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:-

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

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

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

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