My First Question! But it's in two parts.
I am given a Probability Density Function, and the question wants me to find the density of a statistic.

Given pdf:

$$f(x, \theta, \phi)=\frac{1}{\theta \phi}\left(\frac{x}{\theta}\right)^{\frac{1-\phi}{\phi}}, \quad 0 \leq x \leq \theta, \quad 0<\phi<1 $$

Show that the density of the statistic $$T=X_{(n)}$$​ is given by
$$ f_{X_{(n)}}(x)=\frac{n}{\theta \phi}\left(\frac{x}{\theta}\right)^{n / \phi-1} \quad \text { for } 0 \leq x \leq \theta $$otherwise zero.

Part 1:
What Statistic is this?$$T=X_{(n)}$$​ I don't know what it means or what is it called. Is it the nth observation?

Part 2:
How do you find the density of a statistic? If the statistic was the mean, would it possible to find the density of the mean?
Am I making sense? I am not primarily from a Stat Background, but I've been holding on due to my math so far.
But there might be some painfully obvious things I'm unaware about.
Thank You for being patient.

Edit: The question mentions a hint, I am not looking for the solution but rather explanations for my conceptions, so this might be unnecessary but nonetheless.

Hint: You might consider using the following$$P\left(X_{(n)}<x\right)=P\left(X_{1}<x, X_{2}<x, \ldots, X_{n}<x\right)$$

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    $\begingroup$ It must have been mentioned before that $X_{(n)}=\max\{X_1,X_2,\ldots,X_n\}$. See en.wikipedia.org/wiki/Order_statistic. $\endgroup$ – StubbornAtom Jul 31 at 14:21
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    $\begingroup$ See this question/answer. $\endgroup$ – knrumsey Jul 31 at 14:30
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    $\begingroup$ This is a maths rather than stats question in that "statistics" are transforms of a random sample and hence the distribution of a statistic is found by the rules of probability theory regulating the distribution of a transform, eg the Jacobian rule when the transform is one-to-one. $\endgroup$ – Xi'an Jul 31 at 14:46

The given pdf is $$f(x, \theta, \phi) = \frac{1}{\theta\phi}\Big(\frac{x}{\theta}\Big)^{\frac{1 - \phi}{\phi}}, \ \ \ 0 \leq x \leq \theta, 0 < \phi < 1$$ For a independent and identically distributed sample $X_i, i = 1, \dots, n$ from this pdf $f(x, \theta, \phi)$, the statistic $T = X_{(n)}$ is just $$ T = X_{(n)} = \max \{X_{1}, \dots, X_{n} \} $$ To find the density of $T$ you can use the given hint, as $X_{(n)}$ is the maximum, $$ \begin{align*} \mathbb P(X_{(n)} \leq x) =& \ \mathbb {P}(X_{1} \leq x, \dots, X_{n} \leq x) \\ =& \ \prod_{i = 1}^{n}\mathbb {P}(X_{i} \leq x) \ \ \text{As each} \ X_i \ \text{is iid}\\ =& \ \mathbb {P}(X\leq x)^{n} \end{align*} $$ Therefore we know that the CDF of $T = X_{(n)}$ is given by $\mathbb {P}(X\leq x)^{n}$. To find $\mathbb {P}(X\leq x)$ we integrate $f$ $$ \begin{align*} \mathbb {P}(X\leq x) =& \int_{0}^{x}\frac{1}{\theta\phi}\Big(\frac{u}{\theta}\Big)^{\frac{1 - \phi}{\phi}} du\\ =& \frac{1}{\theta}\Big(\frac{u}{\theta}\Big)^{\frac{1}{\phi}}\Bigg|_0^x\\ =& \frac{1}{\theta}\Big(\frac{x}{\theta}\Big)^{\frac{1}{\phi}} \end{align*} $$ Therefore the cdf of $T = X_{(n)}$ is $$ F(x, \theta, \phi) = \mathbb {P}(X\leq x)^n = \frac{1}{\theta}\Big(\frac{x}{\theta}\Big)^{\frac{n}{\phi}} $$ and hence the density is just the derivative of the CDF $$ \begin{align*} f(x, \theta, \phi) = \frac{d}{dx}F(x, \theta, \phi) = \frac{n}{\theta \phi}\Big(\frac{x}{\theta}\Big)^{\frac{n}{\phi}-1} \end{align*} $$

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