Components of the BSM Model

The BSM model for pricing options on a non-dividend-paying stock is given by:

European Call

$$ c_0= S_0N(d_1) – e^{(-rT)}KN(d_2) $$

European Put

$$ p_0=e^{-rT}KN\left({-d}_2\right)-S_0N\left({-d}_1\right) $$

Where:

$$ \begin{align*} d_1 &=\frac{ln{\left(\frac{S}{K}\right)}+\left(r+\frac{1}{2}\sigma^2\right)T}{\sigma\sqrt T} \\ d_2 &=d_1-\sigma\sqrt T \end{align*} $$

\(N(x)\) = Standard normal cumulative distribution function.

\(N(–x) = 1 – N(x)\)

BSM Model has the following variables:

\(T\) = Time to option expiration.

\(r\) = Continuously compounded risk-free rate.

\(S_0\) = Current share price.

\(K\) = Exercise price.

\(\sigma\) = Annual volatility of asset returns.

Interpretation of the BSM Model

The BSM model can be interpreted as the present value of the expected option payoff at expiration. It can be expressed as:

$$ \begin{align*} c_0 &=PV(S_0e^{rT}N\left(d_1\right)– KN(d_2)) \\ p_0 &=PV(KN\left(-d_2\right)-S_0e^{rT}N\left(-d_1\right)) \end{align*} $$

Where the present value factor, in this case, is \(e^{-rT}\).

Alternatively, the BSM model can be described as having two components, a stock component, and a bond component.

The stock component for call options is \(S_0N(d_1)\) while the bond component is \(e^{–rT}KN(d_2)\). Therefore, the BSM model call value is the difference between the stock component and the bond component.

The stock component is \((S_0N(d_1))\) and the bond component is \(e^{rT} KN(-d_2)\) for put options. Therefore, the BSM model put value is the bond component minus the stock component.

An option can be thought of as a dynamically managed portfolio of the underlying stock and zero-coupon bonds. The initial cost of this replicating strategy is given as:

$$ \text{Replicating strategy cost} = n_SS + n_BB $$

Call Options

The equivalent number of underlying shares is \(n_S=N\left(d_1\right) > 0\). \(n_S\) greater than 0 implies that we are buying the stock. On the other hand, the equivalent number of bonds is \(n_B=-N\left(d_2\right) < 0\). \(n_B\) less than 0 implies that we are selling the bond. Note that selling a bond is the same as borrowing money. Therefore, a call option can be viewed as a leveraged position in the stock where \(N(d_1)\) units of shares are purchased using \(e^{–rT}KN(d_2)\) of borrowed money.

Put Options

The equivalent number of underlying shares is \(n_S=-N\left(-d_1\right) < 0.\) This can be interpreted as selling the shares of the underlying stock as \(n_S < 0\).

Further, the equivalent number of bonds is \(n_B=N\left(-d_2\right) > 0\). The bond is being bought here since \(n_B\) is greater than 0. Buying a bond is similar to lending money. Therefore, a put can be viewed as buying a bond where this purchase is partially financed by short selling the underlying stock.

Example: Interpreting BSM Model Components

Consider the following information relating to call and put options on an underlying stock

\(S_0\) = 48

\(K\) = 40

\(r\) = 2.5% (Continuously compounded)

\(T\) = 2

\(\sigma\) = 30%

The current market price of call option = 14

The current market price of put option = 3

The following values have been calculated using the above information:

$$ PV\left(K\right)={40\times e}^{-0.025\times2} = 38.05 $$

\(d_1\) = 0.7597

\(d_2\) = 0.3354

\(N\left(d_1\right)\) = 0.7763

\(N\left(d_2\right)\) = 0.6314

We can determine the replicating strategy cost and arbitrage profits on both options as follows:

According to the no-arbitrage approach to replicating the call option, a trader can purchase \(n_S = N(d_1) = 0.7763\) shares of stock by borrowing \(n_B = –N(d_2)= -0.6314\) shares of zero-coupon bonds priced at \(B = Ke^{–rT} = $38.05\) per bond.

$$ \begin{align*} \text{Replicating strategy cost} & = n_SS + n_BB \\ \text{Replicating strategy cost} & =0.7763\times48+\left(-0.6314\times\ 38.05\right)=$13.24 \end{align*} $$

An arbitrage profit can be realized on the call option by writing a call at the current market price of $14 and purchasing a replicating portfolio for $13.24.

Therefore,

$$ \text{Arbitrage profit} = $14-$13.24 = $0.76 $$

For the put option, we have:

$$ \begin{align*} N(–d_1) & = 1 – N(d_1)= 1 – 0.7763=0.2237 \\ N(–d_2)& = 1 – 0.6314= 0.3686 \end{align*} $$

The no-arbitrage approach to replicating the put option involves:

Purchasing \(n_B =N(-d_2) =0.3686\) shares of zero-coupon bonds priced at \(40e^{-0.025\times 2} =$38.05\) per bond and short-selling \(n_S=–N(-d_1)= –0.2237\) shares of stock resulting in short proceeds of \($48\times0.2237=$10.74\).

Therefore, the replicating strategy cost for the put option is:

$$ \text{Replicating strategy cost} =-0.2237\times$48 + 0.3686\times$38.05=$3.29 $$

A trader can exploit arbitrage profits by selling the replicating portfolio and purchasing puts for an arbitrage profit of $0.29 per put.

$$ \text{Arbitrage profit} = $3.29-$3 = $0.29 $$

Question

Common stock is currently trading at $50. A call option is written on it with an exercise price of $45. Further, the continuously compounded risk-free rate of interest is 4%, the interest rate volatility is 30%, and the time to the option expiry is 2 Years. Using the BSM model, the following components have been calculated:

\(PV(K)\) = $41.54

\(d_1\) = 0.6490

\(d_2\) = 0.2248

\(N\left(d_1\right)\) = 0.7418

\(N\left(d_2\right)\) = 0.5889

\(c_0\) = 12.63

The value of the replicating portfolio is closest to:

1. $9.57.
2. $11.94.
3. $12.63.

Solution

The correct answer is C.

The no-arbitrage approach to replicating the call option involves purchasing \(n_S = N\left(d_1\right) = 0.7418\) shares of stock partially financed with \(n_B = –N(d_2) = –0.5889\) shares of zero-coupon bonds priced at \(B = Ke^{–rT} = $41.54\) per bond.

$$ \begin{align*} \text{Cost of replicating portfolio} & = n_SS + n_BB \\ c &= 0.7418 \left(50\right)+ (–0.5889)41.54 =$12.63 \end{align*} $$

Reading 34: Valuation of Contingent Claims

LOS 34 (g) Interpret the components of the Black–Scholes–Merton model as applied to call options in terms of a leveraged position in the underlying.