Prove Logarithmic function is part of exponential family

Question

The aim is to prove that the logarithmic distribution with parameter $p (0<p<1)$ is part of the exponential family and hence, give its canonical parameter.

To prove a distribution is part of the exponential family, one must express the probability function in the generic form of $$\exp\left(\frac{y\theta -b (\theta )}{\phi }+ c(y,\phi)\right)$$ where $\theta$ is the canonical parameter.

The function for logarithmic is $$f(y;p)=\frac{-1}{\ln(1-p)}\frac{p^{y}}{y}$$ where $y=1,2,\ldots$

I have managed to re-express the function by:

1st step: $$\exp\left(\ln\left(\frac{-p^{y}}{y\ln(1-p)}\right)\right)$$

2nd step: $$\exp\left(-y\ln (p)-\ln (y)-\ln(\ln(1-p)\right)$$

However, I'm stuck at step 2 and can't expand this further to the generic form. So far, I know (rightly or wrongly): $$c(y,\phi ) = -\ln y$$ $$b(\theta ) = \ln(\theta )$$ $$\phi=1$$ $$\theta=??$$

Can anyone please help?

Thanks.

Answer 1

I just finished a similar question and I had $$f(y;p)=\exp\{y \ln(p)-\ln[\ln(1-p)]+\ln(y)\}.$$ It is the same as yours, but I put the minus sign in the denominator of the probability distribution. Note that $\theta$ is multiplying y in the exponential form, so that $\theta= \ln(p)$, and thus $p=e^\theta$. Note also that $b(\theta)=\ln[\ln(1-p)]$, and $\phi=1$ in this case. In the form that I wrote, $c(y, \phi)=\ln(y)$. I hope this helps.

Answer 2

Suppose we write the logarithm function $$f(y;p)-\frac{1}{log(1-p)}\frac{p^y}{y},~~where~y=1,2,~~0<p<1......$$ We can find it is in the exponential family, $$f(y;p)-\frac{1}{log(1-p)}\frac{p^y}{y}=exp\Big[log(\frac{-p^y}{y\cdot log(1-p)})\Big] = exp\Big[log(\frac{p^y}{y\cdot log(\frac{1}{1-p})})\Big]= exp(y\cdot log(p)-log(y)-log(log(\frac{1}{1-p})))$$ Therefore, $c(y, \phi)=-log(y)$, $\theta=-lop(p)$, $b(\theta)=-log(log(\frac{1}{1-p}))=-log(-log(1-p))$, $a(\phi)=1$

$\bullet$ The expectation of logarithm function: taking the first derivative of $b(\theta)$ $$E(y)=b'(\theta)=\frac{d}{d\theta}b(\theta)=-p\cdot \frac{d}{dp}log(-log(1-p))=-p\cdot \frac{1}{-log(1-p)}\frac{1}{1-p}\cdot (-1)=\frac{-p}{log(1-p)\cdot(1-p)}$$ This corresponds greatly to the mean I search on Wikipedia. (https://en.wikipedia.org/wiki/Logarithmic_distribution) However, for the variance, I got stuck on the answer.

I do the variance as follows:

$\bullet$ The variance of logarithm function: taking the second derivative of $b(\theta)$ $$Var(y)=a(\phi)\cdot b''(\theta)=b''(\theta)=\frac{d}{d\theta}b'(\theta)=-p\frac{d}{dp}\Big[\frac{-p}{log(1-p)\cdot(1-p)}\Big]=-p\cdot\Big[\frac{-log(1-p)\cdot(1-p)-(-1-log(1-p)(-p)}{log(1-p)^2\cdot(1-p)^2}\Big]$$

$$=\frac{p\cdot log(1-p)(1-p)-p-p\cdot log(1-p)}{log(1-p)^2\cdot(1-p)^2}=\frac{-p^2\cdot log(1-p)-p}{log(1-p)^2\cdot(1-p)^2}=\frac{-p\cdot(1+p\cdot log(1-p))}{log(1-p)^2\cdot(1-p)^2}$$ which is wrong.

Yet another method:

$$Var(y) = b''(\theta)=\frac{d^2}{d\theta^2}b(\theta)=p^2\frac{d^2}{dp^2}\Big[log(-log(1-p))\Big]=p^2\cdot \frac{d}{dp}\Big[\frac{1}{log(1-p)\cdot(1-p)}\Big]=p^2 \cdot \frac{1+log(1-p)}{log(1-p)^2\cdot(1-p)^2}$$ $$=\frac{p^2\cdot (1+log(1-p))}{log(1-p)^2\cdot(1-p)^2}$$ which is still wrong. Could anyone help?