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p.d.f. of random variable X:

f(x) = \begin{cases} {1\over 2}x & \text{for } 0 < x < 2\\[10pt] 0 &\text{otherwise} \end{cases}

Suppose that $Y = X(2-X)$. How do you determine the c.d.f. and the p.d.f. of $Y$?

Also, how do you determine the p.d.f. of $Y = 4-X^3$

2 Answers 2

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Hints:

Work out the CDF of $X$: $\Pr(X\le x)$ and similarly $\Pr(X\ge x)$.

Transform $y \le x(2-x)$ into expressions of the form $x \le \cdots$ or $x \ge \cdots$ .

Calculate $\Pr(Y\le y)$.

Differentiate.

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For the 2nd one. Let $f(x)$ the pdf of a random variable x. The general rule for finding a PDF for a transformation of the form $y=h(x)$ in our case $Y=X^3$ with $Y∈[0,1], since \ x\in[0,2] \ is:$ $g(y)=f(h^{-1}(y))|\frac{d(h^{-1}(y))}{dy}|$ Applying this rule to your problem will give you the distribution of Y. Note that $h^{-1}(Y)=X=(4-Y)^{\frac{1}{3}}$.

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    Last night, the same incomplete formulation was given in another posted answer, and Kannappan Sampath and I made the essentially the same comment as Byron Schmuland does above. That answer has since been deleted.2012-04-17