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.

  • $\begingroup$ The support of $Y^*$ matters here. What is it? The whole $\mathcal R$, some subset? $\endgroup$ – Alecos Papadopoulos Aug 7 '14 at 13:50
  • $\begingroup$ The whole $\mathcal{R}$. In this setting, usually $Y^*$ follows the normal distribution. $\endgroup$ – Kolibris Aug 7 '14 at 13:53
  • $\begingroup$ Something seems off here. Consider $Y^* \sim N(0,1)$, and suppose we observe $Y^*=-1$. Then, $Y=0$, and $f(0,-1)=\phi(-1) > 0$. However, the joint density you have given would assign a density of zero in this situation, since $Y \neq Y^*$. Am I missing something? $\endgroup$ – ahfoss Aug 7 '14 at 15:01

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 $\{y \geq y^*\}$.

  • $\begingroup$ (+1) Much more straightforward and clear than what I did in my answer. $\endgroup$ – Alecos Papadopoulos Aug 7 '14 at 16:02
  • $\begingroup$ @StéphaneLaurent: what do you mean by $dy$ here? $\endgroup$ – Kolibris Aug 7 '14 at 16:40
  • $\begingroup$ @StéphaneLaurent: I think that there is something wrong with the one-formula, since $I(y = \max(0,y^*)) = 1$ always. It should be $I(y = y^*)$. Could you explain in details how did you get $Pr(Y\in dy,Y^∗\in dy^∗)=I(y=0)Pr(Y^∗\in dy^∗)$? $\endgroup$ – Kolibris Aug 7 '14 at 17:02
  • $\begingroup$ @Kolibris The one-line formula is a correct summary of the two above lines (here $y$ and $y^*$ are just two real numbers satisfying $y \geq y^*$). $\endgroup$ – Stéphane Laurent Aug 7 '14 at 18:37
  • $\begingroup$ @Kolibris The notation $dy$ could be the subject of a discussion :) This is a somewhat intuitive approach but it could be rigorously translated with the help of Radon-Nikodym derivatives. This approach works after a bit of practice :) $\endgroup$ – Stéphane Laurent Aug 7 '14 at 18:38

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 corrected expression for the joint density would be something along the lines of $$ f_{Y,Y^*}(y,y^*) = f_{Y^*}(y^*)\big[I(y=y^*) + I(y=0)\big].$$

  • $\begingroup$ But if $y=y^*=0$, wouldn't that give us $2f_{Y^*}(y^*)$? Since you are using equalities, you should treat this case also. $\endgroup$ – Alecos Papadopoulos Aug 7 '14 at 15:26
  • $\begingroup$ @AlecosPapadopoulos this event has null probability $\endgroup$ – Stéphane Laurent Aug 7 '14 at 15:39
  • $\begingroup$ @StéphaneLaurent This is not an event, I am not using the random variables $Y$ and $Y^*$, I am just considering the pair $\{0,0\}$ from their joint domain. This pair gives non-zero joint probability density function, and equal to $f_{Y^*}(y^*)$, not $\times 2$ that. $\endgroup$ – Alecos Papadopoulos Aug 7 '14 at 15:48
  • $\begingroup$ @AlecosPapadopoulos Yes but I mean $\Pr(Y=0)=0$ hence the value of the density when $y=0$ has no importance. $\endgroup$ – Stéphane Laurent Aug 7 '14 at 15:52
  • $\begingroup$ @AlecosPapadopoulos Sorry I'm not clear. The joint law is absolutely continuous with respect to Lebesgue measure on $\{y \geq y^*\}$, hence the values of the density on a line can be arbitrary. $\endgroup$ – Stéphane Laurent Aug 7 '14 at 15:54

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