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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.
Answer:
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.

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closed as off-topic by noob2, Helin, LocalVolatility, skoestlmeier, phdstudent Sep 24 '18 at 9:30

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  • $\begingroup$ 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 $\endgroup$ – noob2 Sep 22 '18 at 1:54
  • $\begingroup$ When I did the calculations in R, I entered $30.3$ as $30$,$.3$. I now fixed this typo. $\endgroup$ – Bob Sep 22 '18 at 15:08
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Does this sound more reasonable?

enter image description here

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.

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  • $\begingroup$ 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 $14/52$. $\endgroup$ – Bob Sep 22 '18 at 14:52
  • $\begingroup$ 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). $\endgroup$ – noob2 Sep 22 '18 at 15:25
  • $\begingroup$ Thanks @noob2! I have added explanation in the answer. $\endgroup$ – Magic is in the chain Sep 22 '18 at 16:28
  • $\begingroup$ @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? $\endgroup$ – Bob Sep 22 '18 at 16:35
  • $\begingroup$ Yes! 52 weeks in a year, roughly! $\endgroup$ – Magic is in the chain Sep 22 '18 at 16:57
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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 ?

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

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