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Risk-Neutral Probability in a One-Step Binomial Tree

Article Quant Q&A · Author: Charlie P
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Summary

The document asks how to interpret risk-neutral probability in a one-step binomial model. It gives an up factor of 1.2, a down factor of 0.9, and a zero risk-free rate, then applies the standard formula that makes the discounted expected asset value match its current value. With those inputs, the calculated probability is about 0.33.

The question contrasts that value with a proposed probability of 0.5. In this model, the risk-neutral probability is determined by the up and down factors and the risk-free rate; 0.5 is a risk-neutral probability only if it also satisfies the no-arbitrage pricing condition for those inputs. The source is a question rather than a worked answer, so it does not discuss real-world probabilities, the derivation, or broader assumptions such as the model’s one-step structure.

Key ideas

  • The risk-neutral up probability depends on the risk-free growth factor and the model’s up and down factors.
  • For the inputs given, the stated formula produces a probability near 0.33.
  • A candidate probability of 0.5 must be checked against the same pricing condition.
  • The document poses the question but does not provide a detailed derivation or answer.

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# Risk-Neutral Probability in a Binomial Tree


# Risk-Neutral Probability in a Binomial Tree












This question is probably very simple and I'm just missing the easy solution but I'm a bit confused so I thought I might as well try ask here.

I've been given this question:

When I tried to calculate $p$ I got $p = 0.33$ (Let me know if this is incorrect), I obtained this using : $(e^{r_f} - d)/(u-d)$ so I had $(e^0-0.9)/(1.2-0.9)=0.33$

What I don't understand is the next part of the question, when it asks "is $p=0.5$ a risk-neutral probability?". This is confusing me because I've already calculated $p$ to be something else, since my $p$ is different does that mean that $p=0.5$ is not a risk-neutral probability and $p=0.33$ is the risk-neutral probability, right? Or have I got this all wrong?

Any help would be really appreciated just to clarify things for me.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.