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I am reading Options, Futures, and other derivatives by John C. Hull. In the chapter on Binomial trees, he remarks:

A risk-neutral world has two features that simplify the pricing of derivatives:

  1. The expected return on a stock (or any other investment) is the risk-free rate.
  2. The discount rate used for the expected payoff on an option (or any other instrument) is the risk-free rate.

I understood the first point. The second point seems easier and almost obvious, though when I tried to come up with a written justification for it, I was at a lack of words. I am hoping someone could justify the second point for me. Thank you!

Edit: On further thought, it seems that in a world with only two possible investments - a risky stock and a riskless bond - it is the riskless bond that will represent the time value of money, hence we should use that as the discount rate.

But this raises another question: Based on the above argument, the fact that we use the risk-free rate for discounting is not a consequence of the risk neutral world, but this is not what Hull suggests.

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The first statement is kind of clear. If all investors are risk-neutral, they simply do not care about risk and do not pay more or less regardless how risky an asset is. As a consequence, the return of all assets is the risk-free rate.

Regarding the second statement. What does risk-neutral pricing really do? We change the probabilities from the real world, $\mathbb{P}$, to risk-adjusted ones, $\mathbb{Q}$ which capture all this risk. In a way, we take the risk out of the equation. However, as we only need ``real'' discount factors for stochastic, risky cashflows, we can simply discount with the risk-free rate since the risk-adjustment are done in $\mathbb{Q}$. So whilst there is a risk premium in the real worl and hence the need to use an appriopriate discount factor, there is no risk premium in the risk-neutral world - since no one cares about risk in this world. Thus, we may simply discount with the risk-free rate.

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    $\begingroup$ Nice! That makes quite a lot of sense. I tend to ask the silliest questions, just to be 100% sure I am on the right track. You have answered several of my questions in a revealing and insightful way. Many thanks for that! $\endgroup$ – Dhruv Gupta Jul 23 at 13:24

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