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I have been given a pivotal quantity of $2\beta\sum_{i=1}^4X_i$ to determine a confidence interval of random sample $\underline{X}=(X_1,...,X_4)$ from a $\Gamma(4,\beta)$ distribution.

Initially, I want to find the distribution of this pivotal quantity, and why it can be used to construct a confidence interval for $\beta$.

I understand that provided the quantity is a function of the observations and parameter, in this case $g(\underline{x};\beta)$, and the distribution is known and independence of $\beta$ holds, then it can be used as a pivotal quantity. However, I am lost as how to systematically approach such a pivotal quantity to determine its distribution.

I have tried to provide an argument concerning the fact that each $X_i\sim\Gamma(4,\beta)$ and as such $\sum_{i=1}^4X_i\sim\Gamma(16,\beta)$ but this has been to no avail - as I cannot find independence of $\beta$.

Any and all help would be much appreciated. Thank you in advance.

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  • $\begingroup$ What is the pdf of $\Gamma(4,\beta)$? $\endgroup$ Commented Nov 8, 2020 at 15:04
  • $\begingroup$ It is a Gamma distribution with pdf $f(x)=\frac{\beta^4}{3}x^3\exp(-\beta x)$. $\endgroup$
    – duncan
    Commented Nov 8, 2020 at 15:15
  • $\begingroup$ For starters, find the distribution of $Y=2\beta X$ when $X$ has the above pdf. Can you identify this distribution? $\endgroup$ Commented Nov 8, 2020 at 16:10
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    $\begingroup$ So we obtain the transformation $F_Y(y)=P(Y \leq y)=P(2 \beta X \leq Y)=P(X\leq\frac{y}{2\beta})=F_X(\frac{y}{2\beta})$. This gives $f_Y(y)=F_Y'(y)=F_X'(\frac{y}{2\beta})\times\frac{1}{2\beta}=f_X(x)\times\frac{1}{2\beta}=\frac{y^3}{48}\exp(-\frac{x}{2})$. This is a $\chi_8^2$ distribution which is independent of $\beta$. I can now use this to form a confidence interval. Thank you! $\endgroup$
    – duncan
    Commented Nov 8, 2020 at 16:46
  • $\begingroup$ Yes, but note that pdf of $X$ should have $\beta^4/6$ as the normalizing constant. $\endgroup$ Commented Nov 8, 2020 at 16:57

1 Answer 1

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We start our answer by denoting the pivotal quantity by $Y_i=2 \beta X_i$.

As above, each $X_i \sim Gamma(4,\beta)$, so we can obtain that the probability density function (pdf) of $X$ is $f_X(x)=\frac{\beta^4}{6}x^3\exp(-\beta x)$.

We then apply the transformation process, beginning with the cumulative distribution function, $F_Y(y)=P(Y \leq y)=P(2 \beta X \leq Y)=P(X\leq\frac{y}{2\beta})=F_X(\frac{y}{2\beta})$. Applying to densities, we obtain: $f_Y(y)=F_Y'(y)=F_X'(\frac{y}{2\beta})\times\frac{1}{2\beta}=f_X(x)\times\frac{1}{2\beta}=\frac{y^3}{96}\exp(-\frac{y}{2})\textbf{1}_{y>0}$.

Noting that the generic pdf of a $\chi_n^2$ distribution is $\frac{1}{2^{\frac{n}{2}}\Gamma(\frac{n}{2})}x^{\frac{n}{2}-1}e^{-\frac{x}{2}}$. Selecting $n=8$ degrees of freedom allows us to obtain $f_Y(y)$.

We note that $\chi_8^2$ is the distribution for one $X_i$. Therefore, $2\beta\sum_{i=1}^4X_i=\sum_{i=1}^4Y_i$ gives a distribution of $\chi_{32}^2$, by the properties of the distribution.

As above, this is a valid result as $\chi_{32}^2$ is independent of $\beta$ and consists of observations $\underline{X}$.

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  • $\begingroup$ You should write $f_Y(y)=f_X\left(\frac{y}{2\beta}\right)\cdot\frac{1}{2\beta}=\frac1{96}y^3e^{-y/2}\mathbf1_{y>0}$. $\endgroup$ Commented Nov 8, 2020 at 19:46

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