

\begin{array}{|c|l|l|l|}

\hline

{} & Mean & Standard \quad deviation & 95\% \quad CI \\ \hline

A. & 150 & 30 & 110<S_{ T }<330 \\ \hline

B. & 201.38 & 58.12 & 112.30<S_{ T }<336.57 \\ \hline

C. & 5.27 & 0.28 & 4.7212<S_{ T }<5.8188 \\ \hline

D. & 0.35 & 0.2 & 112.30<S_{ T }<336.57 \\ \hline

\end{array}

$$

The correct answer is **B**.

In this case,

\({ S }_{ 0 }=100\),

\(\mu =0.35\),and,

\(\sigma=0.20\)

The probability distribution of the stock price in two years is lognormal, and is given by:

$$ ln{ S }_{ T }\sim \left\{ ln100+\left( 0.35-\frac { { 0.2 }^{ 2 } }{ 2 } \right) 2,{ 0.2 }^{ 2 }\times 2 \right\} \sim \left( 5.27,{ 0.28 }^{ 2 } \right) $$

The mean stock price=\({ S }_{ 0 }{ e }^{ \mu T }=100{ e }^{ 0.35\times 2 }=201.38\)

The standard deviation of the stock price=\(100{ e }^{ 0.35\times 2 }\sqrt { { e }^{ { 0.2 }^{ 2 }\times 2 }-1 } =58.12\)

The 95% confidence interval for \(lnS_{ T }\) is given by

$$ 5.27-1.96\times 0.28<lnS_{ T }<5.27+1.96\times 0.28 $$

$$ 4.7212<lnS_{ T }<5.8188 $$

$$ { e }^{ 4.7212 }<S_{ T }<{ e }^{ 5.8188 } $$

$$ 112.30<S_{ T }<336.57 $$