2021-09-13T04:22:50-04:00
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
2021-09-14T11:23:03-0400
(a)
E ( X ) = ∫ − ∞ ∞ x f ( x ) d x = ∫ 0 1 x ( 3 x 2 ) d x E(X)=\displaystyle\int_{-\infin}^{\infin}xf(x)dx=\displaystyle\int_{0}^{1}x(3x^2)dx E ( X ) = ∫ − ∞ ∞ x f ( x ) d x = ∫ 0 1 x ( 3 x 2 ) d x
= [ 3 x 4 4 ] 1 0 = 3 4 =\big[\dfrac{3x^4}{4}\big]\begin{matrix}
1 \\
0
\end{matrix}=\dfrac{3}{4} = [ 4 3 x 4 ] 1 0 = 4 3
E ( X 2 ) = ∫ − ∞ ∞ x 2 f ( x ) d x = ∫ 0 1 x 2 ( 3 x 2 ) d x E(X^2)=\displaystyle\int_{-\infin}^{\infin}x^2f(x)dx=\displaystyle\int_{0}^{1}x^2(3x^2)dx E ( X 2 ) = ∫ − ∞ ∞ x 2 f ( x ) d x = ∫ 0 1 x 2 ( 3 x 2 ) d x
= [ 3 x 5 5 ] 1 0 = 3 5 =\big[\dfrac{3x^5}{5}\big]\begin{matrix}
1 \\
0
\end{matrix}=\dfrac{3}{5} = [ 5 3 x 5 ] 1 0 = 5 3
V a r ( X ) = E ( X 2 ) − ( E ( X ) ) 2 Var(X)=E(X^2)-(E(X))^2 Va r ( X ) = E ( X 2 ) − ( E ( X ) ) 2
= 3 5 − ( 3 4 ) 2 = 3 80 = 0.0375 =\dfrac{3}{5}-(\dfrac{3}{4})^2=\dfrac{3}{80}=0.0375 = 5 3 − ( 4 3 ) 2 = 80 3 = 0.0375
(b)
E ( Y ) = E ( 3 X − 2 ) = 3 E ( X ) − 2 E(Y)=E(3X-2)=3E(X)-2 E ( Y ) = E ( 3 X − 2 ) = 3 E ( X ) − 2
= 3 ( 3 5 ) − 2 = − 1 5 = − 0.2 =3(\dfrac{3}{5})-2=-\dfrac{1}{5}=-0.2 = 3 ( 5 3 ) − 2 = − 5 1 = − 0.2
V a r ( Y ) = V a r ( 3 X − 2 ) = ( 3 ) 2 V a r ( X ) Var(Y)=Var(3X-2)=(3)^2Var(X) Va r ( Y ) = Va r ( 3 X − 2 ) = ( 3 ) 2 Va r ( X )
= 9 ( 3 80 ) = 27 80 = 0.3375 =9(\dfrac{3}{80})=\dfrac{27}{80}=0.3375 = 9 ( 80 3 ) = 80 27 = 0.3375
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