0
$\begingroup$

Suppose $X$ has uniform $[-1,2]$ distribution. I am trying to find the density of $Z=X^2$.

Here is what I have done thus far:

Range($Z$)$=[0,4]$. I began computing the distribution of $Z$ for $z \in [0,4]$ as follows:

$$\begin{align*} P(Z \le z) &= P(X^2 \le z)\\ &= P(X \in [-\sqrt z, \sqrt z]) \end{align*}$$

I am note sure what to do from here. Any help is greatly appreciated. Once I unravel the distribution of $Z$, I can obtain the density of $Z$ by taking the derivative of its distribution function with respect to $z$.

$\endgroup$
1
$\begingroup$

$$ \mathbb{P}(X\in[-\sqrt{z},\sqrt{z}])=\mathbb{P}(X<\sqrt{z})-\mathbb{P}(X>-\sqrt{z})=\boxplus $$ Now since $X$ in uniform on $[-1,2]$, the CDF of it is: $$ F_X(x)=\chi_{x\in(2,\infty)}+\frac{1}{3}(x+1)\chi_{x\in[-1,2]} $$ THus by taking the range into consideration, we obtain: $$ \boxplus=\chi_{\sqrt{z}\in(2,\infty)}+\frac{1}{3}(\sqrt{z}+1)\chi_{\sqrt{z}\in[0,2]}-\frac{1}{3}(-\sqrt{z}+1)\chi_{-\sqrt{z}\in[-1,0]}= $$ $$ =\chi_{\sqrt{z}\in(2,\infty)}+\frac{1}{3}(\sqrt{z}+1)\chi_{\sqrt{z}\in[1,2]}+\frac{2\sqrt{z}}{3}\chi_{\sqrt{z}\in[0,1]} $$ Basically we handled the cases $z\geq 4$, $z\in[1,4]$, $z\in[0,1]$ and $z<0$ separately. Thus the PDF is: $$ \begin{cases} 0&z\in(-\infty,0)\cup(4,\infty)\\ \frac{1}{3\sqrt{z}}& z\in(0,1]\\ \frac{1}{6\sqrt{z}}& z\in(1,4]\\ \end{cases} $$

$\endgroup$
  • $\begingroup$ I am trying to understand this line: $$ F_X(x)=\chi_{x\in(2,\infty)}+\frac{1}{3}(x+1)\chi_{x\in[-1,2]}. $$ I'm assuming $\chi$ is the indicator function. Where do $1/3$, x+1, and $\chi_{x\in (2, \infty)}$ come from? $\endgroup$ – Marcialo Nov 16 '15 at 8:48
  • $\begingroup$ Since the range of $X$ is $[-1,2]$, thus $\mathbb{P}(X\leq x)=1$, if $x\geq2$. Generally, the uniform distribution on $(a,b)$ has density function $$\frac{x-a}{b-x}\chi_{x\in[a,b]}+\chi_{x>b}$$ As it has uniform density on the interval $[a,b]$, the CDF must be linear, and $\mathbb{P}(X\leq a)=0$ and $\mathbb{P}(X\leq b)=1$. $\endgroup$ – Ákos Somogyi Nov 16 '15 at 8:51
1
$\begingroup$

While Ákos Somogyi's answer is correct, this is not the best approach to solve this kind of problem.

There is a transformation of density formula, which says that if $X$ has density $f_X$ and $h$ is piecewise continuously differentiable and piecewise strictly monotone, then $Y$ has the density $$ f_Y(y) = \sum_{x: h(x) = y} \frac{f_X(x)}{|h'(x)|}. \tag{1} $$

Why is it better to use (1)? The cdf approach goes through: finding the cdf of $X$, solving the inequality $h(x) \le y$ to find cdf of $Y$, differentiating it to get the cdf of $X$. To use (1), one should solve the equation $h(x) = y$, which is much easier; moreover, solving the inequality involves solving the equation anyway. And here you don't make some unnecessary extra operations, like finding the cdf of $X$ and then differentiating the cdf of $Y$ (which are kind of inverse operations, so this is pretty pointless).

$\endgroup$
1
$\begingroup$

$Z$ is nonnegative so $f_{Z}\left(x\right)=0$ if $x<0$.

For $x\geq0$ we have: $$F_{Z}\left(x\right)=F_{X}\left(\sqrt{x}\right)-F_{X}\left(-\sqrt{x}\right)$$

Consequently: $$f_{Z}\left(x\right)=\frac{1}{2\sqrt{x}}f_{X}\left(\sqrt{x}\right)+\frac{1}{2\sqrt{x}}f_{X}\left(-\sqrt{x}\right)=\frac{1}{6\sqrt{x}}\left(\chi_{\left[-1,2\right]}\left(\sqrt{x}\right)+\chi_{\left[-1,2\right]}\left(-\sqrt{x}\right)\right)$$

It remains to work out the RHS.

I leave that to you.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.