Respuesta :
Answer:
[tex]E(X)= n \int_{0}^1 x^n dx = n [\frac{1}{n+1}- \frac{0}{n+1}]=\frac{n}{n+1}[/tex]
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 [tex]x \in (0,1)[/tex] we want this:
[tex]max(U_1, ....,U_n) \leq x[/tex]
And we can express this like that:
[tex]u_i \leq x[/tex] for each possible i
We assume that the random variable [tex]u_i[/tex] are independent and [tex]P)U_i \leq x) =x[/tex] from the definition of an uniform random variable between 0 and 1. So we can find the cumulative distribution like this:
[tex]P(X \leq x) = P(U_1 \leq 1, ...., U_n \leq x) \prod P(U_i \leq x) =\prod x = x^n [/tex]
And then cumulative distribution would be expressed like this:
[tex]0, x \leq 0[/tex]
[tex]x^n, x \in (0,1)[/tex]
[tex]1, x \geq 1[/tex]
For each value [tex]x\in (0,1)[/tex] we can find the dendity function like this:
[tex]f_X (x) = \frac{d}{dx} F_X (x) = nx^{n-1}[/tex]
So then we have the pdf defined, and given by:
[tex]f_X (x) = n x^{n-1} , x \in (0,1)[/tex] and 0 for other case
And now we can find the expected value for the random variable X like this:
[tex]E(X) =\int_{0}^1 s f_X (x) dx = \int_{0}^1 x n x^{n-1}[/tex]
[tex]E(X)= n \int_{0}^1 x^n dx = n [\frac{1}{n+1}- \frac{0}{n+1}]=\frac{n}{n+1}[/tex]
