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:

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

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!

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

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.


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

  • $\begingroup$ Oef painfull mistake, I corrected the first note $\endgroup$ – bladiebla Jun 19 at 10:26
  • $\begingroup$ what does the $1_{y\in[0;1]}$ notation mean? And why do we multiply by that? $\endgroup$ – bladiebla Jun 19 at 10:28
  • $\begingroup$ 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. $\endgroup$ – Graham Kemp Jun 19 at 10:30
  • $\begingroup$ Aha, makes sense. I think I get it now. Thank you! $\endgroup$ – bladiebla Jun 19 at 10:44

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.