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(Solved): \[ f(x)=\frac{3 x(2-x)}{4} \mathrm{I}_{(0,2)}(x) \] where \( \mathbf{I}_{(a, b)}(x) \) is the indic ...
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f(x)=\frac{3 x(2-x)}{4} \mathrm{I}_{(0,2)}(x)
\]
where \( \mathbf{I}_{(a, b)}(x) \) is the indicator function defined on t](https://media.cheggcdn.com/study/00c/00c92a0d-481d-482b-abf3-08c612cf0f37/image)
\[ f(x)=\frac{3 x(2-x)}{4} \mathrm{I}_{(0,2)}(x) \] where \( \mathbf{I}_{(a, b)}(x) \) is the indicator function defined on the interval \( (a, b) \). Define a new random variable as, \( Y=X / 2 \). Derive the probability density function of the random variable, \( Y \). Specify the range of the random variable.
Expert Answer
Given f(x)=3x(2?x)4I(0,2)(x) Where I(a,b)(x) is the indicator function defined on the interval (a,b). Define new random variable Y=X2