3
$\begingroup$

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

enter image description here

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 $p > 1$?

$\endgroup$
3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Using the risk-free rate of return for $\alpha$ gives the risk-neutral probability measure $p^*$. The formula that I'm asking about in the original post is for the true probability measure $p$. How do I interpret that the true probability is greater than 1? $\endgroup$ – user2521987 Feb 11 '16 at 14:30
  • 1
    $\begingroup$ @user2521987 option price does not depend on true probability measure. There are sufficient post on Quant. SE that explain why this is so. The formula provided by you for $p$ is probability of moving up under risk neutral measure. $\endgroup$ – Neeraj Feb 11 '16 at 14:38
  • 1
    $\begingroup$ I'm copying the formula directly from my textbook. Letting $\alpha$ be the expected rate of return for the stock, the formula I have for $p$ is the true probability of an up-move. (letting $\alpha$ be the risk-free rate gives the risk-neutral probability of an up-move, $p^*$) My question is, what is the verbal interpretation that the true probability is greater than 1? $\endgroup$ – user2521987 Feb 11 '16 at 15:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.