Pivot quantity-question

approx · #1 $Old$ October 14th, 2008, 10:16 AM

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

Let Y have a pdf:

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

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

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

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

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

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

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

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

$2/\theta ^2 * 1/2[\theta u\theta -(u\theta) ^2] = 2/2\theta ^2[\theta^2 u -(u\theta) ^2]$

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

$= u-u^2$

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

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

1. a function of the sample measurement, Y and the unknown parameter $\theta$
2. It's probability distribution does not depend on the parameter $\theta$ .

Is this correct? Thanks in advance.

mr fantastic · #2 $Old$ October 14th, 2008, 05:35 PM

Quote:

Originally Posted by approx $View Post$

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

Let Y have a pdf:

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

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

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

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

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

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

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

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

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

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

$= u-u^2$

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

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

1. a function of the sample measurement, Y and the unknown parameter $\theta$
2. It's probability distribution does not depend on the parameter $\theta$ .

Is this correct? Thanks in advance.

Fix the small mistake and you get $F(u) = 2 u - u^2$ .

By the way, you realise that the pdf is non-zero for 0 < u < 1, right? So you require F(1) = 1, right? That was the first thing I checked. Your answer had F(1) = 0, so I knew there was a mistake somewehere.

Nevertheless, nice work, chum.