Originally Posted by

**approx** Hi! I'd be thankful if anyone could check my work to see if I've done this correct:

Let Y have a pdf:

$\displaystyle fy(y) = \frac{2(\theta -y)}{\theta ^2}, 0<y<\theta

$

Show that $\displaystyle Y/\theta $ is a pivotal quantity.

I start out with a new variable: $\displaystyle U = Y/\theta $

$\displaystyle P(U\leq u) = P(Y/\theta \leq u) = P(Y\leq u\theta)$

$\displaystyle \int _0^{u\theta} f(y)dy = \int _0^{u\theta} \frac{2(\theta -y)}{\theta ^2} dy$

$\displaystyle 1/\theta ^2 \int _0^{u\theta} 2(\theta -y) dy = 2/\theta ^2 \int _0^{u\theta} \theta -y dy$

$\displaystyle 2/\theta ^2 [\theta y -\frac{y^2}{2}]_0^{u\theta}$

$\displaystyle 2/\theta ^2 [\theta u\theta -\frac{(u\theta) ^2}{2}]$

$\displaystyle 2/\theta ^2 * 1/2[{\color{red}2} \theta u\theta -(u\theta) ^2] = 2/2\theta ^2[\theta^2 u -(u\theta) ^2]$ Mr F says: Small mistake here. The red 2 needs to be included.

$\displaystyle \frac{2\theta ^2u - 2(u\theta )^2}{2\theta ^2} = \frac{2\theta ^2u -2u^2\theta ^2}{2\theta ^2}$

$\displaystyle = u-u^2$

So this is the distribution function of U. I take the derivative of that to find the pdf of U, which equals $\displaystyle 1 - 2u.$

Then $\displaystyle Y/\theta$ fulfils the conditions for a pivotal quantity. Which is:

1. a function of the sample measurement, Y and the unknown parameter $\displaystyle \theta$

2. It's probability distribution does not depend on the parameter $\displaystyle \theta$.

Is this correct? Thanks in advance.