I'm trying to proof that Kumaraswamy belongs to the exponential family and then get the expect value and it's variance.
The Kumaraswamy density distribution is given by:
$$ f(y;\alpha,\beta) = \alpha\beta y^{(\alpha-1)}(1-y^{\beta-1})$$
We know that a exponential family has the follow format:
$$f(x) = exp{\left(\frac{\theta x- b(\theta)}{a(\phi)} + c(x,\theta)\right)}$$
1) First step is take log of all terms in the equation: $$ln(f(y;\alpha,\beta)) = \left( ln(\alpha) + ln(\beta) + (\alpha - 1) ln(y) + (\beta -1 ) ln(1-y^{\alpha}) \right) $$
2) Second step is take the exponential of the terms:
$$f(y;\alpha,\beta) = exp{\left( ln(\alpha) + ln(\beta) + (\alpha - 1) ln(y) + (\beta -1 )ln(1-y^{\alpha}) \right)}$$
Then, when I expand the equation: $$f(y;\alpha,\beta) = exp{\left( ln(\alpha) + ln(\beta) + \alpha ln(y) - ln(y) + \beta ln(1-y^{\alpha}) - ln(1-y^{\alpha}) \right)}$$
Forward I don't know to adjust the terms to find the exponential family.
Somebody can help me?
Thank you!
I hate to break it to you, but it isn't part of the exponential family.
Typo in q: should be $(1-y^\alpha)^{\beta-1}$ in the PDF.
$\ln f(y; \alpha, \beta) = \ln \alpha + \ln \beta + (\alpha-1)\ln y - (\beta-1)\ln(1-y^{\alpha}) $
The final term has $y, \alpha , \beta$ in it, while not being a linear function of $y$, so it can't be said to be of the form $\frac{\theta y}{a(\phi)}$