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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!

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

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

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  • $\begingroup$ 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

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