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I am reading "Option, Futures and other Derivatives" by John C. Hull, and on Appendix chapter 13, he derives BSM formula from a Binomial Tree.

When he builds U2, I just don't understood how to get equation 13A.5.

Which method is being used to get it, and why "a", that is equal to "J" that is defined as quantity of upward movements, is being probably used as a random variable?

page 299 - Appendix Chapter 13 - "Option, Futures and Other Derivatives - 9th edition"

Regards,

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In the Black Scholes formula the $N(\alpha)$ gives you cumulative probability, i.e, the probability of a randomly selected occurence being below $\alpha$.

To transform the distribution of your variable into the standard normal you subtract the mean and divide by the standard deviation. It is said in the paragraph preceding formula 13A.5 that the mean is $np$ and the standard deviation is $\sqrt{np(1-p)}$.

So:

  • $N(\alpha)$ gives you the cumulative probabily of $\alpha$ in the normal distribution, i.e, probability of a random selection being below $\alpha$
  • $N\left(\frac{\alpha - np}{\sqrt{np(1-p)}}\right)$ gives you the cumulative probabily of $\alpha$ in the standard normal distribution

But because what you want is the probability of the random selection being above $\alpha$ (and not below), i.e. $1 - N(x)$, you can use the fact that the normal distribution is symetric and just use $N(-x)$.

Applying this logic to the case above would give you what you want:

$U_2 = N\left(-\frac{\alpha - np}{\sqrt{np(1-p)}}\right) = N\left(\frac{np - \alpha}{\sqrt{np(1-p)}}\right)$

Also, be aware that the price of the binomial model will only converge to the Black Scholes price for a sufficiently larget number of trials.

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  • $\begingroup$ Perfect, many thanks! $\endgroup$ – DUM03 Jun 16 at 3:33

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