MHB What Is the Expected Value of Y Squared for a Transformed Uniform Variable?

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The discussion revolves around calculating the expected value of Y squared, where Y is defined as 1 minus the square of a uniformly distributed variable X over the interval (0,1). The calculations reveal that E(Y^2) equals 8/15, disproving the options E(Y^2) = 2 and E(Y^2) = 1/2. Variance calculations show that var(Y) is derived from the variance of X squared, leading to a value of 1/12. The participants seek clarification on the formulas for expected values and variance, particularly in relation to transformed variables. The thread emphasizes the importance of applying the correct definitions and formulas in probability and statistics.
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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?
 
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Francobati said:
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 $X\sim U(0,1)$:
$$E(X^2) = \int_0^1 x^2 \cdot 1 \, dx$$
 
$ 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 take it you mean $var(X^2)$?

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