# Option Pricing Class Problem (part 2)

This is the 2nd blog posting about Option Pricing Class Problem that I passed out in class on Thursday, November 7. Here’s the (replicating portfolio) solution to part A, which requires the calculation of the price of a one-year European put option on Ripple, Inc. stock, where = $56, =$60, r = 4%, $\delta t$ = 1, u = 1.3, and d = .9:

According to the Replicating Portfolio Approach for pricing this put option, $\Delta = \displaystyle\frac{{{P_u} - {P_d}}}{{uS - dS}} = \displaystyle\frac{{0-9.60}}{{72.80 - 50.40}} = -.4286\;$ and $B = \displaystyle\frac{{u{P_d} - d{P_u}}}{{{e^{r\delta t}}(u - d)}} = \frac{{1.3(9.6) - .9(0)}}{{1.0408(.4)}} = 29.98$. Then ${V_{RP}} = C = \Delta S + B = -.4286(56) + 29.98 = \5.98.$

As we discussed during yesterday’s Finance 4335 class meeting, if you already know the arbitrage-free price of a call option, then the arbitrage-free price of an otherwise identical (European, same underlying (non-dividend paying) asset, same exercise price, and same time to expiration) put option can also be determined by applying the put-call parity equation: $C + K{e^{ - r\delta t}} = P + S \Rightarrow P = C + K{e^{ - r\delta t}} - S = \4.33 + \ 60{e^{ - .04}} - \ 56 = \5.98.$

As shown above, the put-call parity equation implies that one can create a “synthetic” put option by purchasing an otherwise identical call worth $C and bond worth $K{e^{ - r\delta t}}$, while also shorting one unit of the underlying asset (which generates proceeds worth$S). Now suppose that $P \ne \ 5.98;$ specifically, suppose that actual put is worth more (less) than the synthetic put.  If this were to happen, then one could earn riskless arbitrage profit by selling the actual (synthetic) put and buying the synthetic put (actual put).  Thus, we have reconfirmed that \$5.98 is indeed the arbitrage-free price for the put.

Similarly, “synthetic” versions of the call, the bond, and the underlying asset can be created using the following combinations of the other instruments:

1. Synthetic Call: $C = P + S - K{e^{ - r\delta t}} = \ 5.98 + \ 56 - \ 60{e^{ - .04}} = \ 4.33;$
2. Synthetic Bond: $K{e^{ - r\delta t}} = P + S - C = \ 5.98 + \ 56 - 4.33 = \ 60{e^{ - .04}} = \ 57.65;$ and
3. Synthetic Underlying Asset: $S = C + K{e^{ - r\delta t}} - P = \ 4.33 + \ 57.65 - \ 5.98 = \ 56.$

During tomorrow’s Finance 4335 class meeting, we will 1) discuss the Delta Hedging and Risk Neutral Valuation approaches to pricing calls and put (see pp. 25-34 of the Derivatives Theory, part 1 lecture note),  2) move on to a discussion of multi-period option pricing formulas (which are presented in the Derivatives Theory, part 2 lecture note), and 3) complete Parts C and D of the Option Pricing Class Problem.

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