How do we know the pdf of $X_{max}$?
We have a sample of size $n=6$ from a uniform distribution of the interval $[0,\theta]$. Let $\hat{\theta}=X_{max}$ be the estimator of $\theta$.
(a) Given $\theta=3$ find the probability that $\hat{\theta}$ falls within 0.2 of $\theta$;
(b) Calculate the probability from (a) when sample size $n=3$.
Recall, that $$\text{pdf of $X_{max}$ is $f_{X_{max}}(x)=\frac{n}{\theta^n}x^{n-1}$, $EX_{max} = \frac{n}{n+1}\theta$, and $\operatorname{Var}(X_{max})=\frac{n\theta^2}{n+2}-\left(\frac{n}{n+1}\right)^2\theta^2.$}$$
Original problem image: https://i.stack.imgur.com/74eus.png
We can derive the pdf by calculating the cdf and taking the derivative. The probability that $X_{max}\le x$ for $x$ between $0$ and $\theta$ is the probability that all $n$ measurements fall between $0$ and $x$. But for a single measurement, this is $\frac{x}{\theta}$. The probability $n$ independent measurements fall between $0$ and $x$ is then $\left(\frac{x}{\theta}\right)^n=\frac{x^n}{\theta^n}$. Now we take the derivative to find the pdf, namely $\frac{nx^{n-1}}{\theta^n}$.