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Risk-Neutral Probability Must Fit the Asset’s Up and Down Payoffs

Article Quant Q&A · Author: user2521987
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Summary

The note explains why a probability calculated from a one-period up/down asset model can exceed one. The displayed equation uses the risk-free rate, rather than the stock’s expected return, to determine the risk-neutral probability when pricing under the risk-neutral measure. Substituting the stated 1% rate gives a value within the probability range.

It also checks the inputs by equating the expected terminal stock value to the current value grown at the relevant rate, including the dividend yield. The resulting value of 1.25 signals that the stated assumptions do not support a valid probability for the specified up and down factors. The expected terminal value must lie between the two possible terminal values. This is a basic consistency check for a binomial model; it does not resolve which input in the original exercise is wrong.

Key ideas

  • Risk-neutral probabilities are derived using the risk-free rate, not the stock’s physical expected return.
  • A valid two-outcome probability must fall between zero and one.
  • The expected terminal asset value must lie between the model’s down and up outcomes.
  • An out-of-range probability signals inconsistent inputs or a misapplied formula.

Tags

Derivatives pricingStatisticsOptions
Full text
# Is there an error in this problem on pricing an asset using the true probability of an up move?


# Is there an error in this problem on pricing an asset using the true probability of an up move?












I'm self-studying for an actuarial exam and I encountered the following problem:

The true probability of an up move, $p$, must satisfy: $$p = \frac{e^{{(\alpha - \delta})h} - d}{u - d},$$

where $\alpha$ is the continuously compounded annual expected return of the stock and $\delta$ is the continuous dividend rate.

But with $\alpha = 0.10$, $\delta = 0.03$, $u = 1.04$, and $d = 0.91$, we have $$p = \frac{e^{{(0.10 - 0.03})1} - d}{1.04 - 0.91} = 1.25,$$ which doesn't seem possible since I thought $0 \leq p \leq 1$.

How should I interpret this value of $p$ in terms of a probability, since $p > 1$?

## Answer by Neeraj (score 3)

https://quant.stackexchange.com/a/23216

Your formula for $p$ is $$p = \frac{e^{{(\alpha - \delta})h} - d}{u - d},$$ where $\alpha$ is not expected return on stock but continuous risk free rate, i.e. 1%.

If you use $\alpha$ as 1%, you will get $p=0.009125828 $ which is within $[0,1]$

EDIT: With the information given in the question, it must satisfy following equality: $$S_0e^{\alpha - \delta}=uS_0p+dS_0(1-p)$$ where $S_0$ is initial price. Solving above equation provide $p$=1.25. This indicate that information provided in the question is incorrect. Either, your expected price after 1 year($S_0e^{\alpha - \delta}$) must be within $[S_0u,S_0d]$, so require change in $\alpha$ or $u$ must be sufficiently large such that $uS_0 > S_0e^{\alpha - \delta}$.

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