# joint distribution with indicator function

My question is not about method or computation, I have some doubt about how to deal with indicator function, so if someone could clarify the matter I'll be gratefully!

I have this joint distribution $f_{X,Z}(x,z)=(z-x)e^{-z}1_{\{0\leq x\leq z\}}$

I want to find the marginal distributions and to show if they are or not independent.

I did this:

$$f_X(x)=\int_{-\infty}^{\infty}(z-x)e^{-z}1_{\{0\leq x\leq z\}}dz$$ $$=\int_{x}^{\infty}(z-x)e^{-z}dz$$ with some computation $$=e^{-x}\tag{1}$$

My question is: is it correct? I mean conceptually. Or I have to consider even the condition in the indicator function? In the latter case I would have $$f_X(x)=e^{-x}1_{\{0\leq x\}}\tag{2}$$

I like that since its integral over $\mathbb{R}$ is equal to one, but is it right to do the indicator function appearing suddenly, at the end of computation? The other marginal is $$f_Z(z)=\int_{0}^{x}(z-x)e^{-z}dx=\frac{z^2}{2e^z}\tag{3}$$

or maybe $$f_Z(z)=\frac{z^2}{2e^z}1_{\{0\leq z\}}\tag{4}$$

Also now I like the second one since its integral is equal to one!

For the other point I want just to show that the product of marginals is different from the joint distribution:

$$\frac{z^2}{2e^{z+x}}\neq (z-x)e^{-z} \tag{5}$$

or maybe $$\frac{z^2}{2e^{z+x}}1_{\{0\leq x,0\leq z\}}\neq (z-x)e^{-z}1_{\{0\leq x\leq z\}} \tag{6}$$

or also, (if it means something, I'll change $z$ with $z-x$ to get the condition of the indicator functions the same) $$\frac{(z-x)^2}{2e^{z}}1_{\{0\leq x\leq z\}}\neq (z-x)e^{-z}1_{\{0\leq x\leq z\}} \tag{7}$$

What is the right version, if there is one? :D

• They are clearly not independent since knowing $Z=z$ puts an upper bound on $X$ which would not otherwise be there (similarly, knowing $X=x$ puts a lower bound on $Z$) – Henry Dec 2 '17 at 23:50
• @Henry Indeed. Which is what the indicator functions ...well... indicate. – Graham Kemp Dec 3 '17 at 1:30

\begin{align} f_X(x)&=\int_{-\infty}^{\infty}(z-x)\mathsf e^{-z}~\mathbf 1_{\{0\leq x\leq z\}}~\mathsf dz \\ &= \mathbf 1_{\{0\leq x\}}\cdot\int_{x}^{\infty}(z-x)\mathsf e^{-z}~\mathsf dz \\ & = \mathsf e^{-x}~\mathbf 1_{\{0\leq x\}}\tag 2\\[2ex] f_Z(z)&=\int_{-\infty}^{\infty}(z-x)\mathsf e^{-z}~\mathbf 1_{\{0\leq x\leq z\}}~\mathsf dx \\ &= \mathbf 1_{\{0\leq z\}}\cdot\int_{0}^{z}(z-x)\mathsf e^{-z}~\mathsf dz \\ & = \tfrac 12 z^2\mathsf e^{-z}~\mathbf 1_{\{0\leq z\}}\tag 4\\[2ex] \mathsf e^{-z}~\mathbf 1_{\{0\leq x\leq z\}}&\neq\left(\mathsf e^{-x}~\mathbf 1_{\{0\leq x\}}\right)\left(\tfrac 12 z^2\mathsf e^{-z}~\mathbf 1_{\{0\leq z\}}\right)\tag 6\\[2ex]\hline\therefore\quad f_{X,Z}(x,z)&\neq f_X(x)\cdot f_Z(z) \end{align}