Statistics, constructing upper, lower, and two-sided confidence intervals based on a gamma distribution

Question

anonymous · Answer

We were explained how to prove part a and b, but if you'd like to give an explanation, i won't hold you back, since it didn't make full sense other than utilizing the method of transformations

kirbykirby · Answer

One other way you can do a) is to use the CDF method:
Let $Y=2Xi/\beta$

Then $$\large \begin{align} F_Y(y)=P(Y\le y) &=P\left(\frac{2X_i}{\beta}\le y \right)\
&=P\left(X_i\le\frac{\beta y}{2} \right) \
&=F_{X_i}\left(\frac{\beta y}{2} \right)\end{align}$$

Then differentiate both sides with respect to y:
$$\large f_Y(y)=f_{X_i}\left( \frac{\beta y}{2}\right) \cdot \frac{\beta}{2}, \text{(chain rule)}\
=\large\frac{\beta}{2}  \frac{1}{\Gamma(\alpha)\beta^{\alpha}}\left(\frac{\beta y}{2}\right)^{\alpha-1}e^{\frac{-\beta y}{2}/\beta}\
= \large \frac{1}{\Gamma(\alpha)2^{\alpha}}y^{\alpha-1}e^{-y/2}$$

This is the PDF of $\text{Gamma}(α,2)$

Now just remember that the link between the Gamma and Chi-Square distribution  is:
$\large\text{Gamma}\left(\frac{n}{2},2\right)=\chi^2(n)$, so I hope you see that you get a $\chi^2(2\alpha)$ distribution (note you can write your gamma distribution as $\text{Gam}(2\alpha/2,2)$ ).

For part b), I hope you remember that the sum of independent chi-squared distributions is a chi-square distribution with all the individual degrees of freedom added. You can  easily show this via moment generating functions

kirbykirby · Answer

recall that the MGF for $X\sim\text{Gamma}(\alpha,\beta)$ is $$M_X(t)=\frac{1}{(1-\beta t)^{\alpha}}$$

So for $\chi^2(2\alpha)=\text{Gam}(\alpha,2)$:
$$M_X(t)=\frac{1}{(1-2t)^{\alpha}}$$

Now, using the same variable  Y as defined above, and using the random variable V defined in the question, Recall for independent random variables, $V=Y_1+Y_2+\ldots +Y_n $,
$$\large\begin{align} M_V(t)&=\prod_{i=1}^nM_{Y_i}(t)\
&=\prod_{i=1}^n\frac{1}{(1-2t)^{\alpha}}\
&=\frac{1}{(1-2t)^{n\alpha}}\end{align}$$

This is the MGF of $\text{Gam}(n\alpha,2)=\chi^2(2n\alpha)$, so $V$ has this distribution (I am not sure why the question argue though that you have $n$ degrees of freedom? when I believe it should be $2n\alpha$ degrees of freedom, because the parameter of the Chi-squre distribution are known as degrees of freedom).

kirbykirby · Answer

We can use V as a pivot for $\beta$ because the distribution of V does not have the parameter $\beta$ in it. This is from the definition of a pivot, which is a function whose probability distribution does not depend on unknown parameters (even though $\alpha$ is unknown, it is a pivot for $\beta$ since $\beta$ is not in the distribution of V).

kirbykirby · Answer

For the confidence interval.. first observe that
$\sum_{i=1}^nX_i =n\bar{X}$
So with $n=5$ and $\bar{X}=5.39$, $\sum_{i=1}^5\bar{X}=26.95$

Also, with $\alpha=2$, $\chi^2(2n\alpha)=\chi^2(20)$

So to construct your (exact) confidence interval, you need to use the pivotal quantity (pivot) since it doesn't depend on unknown parameters. A general way of writing a $100p\%$ 2-sided confidence interval with a pivotal quantity $Q(X;\theta)$, where $\theta$ is just the unknown parameter(s) of your distribution of $X$, is first writing
$$P( q_1\le Q(X;\theta)\le q_2)=p$$, then the confidence interval is just $ [q_1, q_2]$, where $q_1$ and $q_2$ are the lower and upper quantiles of your pivotal quantity. 


For a 95% 2-sided confidence interval in your case would be first writing:
$$\large \begin{align}P\left(\chi^2_{0.025}(20)\le\frac{2}{\beta} \sum_{i=1}^5X_i \le\chi^2_{0.975}(20)\right)&=0.95 \
P\left( \chi^2_{0.025}(20)\le \frac{2}{\beta}(26.95)\le \chi^2_{0.975}(20)\right)&=0.95 \
P\left(\chi^2_{0.025}(20)\le \frac{53.9}{\beta}\le \chi^2_{0.975}(20)\right)&=0.95 \
P(\left( \frac{1}{\chi^2_{0.025(20)}}\ge\frac{\beta}{53.9}\ge\frac{1}{\chi^2_{0.975(20)}}\right)&=0.95\
P\left( \frac{53.9}{\chi^2_{0.025(20)}}\ge\beta\ge\frac{53.9}{\chi^2_{0.975(20)}}\right)&=0.95\
P\left(\frac{53.9}{\chi^2_{0.975(20)}}\le \beta\le \frac{53.9}{\chi^2_{0.025(20)}} \right)&=0.95\end{align}$$ 

So a 95% 2-sided CI for $\beta$is: $$\large \left[\frac{53.9}{\chi^2_{0.975(20)}}, \frac{53.9}{\chi^2_{0.025(20)}}  \right] $$
those values are just the quantiles which you can find in a chi-square table with 20 degrees of freedom (just be careful whether the table is showing you CDF values, or tail values). 

Now a one-sided confidence interval is similar as above, but only one inequality is needed and you would need to use quantiles like $\large\chi^2_{0.05}(20)$ and $\large\chi^2_{0.95}(20)$. Although I have rarely used one-sided confidence intervals, 
I believe an upper 95% confidence interval would be based on the result $\large P(Q(X;\theta)\ge \chi^2_{0.05}(20))=0.95$ and the CI should be in the form $[q_1,\infty)$

a lower 95% confidence interval would be based on the result $\large P\left(Q(X;\theta)\le \chi^2_{0.95}(20)\right)= 0.95$ and the CI should be in the form $[0,q_2]$ since the chi-square distribution has support $0\le x < \infty$

If you draw all three confidence intervals, the interval in which all 3 overlap should be $[q_1,q_2]$ when you find those values. 
-----
Also, I just want to add, if you're wondering why doing all the steps in a) and b) were necessary... to find a confidence interval for $\beta$ based on the gamma distribution given can't be done (at least for finding EXACT confidence intervals), because the gamma distribution given is dependent on $\beta$ and is thus not a pivotal quantity. We eventually found a new gamma distribution which was a pivotal quantity because its distribution was not dependent on $\beta$.

And then we converted the gamma into a Chi-square distribution because we can use chi-square tables that are readily available to find the numeric values of the quantiles as we did above.

anonymous · Answer

Thank you so much, i apologize for not commenting or anything when you were answering the first parts, i was in my lectures and did not access the internet during them, and after looking at everything i wanted to see if i really understood so i went and completed another similar problem without issues. So again, thank you so much :D your explanations never fail to help me understand

kirbykirby · Answer

Your welcome, and awesome :)!