Maximum-likelihood estimation of parameter of a distribution function

Question

My Problen:

A Random variable $X$ has the following distribution function: $$ F_X(x)=% \begin{cases} x^\theta, &x\in[0,1]\\ 0, &\text{otherwise}. \end{cases} $$

We have gotten x1 = 0.40 x2 = 0.75 x3 = 0.95 from three independent trials. Determine the Maximum-Likelihood estimation of θ.

Solution
2.39

My attempt of solving the problem
L(θ) = x₁^θx₂^θx₃^θ = (0.40^θ 0.75^θ 0.95^θ)
f(θ) = ln(L(θ)) = ln((0.40^θ 0.75^θ 0.95^θ))
f(θ) = ln((0.40^θ 0.75^θ 0.95^θ)) = ln(0.40^θ) + ln(0.75^θ) + ln(0.95^θ)
At this point I usually make the derivative of f(θ) equal to zero to get θ, but when I try to do so I get
$$\frac{d}{dx}(f(θ)) = (-0.916291) + (-0.287682) + (-0.0512933) = -1.2552663$$
The θ disappears and I get left with -1.2552663 which is not even close to the correct answer.

My Question:
I am clearly getting the wrong answer.
I would be tankful for any help and/or guidance.

Answer 1

You are misunderstanding the question. the function $x^\theta$ is the cumulative distribution function, not the density. In other words, $$\Pr[X \le x] = F_X(x) = x^\theta, \quad x \in [0,1].$$ As such, the density is $$f_X(x) = \theta x^{\theta-1}, \quad x \in [0,1]$$ and this is what you need to use to compute the likelihood: $$\mathcal L(\theta \mid x_1, x_2, x_3) = \theta^3 (x_1 x_2 x_3)^{\theta - 1}.$$ Consequently the log-likelihood is $$\ell(\theta \mid x_1, x_2, x_3) = 3 \log \theta + (\theta - 1) \log(x_1 x_2 x_3)$$ and its critical points satisfy $$0 = \frac{\partial \ell}{\partial \theta} = \frac{3}{\theta} + \log(x_1 x_2 x_3),$$ or $$\hat \theta = -\frac{3}{\log (x_1 x_2 x_3)}.$$ What happens when you substitute the given observations into this expression?