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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.

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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)
= $(0.249, 0.792)$.

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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ @k6adams: I made edits. Does it look better now? $\endgroup$
    – cardinal
    Commented Mar 9, 2014 at 22:07
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    $\begingroup$ @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

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