How to Solve Continuous Random Variables — Univariate RV — distribution of a transform Y = X^2 Questions on Exam P

Exam P Topic: Continuous Random Variables — Univariate RV — distribution of a transform Y = X^2 Verified Procedural Question
Sample Practice Problem ID: #QCO4J
A laboratory's output \(X\) is continuous with density \(f_X\). A calibrated output is \(Y=2X\). Which expression is the density of \(Y\)?
(A)\(F_Y(y)=F_X(2y)\)
(B)\(f_Y(y)=2f_X\!\left(\dfrac{y}{2}\right)\)
(C)\(F_Y(y)=1-F_X\!\left(\dfrac{y}{2}\right)\)
(D)\(f_Y(y)=f_X\!\left(\dfrac{y}{2}\right)\)
(E)\(f_Y(y)=\dfrac{1}{2}f_X\!\left(\dfrac{y}{2}\right)\)
📖 Worked Solution & Strategy
The calibration is a one-to-one increasing transformation because its scale factor is positive. Its CDF follows by transforming an inequality, and differentiating that CDF supplies the required Jacobian factor for the density. Because \(Y=2X\) with \(2>0\), the cumulative distribution function (CDF) of \(Y\) is \(F_Y(y)=\Pr(Y\le y)=\Pr(2X\le y)=F_X\!\left(\dfrac{y}{2}\right)\). Differentiating with respect to \(y\) yields the probability density function \(f_Y(y)=\dfrac{d}{dy}F_X\!\left(\dfrac{y}{2}\right)=\dfrac{1}{2}f_X\!\left(\dfrac{y}{2}\right)\). Thus the density identity is \(f_Y(y)=\dfrac{1}{2}f_X\!\left(\dfrac{y}{2}\right)\), which matches the correct option.
Therefore, the correct answer is option (E).

Final Answer: Option (E)

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