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

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You simply differentiate with respect to both variables, i.e. $$\frac{\partial F(x,y)}{\partial x \partial y}=\begin{cases}0 & y>1 ,\:x>0\\e^{-x} & 0<y<1,\:x > 0\\0 & else\end{cases}$$ 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 $y>1$.

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

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