5
$\begingroup$

I've spent a while looking for an answer to this question and while I feel it is a simple question I have not found an answer.

I know prices of option contracts follow an implied, risk-neutral distribution which is observable and equities follow an unobservable, physical distribution of returns. Now, do futures contracts follow a risk-neutral or physical distribution? Or is my thinking flawed at some point?

I'm still learning so I would greatly appreciate anyone providing me some direction with this.

$\endgroup$
0
$\begingroup$

Perhaps other memeber of qSE are going to correct me, but I think the following rule of thumb is useful. Whenever you have a doubt, try to forget that a pricing measure is a probability measure. This is just a pricing tool: originally for any option/derivative/contingent claim we'd like to know its price, so we introduce a map $\pi:X\to \Bbb R$ such that $\pi(x)$ is the current price of the contingent claim $x$. For example, $x$ can be a call option with maturity of 1 year and ATM strike, or $x$ can be the futures contract expiring in 10 days. Now, it happens that $\pi$ is a linear functional on $X$, and $\pi(1) = 1$ - that is the value of the assets that will pay us $1$ under any circumstance is $1$ (let's assume discount rates are $0$). From that we see that $\pi$ is similar to an expectation operator, so we can define a corresponding probability measure - which we call a pricing (risk-neutral) measure. There are perhaps some deeper thought underlying such a coincidence, but for all philosophical questions: use pricing measure to find the price, to find the Greeks etc. For anything else use the physical measure.

Example: let's say we want to buy an option which we know we can't hedge perfectly, and estimate whether we can afford vacation on Hawaii after expiry. We do the following:

  1. Price the option (use pricing measure)
  2. Compute Greeks to hedge (use pricing measure)
  3. Run Monte-Carlo to estimate our losses/profits from imperfect hedge under the imperfect hedging strategy computed in step 2. (use real measure)
$\endgroup$
3
$\begingroup$

Your question is not clear. What you might want to say is what distribution should the futures price follow, under the risk-neutral or physical probability measure. In this sense, it will depend on your intention. For potential future exposure, you may want to use the physical measure for the price evolution, while the distribution will depend on your model assumption -- could be normal or log-normal. However, for valuation, risk-neutral probability measure is assumed. Moreover, the futures price process is a martingale under the risk-neutral probability measure, and is usually assumed to be log-normal.

$\endgroup$
  • $\begingroup$ Thanks! That actually answers my question for the most part. The way I wrote it was confusing. What I was actually trying to ask was whether or not there is a risk neutral or physical measure used for the valuation of futures contracts. Also I was wondering whether or not the risk neutral probability distribution is implied by the prices across the different expirations if the risk neutral measure is indeed used to value futures contracts. Could you please elaborate on the second part? I'm sorry I didn't clearly state it in the original question, but it would help me if you answered that also. $\endgroup$ – Joe Yurkanin Jul 2 '15 at 20:07
  • $\begingroup$ It may help to know that I'm doing a small project to see if I can use the implied information of the risk neutral probability distribution make trading decisions. Of course that is contingent upon there being an implied distribution to use. I know that I could use the options chain to do so but I was going to compare the different outcomes. $\endgroup$ – Joe Yurkanin Jul 2 '15 at 20:08
  • $\begingroup$ @JoeYurkanin: It is hard to have a short description. I would recommend the book "Commodities and Commodity Derivatives" by Geman. In Chapter 3 of this book, a few plausible models are proposed. $\endgroup$ – Gordon Jul 2 '15 at 20:16
  • $\begingroup$ I think that will help me a great deal and I'll learn a lot along the way. Thanks a lot! $\endgroup$ – Joe Yurkanin Jul 2 '15 at 20:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.