1
$\begingroup$

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 $\sigma^2$.

  • 1
    Calling it "11 samples" rather than "a sample of 11" is a widespread incorrect usage.2012-08-04

2 Answers 2

2

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}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.

2

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.