exponential family help

**Danneedshelp** · March 24th 2010, 12:21 PM

Q: Let $Y_{1},...,Y_{n}$ denote a random sample from the probability density function
$f(y|\theta)=\theta\\y^{\theta\\-1}$ for $0<y<1$ , $\theta>0$ and 0 otherwise.
a) Show that this density function is in the (one-parameter) exponential family and that $\sum_{i=1}^{n}-ln(Y_{i})$ is sufficient for $\theta$ (see previous exercise).
b) If $W_{i}=-ln(Y_{i})$ , show that $W_{i}$ has an exponential distribution with mean $\frac{1}{\theta}$ .

Previews exercise mentioned in (a):

1) Suppose that $Y_{1},...,Y_{n}$ is a random sample from a probability density in the (one-parameter) exponential family so that

$f(y|\theta)=a(\theta)b(y)e^{-[c(\theta)d(y)]}I_{[a,b]}(y)$ , where $I_{[a,b]}(y)$ is the indicator function and a and b do not depend on $\theta$ .

I think I am supposed to define each function in the density directly above for the orginal question; however, I am having a hard time doing so. Some direction would be great. Thanks.

**harish21** · March 24th 2010, 12:37 PM

Originally Posted by Danneedshelp

Q: Let $Y_{1},...,Y_{n}$ denote a random sample from the probability density function
$f(y|\theta)=\theta\\y^{\theta\\-1}$ for $0<y<1$ , $\theta>0$ and 0 otherwise.
a) Show that this density function is in the (one-parameter) exponential family and that $\sum_{i=1}^{n}-ln(Y_{i})$ is sufficient for $\theta$ (see previous exercise).
b) If $W_{i}=-ln(Y_{i})$ , show that $W_{i}$ has an exponential distribution with mean $\frac{1}{\theta}$ .

Previews exercise mentioned in (a):

1) Suppose that $Y_{1},...,Y_{n}$ is a random sample from a probability density in the (one-parameter) exponential family so that

$f(y|\theta)=a(\theta)b(y)e^{-[c(\theta)d(y)]}I_{[a,b]}(y)$ , where $I_{[a,b]}(y)$ is the indicator function and a and b do not depend on $\theta$ .

I think I am supposed to define each function in the density directly above for the orginal question; however, I am having a hard time doing so. Some direction would be great. Thanks.

You can say that the distribution $f(y,\theta)$ belongs to a one parameter exponential family if you can express its pdf in this form:

$f(y,\theta) = a(\theta) \times g(y) \times exp(b(\theta) \times R(y))$

so your pdf is :
$\theta\\y^{\theta\\-1}$

= $\theta \frac{y^{\theta}}{y}$

= $\theta \times \frac{1}{y} \times {y^{\theta}}$

= $\theta \times \frac{1}{y} \times exp(\theta \times log(y))$

compare the above system with $f(y,\theta) = a(\theta) \times g(y) \times exp(b(\theta) \times R(y))$ , you can say that:

$a(\theta) = \theta$

$g(y) = \frac{1}{y}$

$b(\theta) = \theta$

and $R(y) = log(y)$

and

$\sum_{i=1}^{n} R(y)$ is a complete sufficient statistic for $\theta$

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