# Density function $Y := X^2$

Soon I will have my final exam for probability theory. Unfortunately, my university doesn't have a lot of past papers available due to a change of lecturer. Hence I decided to make some exercises on the internet, but some of them don't have answers. I think that this exercise looks quite easy, however I am not sure if I did it right, and it would be nice if someone could verify for me whether I am on the right track or not.

Let $$X$$ be a continuous random variable with pdf:

$$$$f_X(x) = \begin{cases} |x|,& -1\leqslant x \leqslant 1 \\ 0,& \text{otherwise} \end{cases}$$$$

I need to compute the density function of $$Y:= g(X) = X^2$$.

For this I decided to use the following theorem:

"Given that $$g$$ is differentiable and either strictly decreasing or strictly increasing,

$$f_Y(y) = f_X(g^{-1}(y)) |\frac{d}{dy}g^{-1}(y)|$$"

In our case $$g$$ is differentiable, strictly decreasing on $$-1 \leq x < 0$$, and strictly inceasing on $$0 \leq x < 1$$. So my idea was just to apply this theorem over those two intervals.

Computing the inverse of $$g$$ gives us $$g^{-1}(y) = \sqrt{y}$$.

Then, filling this into our formula gives us

\begin{aligned}f_Y(y) &= f_X(g^{-1}(y)) |\frac{d}{dy}g^{-1}(y)|\\ &= |\sqrt{y}||\frac{1}{2\sqrt{y}}|\\ &= |\frac{\sqrt{y}}{2\sqrt{y}}|\\ &= |\frac{1}{2}|\\ &= \frac{1}{2} \end{aligned}

We can observe that $$f_Y(y)$$ will be the same on both intervals, and also, squaring the range of $$x$$ gives us that the range of $$y$$ will be $$0 \leq y \leq 1$$. Hence $$\begin{equation*} f_Y(y) = \begin{cases} \frac{1}{2},& 0\leqslant y \leqslant 1 \\ 0,& \text{otherwise} \end{cases} \end{equation*}$$

Can someone verify for me if this is right? Actually I dont even know if it is okay to use the formula $$f_Y(y) = f_X(g^{-1}(y)) |\frac{d}{dy}g^{-1}(y)|$$, since over the total interval our function is first decreasing and after increasing.

Thanks in advance!

• As a test, check that $\int_0^1 f_Y(y)\mathrm d y=1$. – Graham Kemp Jun 19 at 10:27

## 1 Answer

Note 1: $$\dfrac{\mathrm d \sqrt y}{\mathrm d ~y~~}=\dfrac{1}{2\sqrt y}$$

Note 2: $$x\mapsto x^2$$ folds $$[-1;1]$$ onto $$[0;1]$$, so it effectively has two "inverse" functions.

These are $$y\mapsto +\surd y$$ and $$y\mapsto -\surd y$$, mapping $$[0;1]$$ to $$[0;1]$$ and $$[-1;0]$$ respectively.

Thusly:

$$f_Y(y)=f_X({+}\surd y)\cdot\left\lvert\dfrac{\mathrm d ({+}\surd y)}{\mathrm d~y~}\right\rvert~\mathbf 1_{y\in[0;1]}+ f_X({-}\surd y)\cdot\left\lvert\dfrac{\mathrm d ({-}\surd y)}{\mathrm d~y~}\right\rvert~\mathbf 1_{y\in[0;1]}$$

• Oef painfull mistake, I corrected the first note – bladiebla Jun 19 at 10:26
• what does the $1_{y\in[0;1]}$ notation mean? And why do we multiply by that? – bladiebla Jun 19 at 10:28
• It is called an indicator function; it is a piecewise function that equals one when the indicated condition is realised, and equals zero otherwise.$$\mathbf 1_{y\in[0;1]}=\begin{cases} 1&:& 0\leq y\leq 1\\0&:&\text{elsewhere}\end{cases}$$In this case, it is indicating the support for the probability density function. – Graham Kemp Jun 19 at 10:30
• Aha, makes sense. I think I get it now. Thank you! – bladiebla Jun 19 at 10:44