Rv uniformly distributed

[Francobati]

Hello.
Let $ Y=1-X^2 $, where $ X~ U(0,1) $. What statement is TRUE?
-$ E(Y^2)=2 $
- $ E(Y^2)=1/2 $
- $ var(Y)=1/12 $
- $ E(Y)=E(Y^2) $
-None of the remaining statements.
Solution:
I compute: $ E(Y^2)=E(1-X^2)^2=E(1+X^4-2X^2)=1+E(X^4)-2E(X^2) $, then?

 

[I like Serena]

Hello.
Let $ Y=1-X^2 $, where $ X~ U(0,1) $. What statement is TRUE?
-$ E(Y^2)=2 $
- $ E(Y^2)=1/2 $
- $ var(Y)=1/12 $
- $ E(Y)=E(Y^2) $
-None of the remaining statements.
Solution:
I compute: $ E(Y^2)=E(1-X^2)^2=E(1+X^4-2X^2)=1+E(X^4)-2E(X^2) $, then?

Apply the definition of expected value.
That is:
$$EZ = \int z f_Z(z) \, dz$$
So with $X\sim U(0,1)$:
$$E(X^2) = \int_0^1 x^2 \cdot 1 \, dx$$

 

[Francobati]

$ E(X^2)=\frac{1^3-0^3}{3*1} $
$ E(X^4)=\frac{1^5}{5}$
$ E(Y^2)=1+\frac{1}{5}-2*\frac{1}{3}=1+\frac{1}{5}-\frac{2}{3}=\frac{15+3-10}{15}= \frac{8}{15}\ne2\ne\frac{1}{2} $, so first and second are false.
$ var(Y)=var(1-X^2)=var(1)-var(X^2)=0-var(X^2) $
$ var(X)=\frac{(b-a)^2}{12}=\frac{(1-0)^2}{12}=\frac{1}{12} $
But waht formula I must appky in $E(X^2)$ and in $E(X^4)$ to obtain these values?

 

[I like Serena]

I take it you mean $var(X^2)$?

To find it, apply the definition of variance:
$$var(Z) = E\big((Z-EZ)^2\big) = E\big(Z^2\big) - (EZ)^2$$

 

[Francobati]

Yes and I obtain $E(X^2)=var(X)+(E(X))^2=\frac{(b-a)^2}{12}+(\frac{a+b}{2})^2=\frac{1}{12}+(\frac{1}{2})^2=\frac{1}{12}+\frac{1}{4}=\frac{1}{3}$
This result equal to this $E(X^2)=\frac{1^3-0^3}{3(1)}= \frac{1}{3}$, how I can translate this $E(X^2)=\frac{1^3-0^3}{3(1)}$ in a general formula?