Question

Find the uniform distribution of the continuous type on the interval \((b, c)\) that has the same mean and the same variance as those of a chi-square distribution with 8 degrees of freedom. That is, find \(b\) and \(c\).

Short answer

The uniform distribution that has the same mean and the same variance as those of a chi-square distribution with 8 degrees of freedom is the interval \( (4, 12) \).

Step-by-step solution

Set up the equation for the mean

For a uniform distribution, the mean is \( \frac{b+c}{2} \). Since the mean of a chi-square distribution with 8 degrees of freedom is 8, set these equal to each other: \( \frac{b+c}{2} = 8 \). Solve this equation for one of the variables to obtain \(c = 16 - b\).

Set up the equation for the variance

For a uniform distribution, the variance is \( \frac{(c-b)^2}{12} \). Known that the variance of a chi-square distribution with 8 degrees of freedom is 16, we set \( \frac{(c-b)^2}{12} = 16\). Substitute \(c = 16 - b\) into the equation to obtain \( \frac{(16-2b)^2}{12} = 16 \).

Solve the equation for \(b\)

Solve the equation \( \frac{(16-2b)^2}{12} = 16 \) for \(b\). This will yield two possible values for \(b\), namely 4 and 12.

Calculate \(c\)

Substitute these two possible values of \(b\) into the equation for \(c\) to obtain the corresponding values of \(c\). When \(b = 4, c = 12\), and when \(b = 12, c = 4\).

Determine the valid solution

For the interval \(b, c\), \(b < c\). Therefore, only the solution \(b = 4, c = 12\) is valid.