# What is the density of $Y = X^3$ if $X > 0$ and $Y = X^2$ of $X < 0$?

Let X be a continuous random variable with a density:$f_X(x) = \begin{cases} \frac 2{\pi}(1-x^2)^{1/2}, & \text{if |x| < 1} \\[2mm] 0, & \text{otherwise} \\ \end{cases}$ Find the density of:$Y = \begin{cases} X^3, & \text{if X > 0} \\ X^2, &\text{if X < 0} \\ \end{cases}$

I think that I'm supposed to separate the pdf into when 0 < x < 1 and when -1 < x < 0, but I don't know how to proceed.

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## 1 Answer

Let's compute the distribution function $F_Y$ of $Y$, we will obtain the density by differentiation then. As $1 \ge Y \ge 0$, we consider only $y \in [0,1]$, so let $y \in [0,1]$ be given, we have \begin{align*} F_Y(y) &= P(Y \le y)\\ &= P(Y \le y, X \ge 0) + P(Y\le y, X \le 0)\\ &= P(X^3 \le y, X \ge 0) + P(X^2 \le y, X \le 0)\\ &= P(0 \le X \le y^{1/3}) + P(-y^{1/2} \le X \le 0)\\ &= P(-y^{1/2} \le X \le y^{1/3})\\ &= \int_{-y^{1/2}}^{y^{1/3}} f_X(x)\, dx\\ &= F_X(y^{1/3}) - F_X(-y^{1/2}) \end{align*} So, diffentiating \begin{align*} f_Y(y) &= F'_Y(y)\\ &= f_X(y^{1/3}) \cdot \frac 1{3y^{2/3}} + f_X(y^{1/2})\cdot \frac 1{2y^{1/2}} \end{align*} So we have $f_Y(y) = \begin{cases} 0 & \text{ if y \not\in [0,1]}\\ \frac 2\pi\cdot \left(\frac {(1 - y^{2/3})^{1/2}}{3y^{2/3}} + \frac{(1-y)^{1/2}}{2y^{1/2}}\right) & \text{ if } y \in [0,1] \end{cases}$

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