I have a random variable $X$ and I want to find the probability density function from transforming it through the Heaviside step function. So


$Y = H(X)$


where the $H$ is the Heaviside step function with


$H(0) = 1$


Given the cumulative distribution function of $X$ as $F\_X$, then I have the pdf of $Y$ as


$f\_Y(y) = (1-F\_X(0))\delta(y-1) + F\_X(0) \delta(y)$


However, I think this pdf is for the case when the step function is defined at $0$ as $H(0) = 0$. Do I need to incorporate the point probability


$F\_X(0) - \lim\_{(x \rightarrow 0^-)}F\_X(x)$


into $f\_Y$ to account for the case where $H(0)=1$, or is there a better/nicer way to do it?



Heaviside Step Function of a Random Variable
I don't think there's a much nicer way. You could either say



> 
> Let $G\_X(x)=\mathbb{P}(X <x)=\lim\_{(t\rightarrow x^-)}F\_X(t)$; then
>  $$
> f\_Y(y)=(1-G\_X(0))\delta(y-1)+G\_X(0)\delta(y)
> $$
> 
> 
> 


or change the random variable. Since 
$$\mathbb P(X<x)=1-\mathbb P(X\ge x)=1-\mathbb P(-X\le -x)=1-F\_{-X}(-x),$$
we have



> 
> $$
> f\_Y(y)=F\_{-X}(0)\delta(y-1)+(1-F\_{-X}(0))\delta(y).
> $$
> 
> 
> 


