Mixed joint distribution Let $Y = \max\{0,Y^*\}$, where $Y^*$ follows a continuous distribution. I found in the paper which I am reading that 
$$
f_{Y,Y^*}(Y, Y^*) = I(Y = Y^*)f_{Y^*}(Y^*),
$$
where $I(\cdot)$ is an indicator function. I do not know how to derive this expression. If I use the Bayes rule, I get
$$
f_{Y,Y^*}(Y, Y^*) = f_{Y|Y^*}(Y|Y^*)f_{Y^*}(Y^*),
$$
but I do not know how to deal with this object. 
 A: First note that $(Y,Y^*)$ takes its values on $\{y \geq y^*\}$. 
To derive the density, take $y\geq y^*$ and consider two cases:


*

*if $y^* \leq 0$ then $$\Pr(Y \in dy, Y^* \in dy^*)=I(y=0)\Pr(Y^* \in dy^*)=I(y=0)f_{Y^*}(y^*)$$

*if $y^* \geq 0$ then $$\Pr(Y \in dy, Y^* \in dy^*)=I(y=y^*)\Pr(Y^* \in dy^*)=I(y=y^*)f_{Y^*}(y^*)$$
If you want a one-line formula you could take:
 $$f(y,y^*) = \Pr(Y \in dy, Y^* \in dy^*)=I(y\geq y^*)I(y=\max(0,y^*))f_{Y^*}(y^*)$$
or simply $$f(y,y^*) =I(y=\max(0,y^*))f_{Y^*}(y^*)$$ if you don't forget to mention the support $\{y \geq y^*\}$.
A: For the part of the joint domain of $\{Y, Y^*\}$ , $\big\{\{0\}\times(-\infty,0]\big\}$, we have that
$$P(Y=0 \mid Y^*\leq 0) = 1 = \frac {P(Y=0 , Y^*\leq 0)}{P(Y^*\leq 0)}$$
$$\Rightarrow P(Y=0 , Y^*\leq 0) = F_{Y,Y^*}(0,y^*)=F_{Y^*}(y^*) \tag{1}$$
while for the other part that produces non-zero probability, $\big\{(0,\infty)\times(0,\infty)\big\}$ we have that 
$$P(Y\leq y^* \mid 0<Y^*\leq y^*)= 1 = \frac {P(Y\leq y^* , 0<Y^*\leq y^*)}{P(0<Y^*\leq y^*)} $$
$$\Rightarrow P(Y\leq y^* , 0<Y^*\leq y^*)= F_{Y,Y^*}(y^*,y^*) = F_{Y^*}(y^*) - F_{Y^*}(0) \tag{2}$$
In all,
$$\begin{align} &\{Y, Y^*\} \in \big\{\{0\}\times(-\infty,0]\big\},\qquad &F_{Y,Y^*}(0,y^*)=F_{Y^*}(y^*) \\
&\{Y, Y^*\} \in \big\{(0,\infty)\times(0,\infty)\big\}, \qquad &F_{Y,Y^*}(y^*,y^*) = F_{Y^*}(y^*) - F_{Y^*}(0)\\
\end{align}$$
Note that in the position for $Y$ I have used $y^*$, which stems from how $Y$ is defined.
Differentiate this (with respect to $y^*$ since it is the only one present), and you will get
$$f_{Y,Y^*}(y^*,y^*) = f_{Y^*}(y^*)$$
in both cases (since $F_{Y^*}(0)$ is a constant).  
As for the indicator function: the use of capital $Y$ and $Y^*$ makes the indicator function a random variable, and so the joint density itself a random variable (which I guess, was not what was intended here). So at least it should be written as $I(y=y^*)$. Moreover its existence has some logic if we are referring only to the open interval, that does not include zero. Here, the joint domain is the Cartesian product $\big\{(0,\infty)\times(0,\infty)\big\}$, and this Cartesian product includes pairs of values where $y>y^*$ (in which case the joint density is zero). I.e. instead of writing $y^*$ in place of $y$, the authors used $y$ inside the density and added the indicator function.  
So presumably, what is written in the paper is only one branch of the joint density (after the uppercase-lowercase correction).
A: A corrected expression for the joint density would be something along the lines of $$ f_{Y,Y^*}(y,y^*) = f_{Y^*}(y^*)\big[I(y=y^*) + I(y=0)\big].$$
