Working on a homework question and having some trouble... Any help would be greatly appreciated.

Based on a sample 1.23, 0.36, 2.13, 0.91, 0.16, 0.12 from the GAM$(2,\theta)$ distribution, find an exact 95% CI for parameter $\theta$.

So we know GAM$(\alpha, \lambda)$ has the pdf $f(x)= \dfrac{\lambda^{\alpha}}{\Gamma{(\alpha)}} x^{\alpha - 1} \ e^{-\lambda x} $.

Therefore our random sample is distributed with pdf $f(x)=\theta^{2} x e^{-\theta x}$.

I understand that because the question asks for an "exact" confidence interval, that I need to find the pivotal variable.

The problem I am having is that most examples I find are along the lines of a random sample... $X_1,...,X_n \sim N(\theta, \sigma^{2})$ if $\sigma$ is known then $Z= \dfrac{\bar{X}-\theta}{\frac{\sigma}{\sqrt{n}}}\sim N(0,1)$, is pivotal. And from there finding the CI is relatively simple.

I guess I am at a loss as to how one would go about finding the pivotal variable when things are not normally distributed.

Thank you for your help, any suggestions would be appreciated.


1 Answer 1


Edit: Time to add details, I think. The OP has long since worked it out but hasn't taken the invitation to write up a more complete solution, so I shall, in the interest of having a full answer to the question.

A pivot is a function of the data and the statistic whose distribution doesn't depend on the value of the statistic.

So consider:

(1) what would the distribution of a statistic consisting of the sum of the observations ($T=\sum_i x_i$) be?

A sum of $n$ i.i.d. $\text{gamma}(\alpha,\theta)$ random variables has the $\text{gamma}(n\alpha,\theta)$ distribution (for the shape-rate form of the gamma).

Here $n=6$ and $\alpha=2$, so the sum, $T$ has a $\text{gamma}(12,\theta)$ distribution.

(2) Note that the distribution in (1) does depend on $\theta$ and the form of the statistic doesn't. You need to modify the statistic ($Q=f(T,\theta)$) in such a way that both of those change. (This part is trivial!)

Let $Q=T/\theta$. Then $Q\sim \text{gamma}(12,1)$.

$Q$ satisfies the conditions required for a pivotal quantity.

(3) Once you have a pivotal quantity (i.e. $Q$), write down an interval for the pivotal quantity (in the form of a pair of inequalities, $a< Q< b$) with the given coverage. Since the distribution doesn't depend on the parameter, this interval is always the same (at a given sample size) no matter what the value of $\theta$.

One such interval is $(a,b)$, where $P(a<Q<b)=0.95$, when $a$ is the 0.025 point of the $\text{gamma}(12,1)$ distribution and $b$ is the 0.975 point.

(4) Now write the interval involving the pivotal quantity back in terms of the data and $\theta$. Back out an interval for the parameter, for which the corresponding probability statement must still hold (keeping in mind that the random quantity is not $\theta$ but the interval).

$P(a<T/\theta<b)=0.95$ implies $P(1/b < \theta/T < 1/a)=0.95$, so $P(T/b < \theta < T/a)=0.95$. Therefore $(T/b,T/a)$ is a 95% interval for $\theta$.

Our observed total, $t = 4.91$. The 0.025 point of a gamma(12,1) is 6.2006 and the 0.975 point is 19.682. Hence a 95% interval for $\theta$ is (4.91/19.682,4.91/6.200)
= $(0.249, 0.792)$.

  • 1
    $\begingroup$ If $Y_1$ and $Y_2$ are both $\sim Gamma(\alpha, \beta)$, what's the distribution of $Y_1+Y_2$? I expect you've done it before, so this may be a result you already know. $\endgroup$
    – Glen_b
    Commented Mar 8, 2014 at 0:43
  • 1
    $\begingroup$ Yes. As long as the scale parameter is the same, you can sum the shape parameters. $\endgroup$
    – soakley
    Commented Mar 8, 2014 at 22:00
  • 1
    $\begingroup$ k6adams - Consider exactly the same statement as you wrote about $Y_1$ and $Y_2$ but with $Y_3$ and $Y_4$ in terms of $\alpha_3$ and $\alpha_4$ (that is, the exact same thing with merely a change of dummy variables). Now take your statement about $Y_1$ and $Y_2$ and let $Y_4=Y_1+Y_2$ and $\alpha_4=\alpha_1+\alpha_2$ and substitute those into the statement about $Y_3$ and $Y_4$. Do you see that as soon as you can do it for two, it must be true for as many as you like? $\endgroup$
    – Glen_b
    Commented Mar 8, 2014 at 22:05
  • 1
    $\begingroup$ @k6adams: I made edits. Does it look better now? $\endgroup$
    – cardinal
    Commented Mar 9, 2014 at 22:07
  • 1
    $\begingroup$ @Glen_b Since $\sum{x_i}=T\sim GAM(12,\theta)$, $\theta T\sim GAM(12, 1)$. So, $2\theta T \sim \chi^{2}(24)$. From there one can find upper and lower bounds ($c_1$ and $c_2$ respectively) on a $\chi^{2}$ table, and solve $P(c_1<T<c_2)=0.95$. I think that is it... just leaving this comment as to not leave others hanging. Thanks for the help everyone. $\endgroup$
    – k6adams
    Commented Mar 12, 2014 at 0:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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