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

Question

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 1

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 $[0,1]$

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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 $uS_0 > S_0e^{\alpha - \delta}$.