transformation of single random variables with absolute value ??

Question

integral

I got the final answer to be fy(y)= 1 0< y < 1

I am not sure could anyone correct me if its wrong !

Answer 1

Since the the map $1-|x|$ is not invertable on $[-1,1]$, you cannot apply the general formula, you have to do some calculations instead:

a) Integrate over $[-1,y]$ to get the CDF: $$\int_{-1}^y \frac{1-x}{2}\mathrm{d}x=C+\frac{2y-y^2}{4}$$ As it is a CDF, $C=\frac{3}{4}$, so $$F_X(x)=\frac{3+2x-x^2}{4}$$ b) $|X|\leq 1$, so: $$\mathbb{P}(1-|X|<t)=\mathbb{P}(1-t<|X|)=\mathbb{P}(X<t-1\cup 1-t<X)=F(t-1)+1-F(1-t)=t$$ $$F_Y(t)=t\Rightarrow f_Y(t)=1$$ c) $0.6$ as it is a uniform random variable

d) $\mathbb{E}(Y)=0.5$ and $$\int_{-1}^{1}x\frac{1-x}{2}\mathrm{d}x=-\frac{1}{3}$$

Answer 2

Yes. Your answer is correct.

Ákos Somogyi showed one method to obtain it; by integration, substitution, then derivation.

Another method is to use the change of variables transformation.

$y=1-\lvert x\rvert$ means $x =\pm(1-y)$

The standard formula is modified because the support of $X$ is subdivided into two intervals which both map to the support of $Y$.

$$\begin{align} f_Y(y) & = f_X(1-y)\cdot\left\lvert \frac{\mathrm d (1-y)}{\mathrm d y\qquad}\right\rvert+f_X(y-1)\cdot\left\lvert \frac{\mathrm d (y-1)}{\mathrm d y\qquad}\right\rvert \\[2ex] & = \left[(-0.5(1-y)+0.5)+(-0.5(y-1)+0.5)\right]\cdot\mathbf 1_{y\in[0;1]} \\[2ex] & = \mathbf 1_{y\in[0;1]} \end{align}$$