CEO in Charge of FTX Restructuring Calls Case an ‘Unprecedented’ Mess

“Crypto exchange exhibited a ‘complete failure of corporate controls,’ John Ray, who led Enron in bankruptcy, says in first detailed filing”; this article is the featured Page 1 article in today’s issue of the Wall Street Journal, @ https://www.wsj.com/articles/ceo-overseeing-ftx-restructuring-calls-it-an-unprecedented-mess-11668707836?st=hr94usq7zfpbwz3&reflink=desktopwebshare_permalink.

Problem Set 8 solutions erratum

If you have previously downloaded what you assumed were solutions for Problem Set 8  (between 4:47 pm on Thursday 11:20 am this morning), toss that document and replace it with the corrected version which now appears in its place at http://fin4335.garven.com/fall2022/ps8solutions.pdf.  (Technical note: If you have previously downloaded this document you might have to first clear your browser cache to get the corrected version of the solutions; in most cases, this can also be accomplished by simple reloading the http://fin4335.garven.com/fall2022/ps8solutions.pdf page a few times).

Cox-Ross-Rubinstein (CRR) option pricing spreadsheet

Here’s a spreadsheet that calculates call and put prices using the CRR framework for n = 1, 2, 3, and 4 timesteps, based on the numerical example shown on pp. 3-8 of the http://fin4335.garven.com/fall2022/lecture16.pdf lecture note.

The key idea here involves determining option payoffs without having to recreate the entire stock binomial tree which is shown on page 4 of the above-referenced lecture note.  The way this is done involves determining for each call option the minimum number of up moves required in order to determine the nodes at which the option will be in-the-money.  By rounding up to the nearest integer “a” the value of \ln (K/S{d^n})/\ln (u/d) for n = 1, 2, 3, and 4 (see rows 11 and 12 below), we determine that the 1 timestep call will only be in the money at the u node, the 2 timestep call will only be in the money at the uu node, the 3 timestep call will only be in the money at the uuu and uud nodes, and the 4 timestep call will only be in the money at the uuuu and uuud nodes.  This greatly simplifies the calculations for call prices because we know that the call payoffs will all be zero at all other nodes (see row 13 below).  Having determined the arbitrage-free call prices, then we find the arbitrage-free put prices by applying the put-call parity equation P = C + K{e^{ - rn\delta t}} - S (you can bring up the spreadsheet by clicking on the screenshot below):

Problem Set 9 helpful hints

Here are some helpful hints for Problem Set 9:

Problem 1:

  • Part A: Apply the Cox-Ross-Rubinstein model (as described on pp. 9-11 of the Part 3 option pricing lecture note and pp. 11-13 of the Teaching the Economics and Convergence of the Binomial and Black-Scholes Option Pricing Formulas assigned reading) to determine the call option price.  Since the call option described here is initially out-of-the-money (i.e., since S < K), there will be terminal nodes at which the call expires in-the money and others at which it expires out-of-the-money.  By solving the b = \ln (K/S{d^n})/\ln (u/d) equation and rounding to the nearest integer greater than b (referred to in the above-referenced sources as the parameter “a“), this indicates the minimum number of up moves such that the call option expires inthemoney, thereby simplifying your calculation of the price for the call option.
  • Part B: Apply the put-call parity equation to solve for the put option price.

Problem 2:

  • Scenario A requires solving for call and put option prices using the Black-Scholes-Merton option pricing formulas.  See the Part 3 option pricing lecture note, page 15 for a numerical example of how to do this.
  • Scenario B requires finding the current price of the underlying asset, where the call, put, and exercise prices are all given.  Solve the put-call parity equation (C + K{e^{ - rT}} = P + S) for S.
  • Scenario C requires finding the exercise price, where the call, put, and underlying asset prices are all given.  Solve the put-call parity equation for K.
  • Scenario D requires finding \sigma for a call option worth $2.38 and a put option worth $3.60.  Feel free to use the Black-Scholes spreadsheet from the course website, or better yet, create your own Excel spreadsheet in which you solve for the call and/or the put by varying \sigma (this can be accomplished either via trial and error, or better yet, by using either Solver or Goal Seek).  An important lesson you’ll learn from this part of problem 2 is that call and put option prices are positively related to \sigma .

Put-Call Parity; also known as the “Fundamental Theorem of Financial Engineering”

As we discussed during today’s Finance 4335 class meeting, once you know the arbitrage-free price of either a call or a put 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 or call option can be determined by applying the put-call parity equation:

C + K{e^{ - rT}} = P + S.

Consider the Bitcoin, Inc. numerical example covered in the second problem of the Option Pricing Class Problem.  In that example, Bitcoin, Inc.’s current stock price is S=$56; each year, the stock price can change by a factor of d = 0.9 or u =1.3, and the annual continuously compounded riskless interest rate is 4%. Since the put option’s arbitrage-free price P = $5.98 (as shown in the solution for Problem A.1. in the Option Pricing Class Problem Solutions, it follows from the put-call parity equation that the call option’s arbitrage-free price is C = $4.33, as shown in the calculation below:

C = P + S{\rm{\;}} \Rightarrow C = P + S - K{e^{ - r\delta t}} = \$ 5.98 + 56 - \$ 60{e^{ - .04}} = \$ 4.33.

Thus, by algebraically rearranging the put-call parity equation, one can determine arbitrage-free prices for “synthetic” calls, puts, bonds, and stocks.  In the case of the “synthetic” put option, this is accomplished 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 the 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.