I recently spoke to an options trader that tried to demonstrate option pricing by considering a random walk of balls dropping down a lattice so the underlying stochastic process is a simple random walk of say 100 steps.

The contract considered is $(U_{100}-K)^{+}$ where $U_{100}$ is the number of times the ball goes "up". He states that this is an option. I think he doesn't understand what an option is because there is no underlying market in this case (ie you can't exactly trade the balls to hedge your position and there is no underlying that moves based on what the ball does except for the contract itself). I would say that this is a bet or a game that you would pay for at a casino.

So my question is: Is there a resource that actively explains or demonstrates why a derivative is called a derivative? As in why insurance and bets are fundamentally different from derivatives?

  • $\begingroup$ If markets aren't complete, is there still a difference between a bet and a derivative in your case then? As far as I'm concerned, a derivative is an instrument whose price is derived from another instrument. Whether that instrument is tradeable and whether a replication argumetn can be used to determine a fair value for a derivative is a separate thing for me $\endgroup$
    – Bram
    Commented Jul 3, 2017 at 6:08
  • $\begingroup$ Completeness is stronger than arbitrage free but you only need arbitrage free and some liquidity for a no arbitrage price to exist. I would agree with what you say if there is no liquidity in the market (but is it really a market in this case?). If you want to go into the realm of BSDEs and dynamic programming then super hedging does address this. $\endgroup$
    – RNvsRW
    Commented Jul 16, 2017 at 23:19
  • $\begingroup$ A derivative/bet/insurable contract/etc. can be defined as a function $f(\cdot)$ on a random variable $X$ at a future time $T$: $f(T,X)$. Whether the risk factor $X$ can be (reasonably easily) traded or not will determine how to approach the problem of pricing the payout $f(T,X)$, either by statistical arguments or by no-arbitrage arguments. $\endgroup$ Commented Nov 22, 2017 at 14:54

1 Answer 1


From the introduction (Chapter 1) of Baxter's and Rennie's excellent book Financial Calculus:

With markets where the stock can be bought and sold freely and arbitrarily positive and negative amounts of stock can be maintained without cost, trying to trade forward using the strong law would lead to disaster […].


But the existence of an arbitrage price, however surprising, overrides the strong law. To put it simply, if there is an arbitrage price, any other price is too dangerous to quote.


The strong law and expectation give the wrong price for forwards. But in a certain sense, the forward is a special case. The construction strategy $-$ buying the stock and holding it $-$ certainly wouldn’t work for more complex claims. The standard call option which offers the buyer the right but not the obligation to receive the stock for some strike price agreed in advance certainly couldn’t be constructed this way. If the stock price ends up above the strike, then the buyer would exercise the option and ask to receive the stock – having it salted away in a drawer would then be useful to the seller. But if the stock price ends up below the strike, the buyer will abandon the option and any stock owned by the seller would have incurred a pointless loss.

Thus maybe a strong-law price would be appropriate for a call option, and until 1973, many people would have agreed. Almost everything appeared safe to price via expectation and the strong law, and only forwards and close relations seemed to have an arbitrage price. Since 1973, however, and the infamous Black-Scholes paper, just how wrong this is has slowly come out. Nowhere in this book will we use the strong law again. […] All derivatives can be built from the underlying $-$ arbitrage lurks everywhere.

  • $\begingroup$ You are welcome @RNvsRW. The whole Introduction of their book discusses the statistical vs. the arbitrage approach to pricing if you want more details. $\endgroup$ Commented Jun 20, 2017 at 10:46

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