# problem solving in financial mathematics

## a companion blog for a blog on option pricing models ## Basic practice problem set 1 – one-period binomial pricing model

These practice problems reinforces the concept of binomial option pricing model discussed in the following post

found in this companion blog. The practice problems are basic and straightforward problem to price an option using a binomial tree with arbitrary up factor $u$ and down factor $d$.

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Practice Problems

Practice Problem 1
Suppose that the future prices for a stock are modeled with a one-period binomial tree with $u=1.2$ and $d=0.7$ and having a period of 6 months. The current price of the stock is $80. The stock pays no dividends. The annual risk-free interest rate is $r=$ 4%. • Determine the replicating portfolio and the price of a European 70-strike call option on this stock that will expire in 6 months. Practice Problem 2 The stock is as in Practice Problem 1. Determine the replicating portfolio and the price of a European 85-strike put option on this stock that will expire in 6 months. Practice Problem 3 • Suppose that the price of the call option in Practice Problem 1 is observed to be$16.00. Describe the arbitrage.

Practice Problem 4

• Suppose that the price of the call option in Practice Problem 1 is observed to be $16.90. Describe the arbitrage. Practice Problem 5 • Suppose that the price of the put option in Practice Problem 2 is observed to be$9.80. Describe the arbitrage.

Practice Problem 6

• Suppose that the price of the put option in Practice Problem 2 is observed to be $10.60. Describe the arbitrage. $\text{ }$ _____________________________________________________________________________________ $\text{ }$ $\text{ }$ $\text{ }$ $\text{ }$ $\text{ }$ $\text{ }$ $\text{ }$ $\text{ }$ $\text{ }$ $\text{ }$ $\text{ }$ $\text{ }$ _____________________________________________________________________________________ Answers 1. Replicating portfolio: $\Delta=$ 0.65 shares (long), $B=$ -$35.67923171 (borrowing). $\text{ }$
Call option price = $\Delta S + B$ = $16.32076829 $\text{ }$ 2. Replicating portfolio: $\Delta=$ -0.725 shares (short), $B=$$68.221827665 (lending). $\text{ }$
Put option price = $\Delta S + B$ = $10.22182766 $\text{ }$ 3. Buy low (the option at$16.00) and sell the synthetic option at the theoretical price of $16.32076829. The following table shows the cash flows at expiration. $\text{ }$ $\left[\begin{array}{llll} \text{Expiration Cash Flows} & \text{ } & \text{Share Price = } \ 56 & \text{Share Price = } \ 96 \\ \text{ } & \text{ } \\ \text{Sell synthetic call} & \text{ } & \text{ } & \text{ } \\ \ \ \ \ \text{Short 0.65 shares} & \text{ } & - \ 36.4 & - \ 62.4 \\ \ \ \ \ \text{Lend } \ 35.67923 & \text{ } & + \ 36.4 & + \ 36.4 \\ \text{ } & \text{ } \\ \text{Buy call } & \text{ } & \ \ \ 0 & \ \ \ 26 \\ \text{ } & \text{ } \\ \text{Total payoff} & \text{ } & \text{ } \ \ 0 & \ \ \ 0 \end{array}\right]$ $\text{ }$ The above table shows that the buy low sell high strategy produces no loss at expiration of the option regardless of the share prices at the end of the option period. But the payoff at time 0 is certain:$16.32076829 – $16.00 =$0.32076829.

4. Buy low (the synthetic call option at $16.32076829) and sell high (the call option at the observed price of$16.90). The following table shows the cash flows at expiration. $\text{ }$ $\left[\begin{array}{llll} \text{Expiration Cash Flows} & \text{ } & \text{Share Price = } \ 56 & \text{Share Price = } \ 96 \\ \text{ } & \text{ } \\ \text{Buy synthetic call} & \text{ } & \text{ } & \text{ } \\ \ \ \ \ \text{Long 0.65 shares} & \text{ } & + \ 36.4 & + \ 62.4 \\ \ \ \ \ \text{Borrow } \ 35.67923 & \text{ } & - \ 36.4 & - \ 36.4 \\ \text{ } & \text{ } \\ \text{Buy call } & \text{ } & \ \ \ 0 & - \ 26 \\ \text{ } & \text{ } \\ \text{Total payoff} & \text{ } & \text{ } \ \ 0 & \ \ \ 0 \end{array}\right]$ $\text{ }$

The above table shows that the buy low sell high strategy produces no loss at expiration of the option regardless of the share prices at the end of the option period. But the payoff at time 0 is certain: $16.90 –$16.32076829 = \$0.57923171.

_____________________________________________________________________________________ $\copyright \ 2015 \text{ by Dan Ma}$

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