Estimate bias determination

A statistical question for me went:

Suppose $X_1, X_2, ..., X_n$ are identical and independent variables and follow $uniform(0,\theta)$. Show that $\frac{n+1}{n}X_{(n)}$ is an unbiased estimate of $\theta$.

The answer key does a step which I cannot understand: $$f_{X_{(n)}}(x) = nf_X(x)[F_X(x)]^{n-1} \ =\frac{n}{\theta}[\frac{x}{\theta}]^{n-1}$$

Can anyone explain to me how did $$nf_X(x)[F_X(x)]^{n-1}$$ come about? Why am I multiplying $n$ to a $PDF$ to a $CDF^{n-1}$?

Thank you in advance!

Solution 1:

• The probability density for a particular observation is $f_X(x)=\frac 1\theta$ when $0 \le x \le \theta$

• The probability a particular observation is $x$ or less is $F_X(x)=\frac x\theta$ when $0 \le x \le \theta$, the cumulative distribution function for a particular observation and the integral of the density

• The probability all observations are $x$ or less is $F_{X_{(n)}}(x)=\left(\frac x\theta\right)^n$ when $0 \le x \le \theta$, the cumulative distribution function for the maximum observation and $\left(F_{X}(x)\right)^{n}$

• The probability density function for the maximum observation is then the derivative $$f_{X_{(n)}}(x) = F'_{X_{(n)}}(x)=\frac n\theta \left(\frac x\theta\right)^{n-1} = n f_{X}(x) \left(F_{X}(x)\right)^{n-1} $$ when $0 \le x \le \theta$