###### What Multiple Regression is and How It ...

Example: Multiple Regression in Investment World James Chase, an investment analyst, wants to... **Read More**

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

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

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

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.

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

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:

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

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.

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.

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

:closestto

- $9.57.
- $11.94.
- $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.*