Question

suppose the density function of y is defined as, f(y) =3/2(1-y²), 0≤y≤1 calculate expectation of y
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Answer 1

Given :   the density function of y is defined as, f(y) =(3/2)(1-y²), 0≤y≤1

To Find : calculate expectation of y

Solution:

expectation of y

=  ∫ y. f(y) .dy

f(y) =(3/2)(1-y²), 0≤y≤1

Expected value

$=\int\limits^1_0 {y.\frac{3}{2}(1-y^2) } \, dy$

$=\frac{3}{2}\int\limits^1_0 { (y-y^3) } \, dy$

$=\frac{3}{2} \left [\frac{y^2}{2} - \frac{y^4}{4} \right]_0^1$

= (3/2)  ( 1/2 - 1/4)

= (3/2) ( 1/4)

= 3/8

expectation of y = 3/8

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