# Kumaraswamy - Exponential Family

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 $$\frac{\theta y}{a(\phi)}$$

• How can I obtain the expect value and variance of that density? Apr 2, 2019 at 3:20
• en.m.wikipedia.org/wiki/Kumaraswamy_distribution Apr 2, 2019 at 10:51
• They've done it for you there Apr 2, 2019 at 10:52