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計算Black Scholes的波動率

以下問題來自赫爾的書。我做到了,但我不確定它是對的。我希望這裏有人可以告訴我,如果我做得對,如果不是我錯了。

謝謝 短發

問題:

假設在每個結尾處對股票價格(以美元計)的觀察 連續幾周$ 15美元如下:

$ 30.2,\,\,\,32,\,\,\,31.1,\,\,\,30.1,\,\,\,30.2,\,\,\,30.3,$ $ 30.6,\,\,\,33.0,\,\,\,32.9,\,\,\,33.0,\,\,\,33.5,$ $ \,\,\,33.5,\,\,\,33.7,\,\,\,33.5,\,\,\,\ n,33.2 $

估計股價波動。您的標準錯誤是什麽? 估計?

回答:

$n + 1$ is the number of observations. $S_i$ is the value of the stock at the end of the ith interval, $\tau$ is the length of the intervals in years and $\hat{ {\sigma} }$ is a good estimate of the standard deviation of $u_i$. The standard error of this estimate is about $\frac{ \hat{ \sigma} }{ \sqrt{2n} }$. The relevant formulas are: \begin{eqnarray*} u_i &=& \ln{( \frac{u_i}{u_{i-1}} )} \\ s &=& \sqrt{ \frac{1}{n-1} \sum_{i=1}^{i=n} {(u_i - \bar u)^2} } \\ \hat { \sigma } &=& \frac{s}{\sqrt{\tau} } \\ \end{eqnarray*} Note: $\bar u$ is the mean of the $u_i$ variables. \newline In this case, we have: \begin{eqnarray*} n &=& 14 \\ u_1 &=& \ln {(\frac{32} {30.2} )} = 0.057894 \\ u_2 &=& \ln {(\frac{31.1} {32} )} = -0.02852 \\ u_3 &=& \ln {(\frac{30.1} {31.1} )} = -0.0326826 \\ u_4 &=& \ln {(\frac{30.2} {30.1} )} = 0.003316 \\ u_5 &=& \ln {(\frac{30.3} {30.2} )} = 0.003305 \\ u_6 &=& \ln {(\frac{30.6} {30.3} )} = 0.009852 \\ u_7 &=& \ln {(\frac{33 } {30.6} )} = 0.075507 \\ u_8 &=& \ln {(\frac{32.9} {33} )} = -0.003034 \\ u_9 &=& \ln {(\frac{33 } {32.9} )} = 0.003034 \\ u_{10} &=& \ln {(\frac{33.5 } {33} )} = 0.0150379 \\ u_{11} &=& \ln {(\frac{ 33.5 } {33.5} )} = 0 \\ u_{12} &=& \ln {(\frac{33.7 } {33.5} )} = 0.005952 \\ u_{13} &=& \ln {(\frac{33.5} {33.7} )} = - 0.0059524 \\ u_{14} &=& \ln {(\frac{33.2 } {33.5} )} = - 0.008995 \\ \end{eqnarray*}

\begin{eqnarray*} \sum_{i=1}^{i=5} {u_i} &=& 0.003312 \\ \sum_{i=6}^{i=10} {u_i} &=& 0.1064649 \\ \sum_{i=11}^{i=14} {u_i} &=& -0.0089954 \\ \sum_{i=0}^{i=14} {u_i} &=& 0.003312 + 0.1064649 -0.0089954 = 0.100781 \\ \bar u &=& \frac{1}{n} \sum_{i=1}^{i=n} { u_i } = \frac{1}{14} { \sum_{i=1}^{i=14} { u_i } } = \frac{0.100781}{14} \\ \bar u &=& 0.007198 \\ s &=& \sqrt{ \frac{1}{14-1} \sum_{i=1}^{i=14} {(u_i - 0.007198)^2} } \\ (u_1 - 0.007198)^2 &=& 0.002570 \\ (u_2 - 0.007198)^2 &=& 0.001275 \\ (u_3 - 0.007198)^2 &=& 0.001590 \\ (u_4 - 0.007198)^2 &=& 0.0000151 \\ (u_5 - 0.007198)^2 &=& 0.0000152 \\ (u_6 - 0.007198)^2 &=& 0.0000070 \\ (u_7 - 0.007198)^2 &=& 0.0046661 \\ (u_8 - 0.007198)^2 &=& 0.0001047 \\ (u_9 - 0.007198)^2 &=& 0.0000173 \\ (u_{10} - 0.007198)^2 &=& 0.0000615 \\ (u_{11} - 0.007198)^2 &=& 0.0000518 \\ (u_{12} - 0.007198)^2 &=& 0.0000016 \\ (u_{13} - 0.007198)^2 &=& 0.0001729 \\ (u_{14} - 0.007198)^2 &=& 0.0002622 \\ \end{eqnarray*} \begin{eqnarray*} \sum_{i=1}^{i=5} {(u_i - 0.007198)^2} &=& 0.0054653 \\ \sum_{i=6}^{i=10} {(u_i - 0.007198)^2} &=& 0.0048566 \\ \sum_{i=11}^{i=14} {(u_i - 0.007198)^2} &=& 0.0004885 \\ \sum_{i=1}^{i=14} {(u_i - 0.007198)^2} &=& 0.0054653 + 0.0048566 + 0.0004885 = 0.0108104 \\ s &=& \sqrt{ \frac{1}{14-1} (0.0108104) } = \sqrt{ 0.0108104 } \\ s &=& 0.1039731 \\ \end{eqnarray*} I am figuring that there are $52.1$ weeks in a year. \begin{eqnarray*} \tau &=& \frac{14}{52.1} = 0.268714 \\ \hat { \sigma } &=& \frac{s}{\sqrt{\tau} } = \frac{ 0.1039731 } { \sqrt{ 0.268714 } } = 0.200 \\ \end{eqnarray*} My estimate of the volatility is $0.200$ and the standard error of this estimate is $\frac{.200}{\sqrt{2(14)}} = 0.377$.

最佳答案

第二筆金額應為:

$$ \ sum_ {i = 6} ^ {10} u_i = 0.10039773 $$

這給出了$ 0.0067648 $的平均值和$ \ sigma = .028836 $的標準差。

為了避免這些錯誤,您應該使用一些東西來自動計算。像電子表格這樣的東西。

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