A random sample of size $n$ is taken from the pdf $$f_Y(y;\theta)=\frac{2y}{\theta^2}, \quad 0 \le y \le \theta$$ Find an expression for $\hat{\theta}$, the maximum likelihood estimation for $\theta$.
$$L(\theta) = \prod_{i=1}^{n} \frac{2y_i}{\theta^2}= \frac{2^n \prod\limits_{i=1}^{n}y_i}{\theta^{2n}}\\ \ln(L(\theta)) = n\ln(2) + \ln\big(\prod_{i=1}^{n}y_i\big)-2n\ln(\theta)$$ Normally at this point I would take the derivative. Let $$T = \frac{d\big[\ln(L(\theta))\big]}{d\theta}=\frac{2n}{\theta}$$ But I am not sure if I have done something wrong to get to here because when I let $T=0$, I get $0=\dfrac{2n}{\theta}$.
Can anyone see something that I am missing maybe through ignorance or through silly error? Thanks.
This question is relating to using order statistics as maximum likelihood estimates. I'm reading about it now in the text book, but if someone knows how to do it, please do.
From me reading just now, obviously the differentiation does not work, so I have to actually look at the likelihood function itself and analyse it. $L(\theta)$ is maximised by minimising $\theta$, but from ordered statistics technique, theta cannot be less than than $Y_n$
So let me try this:
$$L(\theta) = \frac{2y}{y_n}$$
When adding the condition that $y\leq \theta$, the log-likelihood becomes \begin{align*} \ln(L(\theta)) &= n\ln(2) + \ln\big(\prod_{i=1}^{n}y_i \mathbb 1(y_i\leq \theta)\big)-2n\ln(\theta)\\ &=n\ln(2) + \ln\big(\mathbb 1(\max_i(y_i)\leq \theta) \prod_{i=1}^{n}y_i \big)-2n\ln(\theta)\\ &=n\ln(2) + \ln(\mathbb 1(\max_i(y_i)\leq \theta)) + \ln\big(\prod_{i=1}^{n}y_i \big)-2n\ln(\theta)\\ &=\begin{cases} -\infty &\text{if }\theta<\max_i(y_i)\\ n\ln(2) + \ln\big(\prod_{i=1}^{n}y_i \big)-2n\ln(\theta)&\text{otherwise} \end{cases} \end{align*}
This means you only consider $\theta\geq \max_{i} (y_i)$ and so since your derivative in this case is $\frac{-2n}{\theta}<0$, then you need to take the smallest value value allowed by the domain, which is $\hat \theta = \max_{i}(y_i)$