# Calculating historical Volatility for the Black Scholes Model [closed]

Below is a problem from the book "Options, Futures, and other Derivatives" by John C. Hull. I did the problem but I am fairly sure that my answer is wrong. I am hoping that somebody can tell me where I went wrong?
Thanks,
Bob

Problem:
Suppose that observations on a stock price(in dollars) at the end of each $$15$$ consecutive weeks are as follows:
$$30.2$$, $$32.0$$, $$31.1$$, $$30.1$$, $$30.2$$, $$30.3$$, $$30.6$$, $$33.0$$,
$$32.9$$, $$33.0$$, $$33.5$$, $$33.5$$, $$33.7$$, $$33.5$$, $$33.2$$
Estimate the stock price volatility.
Let the closing prices be denoted by $$S$$. $$\begin{eqnarray*} u_i &=& \ln{ \bigg( \frac {S_1} {S_{i-1}} \bigg) } \\ \end{eqnarray*}$$ Using R, I find that: $$\begin{eqnarray*} u &=& 0.057893978 \,\, -0.028528084 \,\, -0.032682647 \\ && 0.003316753 \,\, -0.006644543 \,\, -2.302585093 \\ && 2.322387720 \,\, 0.075507553 \,\, -0.003034904 \\ && 0.003034904 \,\, 0.015037877 \,\, 0.000000000 \\ && 0.005952399 \,\, -0.005952399 \,\, -0.008995563 \\ \end{eqnarray*}$$ Now using $$R$$, I find that the standard deviation of $$u$$ is $$0.8744864$$. Call that value $$s$$. I will call the volatility of the stock to be $$\sigma$$. Now let $$\tau$$ be the length of time we observed the value of the stock for. $$\begin{eqnarray*} \sigma &=&\frac{s}{\sqrt{\tau}} \\ \tau &=& \frac{14}{52} = 0.2692308 \\ \sigma &=& \frac{0.8744864}{\sqrt{ 0.2692308}} \\ \sigma &=& 1.6853523 \\ \end{eqnarray*}$$ This number seems way off to me. What did I do wrong?
It also seems strange to me that in the last step you are dividend by $$\sqrt{\tau}$$ but that is the procedure given in the book.

• Assuming your prices are correct, then 3 of your u values are wildly incorrect namely -0.00664, -2.3025, 2.32238. These should be 0.0033058 0.0098523 0.0755076 – noob2 Sep 22 '18 at 1:54
• When I did the calculations in R, I entered $30.3$ as $30$,$.3$. I now fixed this typo. – Bob Sep 22 '18 at 15:08

Does this sound more reasonable?

To annualise the weekly std dev, you need to multiply by the square root of 52.

Let me know if you spot any typo.

Re-comment, 14 is the number of return observations, so it is already incorporated in the calculation of weekly std dev (2.88%). I think the author might have implied that if the observations that you use for stddev calculations have an interval of tau in years (daily=1/252, weekly=1/52, monthly=1/12), then you divide the computed std dev by sqrt(tau), which is sqrt(1/52) in your case. Now dividing by 1/sqrt(52) is same as multiplying by sqrt(52), and that’s exactly the factor you need to multiply 2.88% by to get the answer.

• I follow your calculations up and including the 2.88%. What I do not see is how you go from the 2.88% to the 20.79%. The way I would do it is to divide by the $\sqrt{( \tau )}$ where $\tau$ is $14/52$. – Bob Sep 22 '18 at 14:52
• Multiplying the stdev by $\sqrt{52}$ looks fine. The ratio $14/52$ should only be used when you are starting from the sum of squares (not the stdev). – noob2 Sep 22 '18 at 15:25
• Thanks @noob2! I have added explanation in the answer. – Magic is in the chain Sep 22 '18 at 16:28
• @Magic is in the chain Now I think I get it, you divided by $\sqrt{\tau}$ where $\tau = 7/365$ because we making the observation once a week. Is that right? – Bob Sep 22 '18 at 16:35
• Yes! 52 weeks in a year, roughly! – Magic is in the chain Sep 22 '18 at 16:57

The division is correct but more often you’ll find a multiplication by the inverse of $\sqrt{\tau}$ instead.

Your vector of returns $u$ has gone wrong somehow, there is simply no way you would have numbers over the 0.05 area with the prices provided.

Additionally you would calculate the standard deviation of $\ln{\frac{S_i}{S_{i-1}}}$, do you have a typo there or a logical error ?

• I do have an error there. Thanks for pointing it out. – Bob Sep 22 '18 at 17:59