CDF technique and uniform distribution

Question

Let X have the uniform distribution over$ (-1,3)$. Find the Pdf of $Y=X^2$.

I use CDF technique. $$F_Y(y)=P(Y \leq y)=P(u(X) \leq y)= P(x^2 \leq y)=X^2$$. Then I differentiate and solve for PDF. Am I doing right in terms of steps?

Answer 1

First of all the cdf of X is

$$F_X(x)=\begin{cases} 0, \ x<-1 \\ \frac14 x+\frac14, \ -1\leq x \leq 3 \\ 1, \ x>3 \end{cases}$$

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Now we split it in two cases:

First case: $x\in [-1,1) \Rightarrow y\in [0,1)$

We have $x=y^{1/2} $ for $x\in [0,1)$ and $x=-y^{1/2}$ for $x\in [-1,0)$

Second case: $x\in [1 ,3] \Rightarrow y\in [1,9]$

$\texttt{First case:}$

$$F_Y (y) = P(Y ≤ y) = P(X^2 ≤ y) = P(−y^{1/2} ≤ X ≤y^{1/2})$$

Here you can see symmetry of the limits for X.

$$= P(X ≤ y^{1/2}) − P(X ≤ −y^{1/2}) = F_X(y^{1/2}) − F_X(−y^{1/2})$$.

The cdf of $Y$ is

$$F_Y(y)=F_X\left(y^{1/2}\right)-F_X\left(-y^{1/2}\right)=\frac14 \sqrt y+\frac14-\left(\frac14 (-\sqrt y)+\frac14 \right)=\frac12y^{1/2}$$

$\texttt{Second case:}$

Here $X$ is positive and $Y$ is positive as well. Therefore

$$F_Y(y)=F(X^2<y)=F(X\leq y^{1/2})$$

$$F_Y(y)=\frac14 \cdot y^{1/2}+\frac14=\frac14\cdot y^{1/2}+\frac14$$

Finally you differentiate to obtain the pdf.