Calculating the Variance for X^2: How to Find the Correct Solution

[Addez123]

Homework Statement

X is equally distributed along the intervall [0,1]
Calculate variance for X^2

Relevant Equations

Variance:
$$\int (x-u)^2*fx(x) dx$$

At first I assumed u to be 1/2 since X is equally distributed along 0-1.
$$\int (x-1/2)^2*x^2 dx = 1/30$$
The correct answer should be 4/45.

I would calculate the u but I think I do it wrong.
If fx(x) = x^2 then what is g(x)?

fx is the probability density function, which is the x^2 they supplied, right?
g(x) i have no real definition for.

 

[Orodruin]

fx(x) = x^2

This is not the pdf of a uniform distribution on [0,1]

 

[Addez123]

This is not the pdf of a uniform distribution on [0,1]

So g(x) = x^2, and fx(x) = 1?
Because then you get u = 1/3rd and variance = 1/9th :/

 

[haruspex]

So g(x) = x^2, and fx(x) = 1?
Because then you get u = 1/3rd and variance = 1/9th :/

Show how you are computing the variance.
(I think it is easiest to use E(X²)-(E(X))².)

 

[Addez123]

$$\int (x-1/3)^2 * 1 dx$$
from 0 to 1
EDIT: It should say x^2 since we're donig variance for v(x^2) not v(x). That gives the correct solution.

 

[haruspex]

$$\int (x-1/3)^2 * 1 dx$$
from 0 to 1
EDIT: It should say x^2 since we're donig variance for v(x^2) not v(x). That gives the correct solution.

Yes.