Presuming a Market Maker must delta-hedge, how can it still earn money?

Question

I know that "There are other ways to hedge price risk other than buying the underlying. It's not uncommon but it doesn't "always" happen". "You can buy/sell other options to reduce price risk, or be long a different underlying, or have a natural long/short position (e.g. refiners are naturally short crude oil and naturally long products like gasoline)".

None the less, assume that you bought 1 call and in your case, the MM delta-hedges by buying the underlying asset (stock in this case).

Case 1. $S_0$ (current stock price) $> K$ (strike price) $+ c$ (call premium). Presume you sell your option before expiration.

Then you profit and the MM loses money. To unwind the delta hedge, now the MM must sell its underlying stock at a higher price. It's unclear if the MM profited. I don't know if the MM's profit from selling the stock > MM's loss from buying your call, or vice versa.

Case 2. $S_0 ≤ K + c$. Presume you let your call expire worthless.

Then you lost your call premium that the MM earned. To unwind the delta hedge, now the MM must sell its underlying stock at a lower price. Again it's unclear if the MM profited. I don't know if MM's profit from selling you the call > MM's loss on selling the stock, or vice versa.

In both cases, the MM can lose money! Correct? If so, why would MM ever delta-hedge?

Answer 1

Here's a bit of an iterative answer. If market makers couldn't make money then why would any of them go to all of the effort and cost to become a Designated Primary Market Maker?

Not all market making involves delta neutral hedging. There are arbitrages available where the MM has zero risk (conversions, reversals, dividend arbitrage, box spreads, etc.), other than maybe pin risk

For example, if options are overpriced relative to the underlying stock, the MM buys the underlying stock and offsets it with an equivalent synthetic short stock (long put + short call) position. The trader who is long the call and the trader short the put bear the risk. The MM has a riskless profit.

Answer 2

Option MMing is trying to earn the spread, i.e. an option MM will price the option (at P) and then work bid at P-a and offer at P+a, if he gets filled he theoretically earned a. This is of course massively simplified, but let's continue, now you have a position in options, you have 2 choices, close it with options, or hedge it with the underlying, since the bid-ask spread in options is 2*a wide, closing it with options will cost you your earnings and you just lost commissions, so if your model is right and your hedging expertise are good, you can unwind the option at the price P by hedging it all the way till expiration and earn your a.

potentially you can get filled at bid and offer and earn 2*a, that is why in markets with 2 way, high frequency flow, you will see very tight spreads, as MMs are happy with a very low a. and in market with one directional flow or low liquidity, you will see MMs quoting wide spreads (high a)

There is a lot more to market making modeling, but that's the base

Answer 3

First off, note that the hedging is not a one-off affair that takes place along with the option sale. Instead, the hedges are adjusted (to match changes in delta) as the underlying price moves around (i.e. as the underlying experiences realized volatility $\sigma_{r}$).

The options will have been sold for some particular price that can be converted to implied an "implied volatility" $\sigma_{impl}$, which is directly comparable to realized volatility. Of course, at the time of sale, the future realized volatility is not yet known!

Within the Black Scholes model, if our market-maker hedges delta (on those calls that they sold) about as well as the B-S model assumes, then the market-maker will make a small profit when time marches forward and realized volatility is less than the implied volatility ($\sigma_{r} < \sigma_{impl}$), and lose a bit of money if $\sigma_{r} > \sigma_{impl}$.

(I deliberately ignore the bid-offer spread on the options here, which is also a small source of profit).

Now, the Black-Scholes model assumes infinitely divisible prices, infinitesimally small hedge increments, etc etc. So it is not perfect. But the way the classical B-S proof works, one finds that the option price for a particular volatility emerges as the limit when hedge increments go to zero. So it sort of "proves" the above claim about when market makers profit.

If you want to see for yourself, it is very easy to code up a Monte Carlo simulation of the Black-Scholes SDE, along with a simulation of market-maker behavior, and prove the profit/loss proposition numerically.

Answer 4

Nice question. In a sense any market participant is a market maker (MM). This is because what's relevant is the execution (either buying or selling) at a specific price - which leads to price discovery. This is in the spirit of the old adage "a house is worth as much as someone is willing to pay for it". Now, if you're specifically referring to market making as showing a two way price then the above answers have addressed this already. I would just add that two-way prices shown in this manner are also a function of factors such as the current position of the quoter's book, their own views, risk appetite...etc - hence the "skewing" of quotes to bid or offer to reflect these factors. So while there may be such a concept as a "fair price", bid/offers (and hence mids) can have material variability between market makers. In a nutshell, market making is a lot more shades of grey than black and white.