# Talk:Kumaraswamy distribution

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## Max likelihood estimates of a and b

Can anyone get the log likelihood estimates of a & b? b as a function of a is pretty easy, but a doesn't seem to have a closed form solution. Martin (talk) 11 November 2006

I think you are right - about the best I can do is this - Define

$f(a)=\ln\left(\prod(1-x_i^a)\right)$
$g=\ln\left(\prod x_i\right)$

Then the likelihood function is

$L=a^Nb^Ne^{(a-1)g+(b-1)f(a)}\,$

and the log likelihood function is

$\ell = N\ln(a)+N\ln(b)+(a-1)g+(b-1)f(a)\,$

The maximum occurs at

$\frac{\partial \ell}{\partial a}=0=N/a+g+(b-1)\frac{df}{da}$
$\frac{\partial \ell}{\partial b}=0=N/b+f$

Eliminating b we get

$\left(N/a+g\right)da=\left(N/f+1\right)df$

Integrating, we get

$N\ln(a)+ga+K=N\ln(f)+f\,$

where K is some constant (I think it can be determined). That equation cannot be analytically solved for f(a). If it could, we would still have to substitute that into the defining equation for f(a) and solve for a. PAR 16:57, 11 November 2006 (UTC)

Actually, the best way would be to forget the integration, and match the derivatives instead. The derivative at maximum would be

$\frac{df}{da}=\frac{N/a+g}{N/f(a)+1}\,$

and you would equate that to the derivative of the defining equation for f(a) and solve numerically for a. PAR 23:49, 11 November 2006 (UTC)

## the moments

I tried to find the variance of this distribution from the moments given in this paper, but it became negative! Can someone check their accuracy? --Pejman47 (talk) 15:38, 29 June 2008 (UTC)