Question

Let U1, ..., Un be i.i.d. Unif(0, 1), and X = max(U1, ..., Un). What is the PDF of X? What is EX? Hint: Find the CDF of X first, by translating the event X < x into an event involving U1,...,Un.

Answer 1

Answer:

$E(X)= n \int_{0}^1 x^n dx = n [\frac{1}{n+1}- \frac{0}{n+1}]=\frac{n}{n+1}$

Step-by-step explanation:

A uniform distribution, "sometimes also known as a rectangular distribution, is a distribution that has constant probability".

We need to take in count that our random variable just take values between 0 and 1 since is uniform distribution (0,1). The maximum of the finite set of elements in (0,1) needs to be present in (0,1).

If we select a value $x \in (0,1)$ we want this:

$max(U_1, ....,U_n) \leq x$

And we can express this like that:

$u_i \leq x$ for each possible i

We assume that the random variable $u_i$ are independent and $P)U_i \leq x) =x$ from the definition of an uniform random variable between 0 and 1. So we can find the cumulative distribution like this:

$P(X \leq x) = P(U_1 \leq 1, ...., U_n \leq x) \prod P(U_i \leq x) =\prod x = x^n$

And then cumulative distribution would be expressed like this:

$0, x \leq 0$

$x^n, x \in (0,1)$

$1, x \geq 1$

For each value $x\in (0,1)$ we can find the dendity function like this:

$f_X (x) = \frac{d}{dx} F_X (x) = nx^{n-1}$

So then we have the pdf defined, and given by:

$f_X (x) = n x^{n-1} , x \in (0,1)$  and 0 for other case

And now we can find the expected value for the random variable X like this:

$E(X) =\int_{0}^1 s f_X (x) dx = \int_{0}^1 x n x^{n-1}$

$E(X)= n \int_{0}^1 x^n dx = n [\frac{1}{n+1}- \frac{0}{n+1}]=\frac{n}{n+1}$