What is the MLE $\theta^*$ of $\theta$?

Question

I have that $x_1, x_2,...,x_n$ are from a rv $X$ that has the density function $f_X(x)=\frac{2x}{\theta^2} \quad$ for $0 \le x \le \theta \quad$ and $f_X(x)=0 \quad$ otherwise. Ihave to determine the MLE of $\theta^*$ of $\theta$
Here is how I have done it:
$L(\theta)= \frac{2}{\theta^{2n}}\prod_{i=1}^nx_i$
$\frac{\partial L(\theta)}{\partial \theta} =...=\frac{-4n}{\theta^{2n+1}}\prod_{i=1}^nx_i + \frac{2}{\theta^{2n}}\frac{\partial(\prod_{i=1}^nx_i)}{\partial \theta}$
Is this correct? and also how do I calculate the CDF $F_{\theta^*}$, the pdf $f_{\theta^*}$ and the expectation $E[\theta^*]$ of the maximum likelihood estimator $\theta^*$?

Answer 1

The likelihood you wrote is wrong. You did not consider that the support of $X$ depends on $\theta$

The correct likelihood is

$$L(\theta)\propto \frac{1}{\theta^{2n}}\cdot\mathbb{1}_{[x_{(n)};\infty)}(\theta)$$

now, looking at your L, it is evident that it is strictly decreasing, thus

$$\hat{\theta}_{ML}=X_{(n)}$$

where $X_{(n)}$ is the max of the observations

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Distribution of Max

Set $Z=max(X_i)$

$$F_Z(z)=P(Z\leq z)=P(X_1\leq z,\dots X_n\leq z)=P(X_1\leq z)\dots P(X_n \leq z)=$$

$$=[P(X_1\leq z)]^ n$$

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To find the distribution of $\hat{\theta}_{ML}=X_{(n)}$ consider that the CDF of Max is the product of the single CDF's thus

$$F_X(x)=\int_{-\infty}^x f(t) dt=\frac{1}{\theta^2}\int_0^x 2t dt=\frac{x^2}{\theta^2}$$

Thus, setting $Z=\hat{\theta}$ you get

$$F_Z(z)=\frac{z^{2n}}{\theta^{2n}}$$

derivating you get the density and then calculate the expectation with the usual definition

$$f_Z(z)=\frac{d}{dz}F_Z=\frac{2n z^{2n-1}}{\theta^{2n}}$$

$$E(Z)=\int_0^\theta z f_Z(z)dz$$