What does this Statistic mean? And how to find a density of a statistic? My First Question! But it's in two parts.
Context:
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$$P\left(X_{(n)}<x\right)=P\left(X_{1}<x, X_{2}<x, \ldots, X_{n}<x\right)$$

 A: 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
$$
\begin{align*}
f(x, \theta, \phi) = \frac{d}{dx}F(x, \theta, \phi) = \frac{n}{\theta \phi}\Big(\frac{x}{\theta}\Big)^{\frac{n}{\phi}-1}
\end{align*}
$$
