# probability density of the maximum of samples from a uniform distribution

Suppose

$$X_1, X_2, \dots, X_n\sim Unif(0, \theta), iid$$

and suppose

$$\hat\theta = \max\{X_1, X_2, \dots, X_n\}$$

How would I find the probability density of $\hat\theta$?

Thank you!

#### Solutions Collecting From Web of "probability density of the maximum of samples from a uniform distribution"

\begin{align}
P(Y\leq x)&=P(\max(X_1,X_2 ,\cdots,X_n)\leq x)\\&=P(X_1\leq n,X_2\leq n,\cdots,X_n\leq x)\\
&\stackrel{ind}{=} \prod_{i=1}^nP(X_i\leq x )\\&=\prod_{i=1}^n\dfrac{x}{\theta}\\&=\left(\dfrac{x}{\theta}\right)^n
\end{align}

Let random variable $W$ denote the maximum of the $X_i$. We will assume that the $X_i$ are independent, else we can say very little about the distribution of $W$.

Note that the maximum of the $X_i$ is $\le w$ if and only if all the $X_i$ are $\le w$. For $w$ in the interval $[0,\theta]$, the probability that $X_i\le w$ is $\frac{w}{\theta}$. It follows by independence that the probability that $W\le w$ is $\left(\frac{w}{\theta}\right)^n$.

Thus, in our interval, the cumulative distribution function $F_W(w)$ of $W$ is given by
$$F_W(w)= \left(\frac{w}{\theta}\right)^n.$$
Differentiate to get the density function of $W$.