User:Yoctobarryc

Hi.

Some silly maths fiddling

If the distribution function of a distribution is:

$f(x)=1-(1-x^{a})^{b}$

Then: ${\frac {\partial f(x)}{\partial a}}=x^{a}\ln(x)b(1-x^{a})^{b-1}$

${\frac {\partial f(x)}{\partial b}}=-(1-x^{a})^{b}\ln(1-x^{a})$

${\frac {\partial f(x)}{\partial x}}=ab(1-x^{a})^{b-1}x^{a-1}$

If we take the logarithm of the likelihood function:-

$\ln(f(x))=\ln(1-(1-x^{a})^{b})$

${\frac {\partial \ln(f(x))}{\partial a}}={\frac {-x^{a}b(1-x^{a})^{b-1}\ln(x)}{(1-x^{a})^{b-1}}}=-x^{a}b\ln(x)$

${\frac {\partial \ln(f(x))}{\partial b}}={\frac {(1-x^{a})^{b}\ln(1-x^{a})}{(1-x^{a})^{b-1}}}=(1-x^{a})\ln(1-x^{a})$

${\frac {\partial \ln(f(x))}{\partial x}}={\frac {-ab(1-x^{a})^{b-1}x^{a-1}}{(1-x^{a})^{b-1}}}=-abx^{a-1}$