2
$\begingroup$

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

Therefore the answers should be:

(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

$\endgroup$
1
$\begingroup$

$\dfrac{(n-1)S^2}{\sigma^2}$ 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.

$\endgroup$
  • $\begingroup$ 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... $\endgroup$ – lstbl Mar 14 '16 at 19:51
  • 1
    $\begingroup$ $$ \phantom{mnnn}\frac 1 {\sigma^2} \sum_{i=1}^n (X_i - \bar X)^2 \sim \chi^2_{n-1}, $$ $$\text{but } \frac 1 {\sigma^2} \sum_{i=1}^n (X_i - \mu)^2 \sim \chi^2_n,$$ where $\mu$ is the population mean and $\bar X$ is the sample mean. $\qquad$ $\endgroup$ – Michael Hardy Mar 14 '16 at 19:56
  • 1
    $\begingroup$ $\ldots\,$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. $\qquad$ $\endgroup$ – Michael Hardy Mar 14 '16 at 20:02
  • $\begingroup$ 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! $\endgroup$ – lstbl Mar 14 '16 at 20:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.