# How do we identify the distribution of a pivotal quantity?

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.

• What is the pdf of $\Gamma(4,\beta)$? Commented Nov 8, 2020 at 15:04
• It is a Gamma distribution with pdf $f(x)=\frac{\beta^4}{3}x^3\exp(-\beta x)$. Commented Nov 8, 2020 at 15:15
• For starters, find the distribution of $Y=2\beta X$ when $X$ has the above pdf. Can you identify this distribution? Commented Nov 8, 2020 at 16:10
• 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! Commented Nov 8, 2020 at 16:46
• Yes, but note that pdf of $X$ should have $\beta^4/6$ as the normalizing constant. Commented Nov 8, 2020 at 16:57

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 $$\underline{X}$$.

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