Question #236605
Let X be a r.v with pdf f(x) = 3x^2, 0 < x < 1. Then

(a) Calculate the quantities E(X), E(X^2) and Var(X).

(b) If the r.v.Y is denoted by Y = 3X − 2, calculate the E(Y ) and Var(Y).
1
Expert's answer
2021-09-14T11:23:03-0400

(a)


E(X)=xf(x)dx=01x(3x2)dxE(X)=\displaystyle\int_{-\infin}^{\infin}xf(x)dx=\displaystyle\int_{0}^{1}x(3x^2)dx

=[3x44]10=34=\big[\dfrac{3x^4}{4}\big]\begin{matrix} 1 \\ 0 \end{matrix}=\dfrac{3}{4}



E(X2)=x2f(x)dx=01x2(3x2)dxE(X^2)=\displaystyle\int_{-\infin}^{\infin}x^2f(x)dx=\displaystyle\int_{0}^{1}x^2(3x^2)dx

=[3x55]10=35=\big[\dfrac{3x^5}{5}\big]\begin{matrix} 1 \\ 0 \end{matrix}=\dfrac{3}{5}

Var(X)=E(X2)(E(X))2Var(X)=E(X^2)-(E(X))^2

=35(34)2=380=0.0375=\dfrac{3}{5}-(\dfrac{3}{4})^2=\dfrac{3}{80}=0.0375

(b)


E(Y)=E(3X2)=3E(X)2E(Y)=E(3X-2)=3E(X)-2

=3(35)2=15=0.2=3(\dfrac{3}{5})-2=-\dfrac{1}{5}=-0.2

Var(Y)=Var(3X2)=(3)2Var(X)Var(Y)=Var(3X-2)=(3)^2Var(X)

=9(380)=2780=0.3375=9(\dfrac{3}{80})=\dfrac{27}{80}=0.3375


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