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The likelihood function is given by \(L(\theta) = \prod_{i=1}^{n} f(x_{i} ; \theta)\). Substituting the probability density function, \(f(x;\theta)\), gives \(L(\theta) =\prod_{i=1}^{n} 2x_{i}/\theta^{2}\). Multiply \(2x_i/\theta^2\) n times for all \(x_i\), then the likelihood function becomes \(L(\theta)=(2/\theta^{2})^{n} \prod_{i=1}^{n} x_{i}\) which is valid when \(0<x_i \leq \theta\) for \(1\leq i \leq n\). Otherwise, the likelihood is zero.To find the MLE, we need to maximize this function. But the problem is easier to solve by instead maximizing the log likelihood, \(l(\theta) = \log(L(\theta))\), because the log is a monotonically increasing function. The log likelihood function is \(l(\theta) = n\log(2) - 2n\log(\theta) + \log\left(\prod_{i=1}^{n} x_{i}\right)\) and its derivative with respect to \(\theta\) equals \(-2n/\theta\).Setting the derivative to zero and solving for \(\theta\) gives rise to no solution. Remember the constraints, the likelihood function equals \(0\) when \(0<x_i< \theta\), and \(\theta\) equals \(0\) when \(x_{i}=\theta\) for some \(i\). Therefore \(x_i= \theta\) for some \(i\) maximizes the likelihood function. So, \(\hat{\theta}= \text{max} \{X_1, ..., X_n\}\) assuming at least one \(x_i\) equals \(\theta\).

The problem asks for constant \(c\) such that \(E(c\hat{\theta})= \theta\). Since \(E(x)= \text{mean of } x\), let's first find out the expectation \(E(\hat\theta)\).Again, remember that \(\hat{\theta}= \text{max} \{X_1, ..., X_n\}\). Now, \(P(\hat{\theta}\leq x)= P(X_1\leq x, ..., X_n\leq x)= P(X_n\leq x)\). And each \(X_i\) has a distribution with \(f(x;\theta)= 2x/\theta^{2}\) for \(0<x \leq \theta\) and \(0\) elsewhere. So the cumulative distribution function (CDF) is \(P(X_i\leq x)= F(x)= x^{2}/\theta^{2}\) for \(0<x \leq \theta\).Then, the CDF of \(\hat{\theta}\), \(P(\hat{\theta}\leq x)\) equals \(x^{2n}/\theta^{2n}\) for \(0<x \leq \theta\). The expectation of \(\hat{\theta}\) can be found by integrating \((1-{x^{2n}/\theta^{2n}})\) from \(0\) to \(\theta\), which equals \(\theta* (1-1/(2n+1))\). Therefore, \(E(c\hat{\theta})= c\theta(1-1/(2n+1))\), and it equals \(\theta\) when \(c= 1/(1-1/(2n+1))\). In other words, \(c= (2n+1)/2n= 1+1/(2n)\).

For this given distribution \(f(x;\theta)=2x/\theta^{2}\) for \(0<x \leq \theta\), the cumulative distribution function is \(F(x)= x^{2}/\theta^{2}\) for \(0<x \leq \theta\). The median \(m\) is the value that makes \(F(m)=1/2\). Solving \(m^{2}/\theta^{2}=1/2\) for \(m\) gives \(m= \theta/\sqrt{2}\). As the median \(m\) equals \(\theta/\sqrt{2}\), which doesn't depend on the distribution of the data (doesn't have terms like \(X_1, ..., X_n\)), the MLE for \(m\), \(\hat{m}\) equals the MLE of \(\theta\), \(\hat{\theta}/\sqrt{2}= \text{max}\{X_1, ..., X_n\}/\sqrt{2}\).