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Let X be a continuous random variable with probability density function (PDF)

$$ f_X(x) = begin{cases}
cx^2+|x|, & -1/2 < x < 1/2 \
0, & otherwise
end{cases} $$

1. Determine the value of the c constant.


2. Obtain the cumulative distribution function (CDF) of X.

Solution

1. Having to be
   $$ int_{-infty}^{infty} f(x) , mathrm{d}x = 1, $$
   it is
   $$1=int_{-1/2}^{1/2} f(x) , mathrm{d}x=int_{-1/2}^{1/2} cx^2+|x| , mathrm{d}x=cint_{-1/2}^{1/2}x^2, mathrm{d}x+int_{-1/2}^{1/2}|x| , mathrm{d}x$$
   

By integrating it, we get:

$$1=cleft[frac{x^3}{3}right]_{-1/2}^{1/2}+left[frac{x|x|}{2}right]_{-1/2}^{1/2}=cleft[frac{(frac{1}{2})^3-(-frac{1}{2})^3}{3}right]+left[frac{frac{1}{2}|frac{1}{2}|-(-frac{1}{2}|-frac{1}{2}|)}{2}right]=$$
$$=cfrac{frac{1}{8}+frac{1}{8}}{3}+frac{frac{1}{4}+frac{1}{4}}{2}=cfrac{2}{8}timesfrac{1}{3}+frac{2}{4}timesfrac{1}{2}=frac{c}{12}+frac{1}{4}$$

So we can conclude that:

$$c=12timesleft(1-frac{1}{4}right)=12timesleft(frac{3}{4}right)=3times3=9$$
that matches with the solution provided by the book.

2. Now, for this step the book provides the following solution:
   
   Probably here the integral of |x| has been splitted into 2 integrals (positive and negative numbers), however I'm not getting how it is possible to turn a definite integral with x variables to an expression of x variables, by passing from another integral by substituting y=x.

By integrating
$$9int_{-1/2}^{1/2}x^2, mathrm{d}x+int_{-1/2}^{1/2}|x| , mathrm{d}x$$

I would get only a number.

How do I turn this integral to an expression in x variables?

The CDF $F_X(x)$ of a random variable $X$ is the probability that $X$ takes values smaller than $x$, so
$$
F_X(x)=int_{-infty}^x f_X(x')mathrm{d}x' .
$$
Everything boils down to computing the following integral:
$$
F_X(x)=int_{-infty}^x mathrm{d}x'(9x'^2+|x'|)mathbb{1}_{-1/2leq x'leq 1/2}.
$$
Clearly, this is a function of $x$, not just a number!

To compute the integral, we first observe that, if $x<-1/2$ the integral is zero (because the integrand is zero), and if $x>1/2$ the integral is identical to $int_{-1/2}^{1/2} mathrm{d}x'(9x'^2+|x'|)$ and thus is $=1$ by normalization.

Since the integrand is zero if $x'<-1/2$, we can truncate the range of integration as follows:
$$
F_X(x)=int_{-1/2}^x mathrm{d}x'(9x'^2+|x'|)mathbb{1}_{x'leq 1/2}.
$$

Now, it is convenient to consider the cases $-1/2<x<0$ and $0<x<1/2$ separately. If $-1/2<x<0$, you are integrating the function $9x'^2+|x'|$ over a negative range, which means that $|x'|= -x'$, and your integral is
$$
F_X(x)=int_{-1/2}^x mathrm{d}x'(9x'^2-x')=9x'^3/3-x'^2/2Big|_{-1/2}^x
=9x^3/3-x^2/2-frac{9}{3}(-1/2)^3+frac{1}{2}(-1/2)^2=
$$
$$
3 x^3-frac{x^2}{2}+frac{1}{2} ,
$$
which coincides with the result given.

If $0<x<1/2$ instead, one needs to be more careful, because the function is now integrated over both a negative and positive range, therefore it is convenient to split the integration range $(-1/2,x)$ into $(-1/2,0)$ and $(0,x)$ - the difference between the two resulting integrals being that the absolute value $|x'|$ is equal to $-x'$ in the first case, and $x'$ in the second case.