How Forward Curves Inform Option Pricing and Put–Call Parity

# Relationship between forward and option prices


# Relationship between forward and option prices












Do forward prices factor into option prices at all? It seems to me from Black-Scholes that you just need a spot price and interest rate r.

I understand that $F_t = S_0 e^{r t}$, but I don't know if this means that option prices actually rely on traded futures to derive $r$ and then plug it into Black-Scholes or not..?

Is there a good resource to read on the relationship between options and forwards?

## Answer by Quantuple (score 7, accepted)

https://quant.stackexchange.com/a/35934

At the heart of the (relative) pricing theory is the concept of no arbitrage and replication. I'll focus on equities here because as stated in the comments it may be more complicated for commodities.

Forwards deliver a payout linear in the future value of the underlying asset. Hence they can be replicated statically by a simple cash & carry replication strategy (REM: this is where the argument gets more elaborate for commodities such as electricity or perishables). Using the absence of arbitrage argument yields the famous $$F(0,T)=S_0 e^{f T}$$ where $f$ reflects the cost of funding the equity purchase and carrying it (borrowing cash + having the equity in your balancesheet, placing the stock in repo + reinvesting the divs to decrease the overall cost).

This term structure of funding cost $f(t)$ (i.e. it is clear that $f$ is not constant in practice), also factors in when pricing non-linear products such as options, since their pricing equations also stem from some replication principle - this time dynamic instead of static - where one purchases the relevant quantity of stocks to replicate the target instrument's behaviour. If you assume proportional dividends and no jumps, then the risk-neutral dynamics of an asset writes: $$ dS_t/S_t = f(t) dt + \sigma(t) dW_t $$ or equivalently $$ dS_t/S_t = \frac{\partial \ln\left(F(0,t)/S_0\right)}{\partial t} dt + \sigma(t) dW_t $$ So you see that the forward curve indeed intervenes. You could therefore see the market forward prices as a way of extracting the implied funding cost and hence (partly) marking your model to the market. This is independent of the volatility model used $\sigma(t)$ (it could be BS, Heston, Local volatility à la Dupire). Of course if there is arbitrage opportunities or the above argument fails to apply, everything breaks down.

Mathematically, this is also understandable as follows: the forward curve characterises the first moment of $S_t$ under the risk-neutral measure knowing the current information, so obviously it'll intervene somewhere when pricing since it's part of the characterisation of the pdf of $S_t$. If you assume BS, then adding the volatility surface to the forward curve gives you the full representation of the future distributions of $S_t \vert S_0$ under the risk-neutral measure. This is enough for European options but not for some more exotic options (e.g. forward starts) that depend on the conditional densities $S_{t_2} \vert S_{t_1}$).

Regarding your question:

> But do futures prices on the market have any effect on option prices in the model. It seems to me that S=Fe^(r-d)T is just a theoretical equation by no-arbitrage. Do futures actually follow this equation?

You could actually say the same about European options (quoted using the BS model for instance). You should see it the other way around: these are inputs that allow you to calibrate your model so that you could apply relative valuation principles afterwards, that is, some sort of ketchup economics: knowing that basic building blocks are priced at X in the market, I know some combination of these blocks should be worth Y in my model.

## Answer by vonjd (score 3)

https://quant.stackexchange.com/a/35907

In developed equity markets you have at least those three entities: Stocks, futures and options.

As far as my experience goes you indeed often hedge options with futures.

What is more interesting is the relationship concerning stocks and futures: In theory stocks are the underlying and futures are the derivatives. In practice it is, interestingly enough, often the other way around: Futures are the more liquid instruments and get traded more often, stocks follow suit due to arbitrage conditions.

So, all in all the instruments that are often the most important in modern financial markets - for options and stocks - are indeed futures!

NB: There seem to be regional differences, e.g. in Europe single-stock futures play a much bigger role than in the US, e.g. EUREX trades futures on most of the EURO STOXX® 600 Index members. Yet the above is true for index futures in any case.

## Answer by Pavel (score 2)

https://quant.stackexchange.com/a/36150

There is absolutely an important relationship between forward contracts and options that doesn't rely on BS (or any model). Although noob2 already mentioned put-call parity being written in terms of forward price, I wanted to shed more light on this. The following derivation applies to any asset that doesn't pay dividends, the ability to take long and short positions on forward contracts, and the ability to long and short European puts and calls. I will use the notation $\alpha^+ = max ( 0, \alpha)$.

Suppose we are at present time $t=0$ and there are European put $P$ and European call $C$ options on an asset $S$ with expirations $T > 0$ and strikes $K > 0$. Also suppose that the asset $S$ has a forward price for delivery at $T$ of $F$. Consider the strategy at $t = 0$:

- Take the long position on a forward contract on $S$ for delivery at $T$

- Long one put at price $P_0$

- Short one call at price $C_0$

It follows that the terminal capital of this strategy is: $$X_T = P_T - C_T + S_T - F = (K - S_T)^+ - (S_T - K)^+ + S_T - F$$ Then simplifying the difference in the two option payouts yields: $$X_T = K - F$$ Investing $d(T)(K - F)$ at $t=0$ would result in the same terminal capital, where $d(T)$ is the discount factor from $T$ to $0$. Assuming no arbitrage, at $t=0$ our two strategies have the same initial capital and it follows that: $$P_0 - C_0 = d(T)(K - F)$$ It is important to note that we have made no assumptions about the behavior of the underlying.