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The log return $X$ on a certain stock investment is an $N(\mu,\sigma^2)$ random variable.

A financial analyst has claimed that the volatility $\sigma$ of the log return on this stock is less than $3$ units. A random sample of $11$ returns on this stock gave an estimated variance of the log-returns as $s^2 = 16$.

  1. Assess the analyst's claim by using a significance test at level $\alpha = 0.05$ to test

    $H_0: \sigma^2 \leq 9 \text{ against } H_1: \sigma^2 > 9.$

  2. Find a two-sided $95\%$ confidence interval for $\sigma^2$.

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what do you mean by sample of 11 returns.. you mean you have 11 samples in total and you estimated the variance as 16? how kind of estimator gives such an estimate? –  Seyhmus Güngören Aug 4 '12 at 12:14
1  
Calling it "11 samples" rather than "a sample of 11" is a widespread incorrect usage. –  Michael Hardy Aug 4 '12 at 18:24

2 Answers 2

up vote 2 down vote accepted

Since $$ \frac{(n-1)S^2}{\sigma^2} = \frac{10 S^2}{\sigma^2} \sim \chi^2_{11-1} = \chi^2_{10}, $$ you have $$ \Pr\left( A < \frac{(n-1)S^2}{\sigma^2} < B\right) = 0.95 $$ where $A$ and $B$ are so chosen that $$ \Pr(\chi^2_{10}<A) = 0.025 = \Pr(\chi^2_{10}>B). $$ (You can get $A$ and $B$ from tables or software.)

Via simple algebra it follows that $$ \Pr\left( \frac 1 B < \frac{\sigma^2}{10S^2} < \frac 1 A \right) = 0.95, $$ whence $$ \Pr\left( \frac {10S^2}{B} < \sigma^2 < \frac{10S^2}{A} \right) =0.95. $$ So there's your confidence interval.

If $9$ is less than the lower end of this confidence interval, you'd reject the null hypothesis at the 2.5% level. So instead of half of 5% that we used above, use half of 10%, and that will tell you whether to reject the null hypothesis at the 5% level.

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Since log returns is normally distributed the sample variance $S^2$ has a distribution that is proportional to a chi-square with $n-1$ degrees of freedom. So precisely stated

$(n-1)S^2/σ^2$ has a chi square distribution with $n-1$ degrees of freedom. So for the above test use

$(n-1)S^2/9$ as the test statistic and use the chi square distribution with $n-1$ degrees of freedom to get the test results.

In your case n-1 =10.

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