Maximum likelihood estimate of $a$ from a single observation with CDF $F(x)=\frac{ax^2}{1+ax^2}$ on $x>0$

Question

For this question, I think the approach that I'm using is correct but I'm not able to get to the right answer. Since we are given F(x), I first differentiated it to get f(x). Since one observation is drawn, L(a) would just be f(x). I then took the log of L(a) to get l(a) which I then differentiated and set to 0 to get an estimate for a but when I set my answer to 0 I get 3=0 which is obviously wrong.

Answer 1

The probability distribution function becomes $$F'(x) = f(x) = \frac{2ax(1 + ax^2) - 2a^2x^3}{(1+ax^2)^2} = \frac{2ax}{(1+ax^2)^2}.$$

Then, taking ln, we get $$\ell f(a) = ln(2ax) - 2ln(1 + ax^2), $$ $$\ell f'(a) = \frac{1}{a} - \frac{2x^2}{1 + ax^2}. $$

Then you get $$\ell f'(a) = 0 \Longleftrightarrow 2x^2a = 1+ax^2 \Longleftrightarrow a = \frac{1}{x^2}.$$

Plugging in, you get $\hat{a} \approx 0.11$.