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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.

Glen_b
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k6adams
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1 Answers1

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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.

$P(a<T/\theta<b)=0.95$ 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)
= $(0.249, 0.792)$.

Glen_b
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  • thanks for responding. I am going to dwell on this and see if I can come up with an answer. – k6adams Mar 07 '14 at 23:11
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    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. – Glen_b Mar 08 '14 at 00:43
  • If $Y_1$ is $\sim GAM(\alpha_1, \beta) $ and $Y_2$ is $\sim GAM(\alpha_2, \beta)$ then $Y_1 + Y_2 \sim GAM(\alpha_1 + \alpha_2, \beta)$. Does this extend to $Y_3,...Y_n$? – k6adams Mar 08 '14 at 01:28
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    Yes. As long as the scale parameter is the same, you can sum the shape parameters. – soakley Mar 08 '14 at 22:00
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    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? – Glen_b Mar 08 '14 at 22:05
  • @Glen_b I have to admit I am a bit over my head on this one, but here it goes... It is given that $X_1..X_6$ are iid $GAM(2,\theta)$, and if $T= \sum {x_i}$, then $T \sim GAM(12,\theta)$. It is known if $Q \sim \chi^2(v)$ and $c$ is a positive constant, then $cQ\sim GAM(\dfrac{v}{2}, 2c)$. I am a little unsure about how to interpret the $\chi^2$ transformation. I would assume that $\theta$ is the constant $c$ referred to in the theorem, so $Q= 2T\sim\chi^2(24)$ and $P(2c_1 – k6adams Mar 09 '14 at 20:04
  • for some reason the rendering of the right hand column of the webpage is overlapping my comments and is preventing me from editing or deleting my post... – k6adams Mar 09 '14 at 20:09
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    @k6adams: I made edits. Does it look better now? – cardinal Mar 09 '14 at 22:07
  • k6adams: If the $\chi^2$ is giving you any confusion, you could stick to leaving it as a gamma. Most decent stats packages will give you gamma quantiles (or must you use tables?). – Glen_b Mar 09 '14 at 22:48
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    @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 – k6adams Mar 12 '14 at 00:27
  • That sounds like it's right. If you'd like to put the details in an answer, I'd upvote it. If you'd rather edit my answer, that would be fine (assuming you have enough rep for the privilege, I don't recall). If you'd rather I edit my answer, I could do that ... any of them are good as long as it ends up in an answer, but I'm happy for you to get some reputation out of it. – Glen_b Mar 12 '14 at 01:01
  • Actually it looks like anyone - even anonymous users - can suggest edits (perhaps with a small delay from the time of the posting). – Glen_b Mar 12 '14 at 01:08