# Determine the joint density function $f(x,y)$ for $0<y<1$ and $x>0$

The continuous random variables X and Y have the joint distribution function $$F(x,y)=\begin{cases}1-e^{-x} & y>1 ,\:x>0\\y(1-e^{-x}) & 0 0\\0 & else\end{cases}$$ Determine the joint density function $$f(x,y)$$ for $$0 and $$x>0$$

Usually to go from the distribution function to the density function I differentiate the distribution function, but in this case I have the joint so it depends on $$x$$ and $$y$$, how can I calculate the joint density function? Also because the interval $$x>0$$ is present in $$2$$ cases, how can I deal with?

## 1 Answer

You simply differentiate with respect to both variables, i.e. $$\frac{\partial F(x,y)}{\partial x \partial y}=\begin{cases}0 & y>1 ,\:x>0\\e^{-x} & 0 0\\0 & else\end{cases}$$ The answer might seem counter-intuitive but inspect the joint CDF. When $$y>1$$, $$F(x,y)=P(X\leq x, Y \leq y)$$ doesn't change with respect to $$y$$, i.e. there is no volume for $$y$$, when $$y>1$$.

• You are right, thank you a lot for the answer! :) – Luke Marci Jan 26 at 18:44