# confidence about range of standard deviation

Suppose $X_1, X_2$ is a random sample of size 2 from a normal population known to have mean $0$ and variance $\sigma^2$; further assume $x_1 = -0.75$, $x_2=0.16$. How sure (or confident) would you be that:

(a) $\sigma^2 \ge 1.0221$ ?

(b) $\sigma^2 \ge 0.4242$ ?

(c) $\sigma^2 \ge 2.7908$ ?

I reason that this problem should be able to be solved using the following logic:

We know that if $Z_i$ is a standard normal random variable, $(Z_i - \bar{Z})$ is also an independent random variable with normal distribution, and therefore $(Z_i - \bar{Z})^2$ is $\chi^2$. However, we can use the definition of sample variance $S^2 = \frac{1}{n-1}\sum{(X_i - \bar{X})^2}$ to show that:

$$\sum_{i=0}^n{(Z_i - \bar{Z})}^2 = \frac{(n-1)S^2}{\sigma^2}$$

So, $\frac{(n-1)S^2}{\sigma^2}$ is $\chi^2$ with $n-1$ degrees of freedom.

I used the above reasoning to try and solve this problem in the following manner:

$$P(\sigma^2 \ge t) = P\left(\frac{\sigma^2}{S^2} \ge \frac{t}{S^2}\right) = P\left(\frac{S^2}{\sigma^2} \le \frac{S^2}{t}\right) = F_{\chi^2_{df=1}} \left(\frac{S^2}{t}\right)$$

Using the values in the problem to solve for $S^2$:

$$S^2 = \frac{1}{2-1}\left((-0.75-\bar{x})^2 + (0.16 - \bar{x})^2\right) = 0.41405$$

(since $\bar{x} = \frac{0.16-0.75}{2} = -0.295$ )

(a) $F_{\chi^2_{df=1}}(\frac{0.41405}{1.0221}) = F_{\chi^2_{df=1}}(0.4051) = 0.48$

(b) $F_{\chi^2_{df=1}}(\frac{0.41405}{0.4242}) = F_{\chi^2_{df=1}}(0.976) = 0.68$

(c) $F_{\chi^2_{df=1}}(\frac{0.41405}{2.7908}) = F_{\chi^2_{df=1}}(0.148) = 0.3$

However, the solutions in the back of the book gives the following answers:

(a) 0.25

(b) 0.5

(c) 0.1

$\dfrac{(n-1)S^2}{\sigma^2}$ is $\chi^2$ with $n-1$ degrees of freedom.
That much is true. But at the outset you said this is from a population that is KNOWN to have expectation $0$. Unrealistic perhaps, but it is clearly intended that you would use that information, and you did not.
You defined the sample variance: $\displaystyle S^2 = \frac{1}{n-1}\sum{(X_i - \bar{X})^2}$. This uses the sample mean $\bar X$ as an estimator of of the population mean. There is no reason to do that when the population mean is known with certainty. Instead, the appropriate thing is $$S^2 = \frac 1 n \sum_{i=1}^n (X_i - 0)^2,$$ so you have $$\frac{nS^2}{\sigma^2} \sim \chi^2_n.$$ Your confidence intervals should be based on that.
• Very good point, thank you. The answer comes out to be correct when you use $\bar{X} = 0$ and assume 2 degrees of freedom. I'm very confused by the latter point, however. Since there are only 2 values, the degrees of freedom should only be 1... – lstbl Mar 14 '16 at 19:51
• $$\phantom{mnnn}\frac 1 {\sigma^2} \sum_{i=1}^n (X_i - \bar X)^2 \sim \chi^2_{n-1},$$ $$\text{but } \frac 1 {\sigma^2} \sum_{i=1}^n (X_i - \mu)^2 \sim \chi^2_n,$$ where $\mu$ is the population mean and $\bar X$ is the sample mean. $\qquad$ – Michael Hardy Mar 14 '16 at 19:56
• $\ldots\,$so you reduce the number of degrees of freedom from $2$ to $2-1$ when you use $\bar X$ as an estimator $\text{of } \mu$, but not when $\mu$ is known and used. $\qquad$ – Michael Hardy Mar 14 '16 at 20:02
• I see. Essentially because we know the value for $\mu$, the $\chi^2$ distribution is just defined as a linear combination of each random variable $X$. Therefore, the degrees of freedom just follows the typical pattern of how many normal random variables are squared (i.e. 2). THANKS! – lstbl Mar 14 '16 at 20:03