Transformation of probability density function under indicator function

Question

I have to calculate the pdf $f_Y(y)$ where $y=\mathbb{I}_{\left[-c,c\right]}( x )$ where the pdf of $x$ is known and denoted by $f_X(x)$ and $c$ is a constant. In this case, $\mathbb{I}_{\mathcal{A}} ( x )$ denotes the indicator function and is $x$ if $x \in \mathcal A$ and $0$ otherwise.

So far I tried to apply the standard method $f_Y(y)=f_X(x)\vert \frac{dx}{dy}\vert$ but I think this will fail, because it requires in this case the existence of an inverse function of the indicator function, right?

Answer 1

As usual, identifying the distribution by computing expectations simplifies everything. One looks for the measure $\mu$ such that, for every bounded measurable function $u$, $$ E(u(Y))=\int u(y)\mathrm d\mu(y). $$ Then one can be sure that $\mu$ is the distribution $P_Y$ of $Y$. In your case, $Y=h(X)$ where $h=\mathbb 1_{[-c,c]}$ hence, by definition of the density $f_X$ of the distribution of $X$, $$ E(u(Y))=E(u(h(X)))=\int u(h(x))f_X(x)\mathrm dx. $$ Thus, $$ E(u(Y))=u(0)P(|X|\gt c)+u(1)P(|X|\leqslant c). $$ This shows that $\mu$ is a Bernoulli distribution with parameter $P(|X|\leqslant c)$, that is, considering the CDF $F_X$ of $X$, $$ P(Y=0)=P(|X|\gt c)=F_X(-c)+1-F_X(c), $$ and $$ P(Y=1)=P(|X|\leqslant c)=F_X(c)-F_X(-c). $$ Note that $Y$ is a purely discrete distribution hence the density $f_Y$ does not exists. If need be, one can write down the distribution $P_Y$ as $$ P_Y=(1-p)\delta_0+p\delta_1,\qquad p=P(Y=1). $$

Answer 2

Consider a bounded continuous function $F$.

$$E(F(Y)) = E(F(1_{[-c,c]}(X))) =\int {F(1_{[-c,c]}(x))}f_X(x)dx \\= \int_{(-\infty,-c)\cup(c,\infty)}F(0) f_X(x)dx + \int_{-c}^c F(1)f_X(x)dx \\= \int [P(|X|>c)\delta_0(x) + P(|X|\le c)\delta_1(x)] F(x) dx $$ So the pdf is definied only in a weak sense (this is a first order distribution), and is $$ f_Y=P(|X|>c)\delta_0 + P(|X|\le c)\delta_1 $$