Thread: Logarithmic distribution: Find the canonical parameter

1. Logarithmic distribution: Find the canonical parameter

Hi,

The logarithmic's probability function is:

$f(y;p)=\frac{-1}{ln(1-p)}\frac{p^{^{y}}}{y} where y=1,2...$

To find the canonical parameter, one must re-express the above to the generic pdf of exponential distribution:

$exp(\frac{y\theta -b (\theta )}{\phi }+ c(y,\phi))$

I have managed to:

1: take log then exp:

$exp(ln(\frac{-p^{y}}{yln(1-p)}))$

2. apply log rules:

$exp(-ylnp-lny-ln(ln(1-p))$

Unfortunately, this is where I could not expand further to get the expression to be in the generic form. So far, I know (rightly or wrongly):
$c(y,\phi ) = -lny$
$b(\theta ) = ln(\theta )$
$\phi=1$
but $\theta$= ???

Thanks.

2. Re: Logarithmic distribution: Find the canonical parameter

Originally Posted by meeksoup
Hi,

The logarithmic's probability function is:

$f(y;p)=\frac{-1}{ln(1-p)}\frac{p^{^{y}}}{y} where y=1,2...$

To find the canonical parameter, one must re-express the above to the generic pdf of exponential distribution:

$exp(\frac{y\theta -b (\theta )}{\phi }+ c(y,\phi))$

I have managed to:

1: take log then exp:

$exp(ln(\frac{-p^{y}}{yln(1-p)}))$

2. apply log rules:

$exp(-ylnp-lny-ln(ln(1-p))$

Unfortunately, this is where I could not expand further to get the expression to be in the generic form. So far, I know (rightly or wrongly):
$c(y,\phi ) = -lny$
$b(\theta ) = ln(\theta )$
$\phi=1$
but $\theta$= ???

Thanks.
you're pretty close

$\ln\left(\dfrac{-1}{\ln(1-p)}\dfrac{p^y}{y}\right)=\ln\left(\dfrac{-1}{\ln(1-p)}\right)+y\ln(p)-ln(y)$

by inspection

$\phi=1$

$\theta=\ln(p)$

$b(\theta)=-\ln\left(\dfrac{-1}{\ln(1-p)}\right)=-\ln\left(\dfrac{-1}{\ln(1-e^\theta)}\right)$

$c(y,\phi)=-\ln(y)$

Thanks.