Sufficiency question

**lllll** · December 7th 2008, 12:00 AM

Let $Y_1, \ Y_2, \ \dotso,\ Y_n$ denote a random sample from a Weibull dist. with known $m$ and unknown $\alpha$ . Show that $\sum^n_{i=1} Y^m_{i}$ is sufficient for $\alpha.$

Attempt:

$\frac{1}{\alpha} m y^{m-1} \exp \left( -\frac{y^m}{\alpha} \right)$ is my function.

so the likelihood is:

$L(y_1, \ y_2, \ \dotso, \ y_n|\alpha) = f(y_1, \ y_2, \ \dotso, \ y_n|\alpha) = f(y_1|\alpha) \cdot f(y_2|\alpha) \cdot \dotso \cdot f(y_n|\alpha)$

$=\prod^n_{i=1} \frac{1}{\alpha} m y^{m-1} \exp \left( -\frac{y^m}{\alpha} \right)$

$=\bigg{[}\frac{1}{\alpha} m y_1^{m-1} \exp \left( -\frac{y_1^m}{\alpha} \right)\bigg{]}$ $\times \bigg{[}\frac{1}{\alpha} m y_2^{m-1} \exp \left( -\frac{y_2^m}{\alpha} \right)\bigg{]} \times \dotso \times$ $\bigg{[}\frac{1}{\alpha} m y_n^{m-1} \exp \left( -\frac{y_n^m}{\alpha} \right)\bigg{]}$

$=\bigg{(}\frac{1}{\alpha} m\bigg{)}^n \prod^n_{i=1} y_i^{m-1} \exp \left( -\sum^n_{i=1}\frac{y_i^m}{\alpha} \right)$

at which point I get stuck, I know I'm supposed to factor it out and have a second function that isn't dependent on $\alpha$ , but I'm not quite sure on how to separate it out.

**mr fantastic** · December 7th 2008, 02:29 AM

Originally Posted by lllll

Let $Y_1, \ Y_2, \ \dotso,\ Y_n$ denote a random sample from a Weibull dist. with known $m$ and unknown $\alpha$ . Show that $\sum^n_{i=1} Y^m_{i}$ is sufficient for $\alpha.$

Attempt:

$\frac{1}{\alpha} m y^{m-1} \exp \left( -\frac{y^m}{\alpha} \right)$ is my function.

so the likelihood is:

$L(y_1, \ y_2, \ \dotso, \ y_n|\alpha) = f(y_1, \ y_2, \ \dotso, \ y_n|\alpha) = f(y_1|\alpha) \cdot f(y_2|\alpha) \cdot \dotso \cdot f(y_n|\alpha)$

$=\prod^n_{i=1} \frac{1}{\alpha} m y_{\color{red}i}^{m-1} \exp \left( -\frac{y_{\color{red}i}^m}{\alpha} \right)$ Mr F says: Note the small errata.

$=\bigg{[}\frac{1}{\alpha} m y_1^{m-1} \exp \left( -\frac{y_1^m}{\alpha} \right)\bigg{]}$ $\times \bigg{[}\frac{1}{\alpha} m y_2^{m-1} \exp \left( -\frac{y_2^m}{\alpha} \right)\bigg{]} \times \dotso \times$ $\bigg{[}\frac{1}{\alpha} m y_n^{m-1} \exp \left( -\frac{y_n^m}{\alpha} \right)\bigg{]}$ Mr F says: I'll pick it up from here because the line below is wrong and not needed.

$=\bigg{(}\frac{1}{\alpha} m\bigg{)}^n \prod^n_{i=1} y_i^{m-1} \exp \left( -\sum^n_{i=1}\frac{y_i^m}{\alpha} \right)$

at which point I get stuck, I know I'm supposed to factor it out and have a second function that isn't dependent on $\alpha$ , but I'm not quite sure on how to separate it out.

$= \frac{m^n}{\alpha^n} \, (y_1 \, y_2 \, .... \, y_n)^{m-1} \exp \left(-\frac{1}{\alpha} \sum_{i=1}^n y_i^m\right)$

$= m^n \, (y_1 \, y_2 \, .... \, y_n)^{m-1} \, \frac{\exp \left(-\frac{1}{\alpha} u\right)}{\alpha^n}$

where $U = \sum_{i=1}^n Y_i^m$

$= \left(\frac{\exp \left(-\frac{u}{\alpha} \right)}{\alpha^n}\right) \cdot \left(m^n \, (y_1 \, y_2 \, .... \, y_n)^{m-1} \right)$

$= g\left(\alpha, \, u \right) \cdot h(y_1 \, y_2 \, .... \, y_n)$ .

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