The risk neutral probability measure

[Matthew H]

Please could you help me understand why we need the risk neutral probability measure and why we need to mention no arbitrage.

I'm thinking of a one-step binomial model to price a derivative. The set up for this is:
- The derivative pays Cu if the share price goes up, and Cd if the share price goes down.
- We set up a replicating portfolio - we set the value of the replicating portfolio equal to Cu if the share price goes up, and equal to Cd if the share price goes down.
- That gives us two simultaneous equations we can solve to find the values of phi (units of the share) and psi (units of cash).
- Since the value of the portfolio replicates the derivative payoff, the value of the derivative at time 0 (i.e. V0) must equal the value of the portfolio at time 0.
- That is, V0 = phi*S0 + psi
- Plugging in phi and psi we get:
- V0 = exp(-r) * [q*Cu + (1-q)*Cd]
- where q = [exp(r) - d] / [u-d]

Here's where I get confused:
- We have just established the value of the derivative without intentionally creating a new probability measure, without explicitly assuming that people are risk neutral, without explicitly mentioning no-arbitrage and without explicitly trying to ensure that q lies between 0 and 1.
- Why do we even need to acknowledge that (if we assume no arbitrage) q is between 0 and 1?
- We've found the price of the derivative (which was presumably the goal) so why don't we just stop there? Why do we need to pick up of the fact that q lies between 0 and 1 (as do probabilities) and that it only does so if we assume no-arbitrage and why do we need to formalise a risk neutral probability measure?

- Maybe it's because if we formalise the risk neutral probabilities then we can use those q's and (1-q)'s elsewhere? But I don't see why we'd think we can use them elsewhere given that (to me) they just seem to be the coincidental numbers that fall out of the maths (in the above specific scenario) - the q doesn't actually represent the probability of an increase, and (1-q) doesn't actually represent the probability of a fall, they're just number relating to a specific scenario, so why would we think they can be used elsewhere?


Also, another question I thought of when writing this -> when we set up the replicating portfolio, why did we think to choose a share and cash as the constituents of the portfolio? (I can't think of a better suggestion but how do we know that just those two assets would be able to replicate everything?)

 

[Matthew H]

A related follow up to my previous message:

Is this the way to prove that Eq[S1 | F0] = S0*exp(r)?
- Eq[S1 | F0] = q*S0*u + (1-q)*S0*d, where q = [exp(r) - d] / [u-d]
- Eq[S1 | F0] = [exp(r) - d]/[u-d] *S0*u + [u - exp(r)]/[u-d] *S0*d
- Eq[S1 | F0] = S0*exp(r) * [(u-d)/(u-d)]
- Eq[S1 | F0] = S0*exp(r)

If the above is correct, I don't really understand why because in the previous message we noted that q and (1-q) aren't really probabilities (they're just coincidentally between 0 and 1) and they seemed specific to that scenario, so why would we think so set up Eq[S1 | F0] using those q's and (1-q)'s?

I think, in general, i'm struggling to see how we go from deriving the answer to the specific scenario considered in my previous message, to assuming that part of answer (the q's and the structure of the formula) apply more generally?

 

[Steve Hales]

Here's where I get confused:
- We have just established the value of the derivative without intentionally creating a new probability measure, without explicitly assuming that people are risk neutral, without explicitly mentioning no-arbitrage and without explicitly trying to ensure that q lies between 0 and 1.
- Why do we even need to acknowledge that (if we assume no arbitrage) q is between 0 and 1?
- We've found the price of the derivative (which was presumably the goal) so why don't we just stop there? Why do we need to pick up of the fact that q lies between 0 and 1 (as do probabilities) and that it only does so if we assume no-arbitrage and why do we need to formalise a risk neutral probability measure?

Hi.
Thanks for your questions, they get right to the heart of risk-neutral valuations!

You say, "Since the value of the portfolio replicates the derivative payoff, the value of the derivative at time 0 (i.e. V0) must equal the value of the portfolio at time 0." This is only true because there is no arbitrage. If arbitrage were possible, then two assets with the same cashflows in the future might have different values now. So this is where no arbitrage needs to be explicitly stated.

There is only one way of pricing derivates; you need to build a replicating portfolio. However, that can be really cumbersome, so there's a computational shortcut we can take by making up a quantity called q. The value of q is carefully chosen so that the expression exp(-r*T)*Eq[XT|F0] matches the replicating portfolio approach. Finding q is an exercise in algebraic manipulation. I'm not even going to call q a probability at this point, all I'm doing is using q in the expectation notation.

The binomial tree you've mentioned has up and down multiplicate factors of u and d. So that arbitrage is not possible by trading the share on its own we require d < exp(r) < u. This demands that 0 < q < 1. Now that q looks as though it sits on the unit interval we can pretend that it's a probability, but that's only for convenience.

Now that we've got a fictional probability q, let's look at what happens if it's applied to finding the expected future value of the simplest of all derivatives; the share price itself. Under q, the share price is expected to grow like the risk-free rate - so let's call q the risk-neutral probability of an up-step because investors aren't expecting any additional return for accepting risk.

Hope that starts to help.

 

[Steve Hales]

If the above is correct, I don't really understand why because in the previous message we noted that q and (1-q) aren't really probabilities (they're just coincidentally between 0 and 1) and they seemed specific to that scenario, so why would we think so set up Eq[S1 | F0] using those q's and (1-q)'s?

Take a look at the previous post, and come back with follow-up questions.

Thanks

Steve