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Thread: Pivot quantity-question

  1. #1
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    Pivot quantity-question

    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 $

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

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

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

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

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

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

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

    $\displaystyle = u-u^2$

    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.
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  2. #2
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    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:

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

    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.
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