Given a probability distribution with joint CDF:

$F(x,y)= 1 - e^{-x}-e^{-y}+e^{-(x+y + \alpha x y)} \quad \text{for} \quad x >0 ,\quad y>0, \quad \alpha \in [0,1]$

We can find the corresponding PDF through the differentiation of both variables and after simplifying we obtain:

$f(x,y) = -\alpha e^{-x-y-\alpha x y} + e^{-x-y-\alpha xy}(-1-\alpha x)(-1-\alpha y)$

The next step would be to obtain the marginal PDFs from $f(x,y)$ , however as the function is clearly non-separable into $f(x,y) = g(x)h(y)$ along with the variables clearly being dependent, this seems to be a quite nontrivial. So far I have concluded that given the form of the PDF, the marginals must be distributed exponentially for some parameter $\lambda(\alpha)$, however I am unsure of how to proceed to show this to be the case without having to perform excessive computations.

  • $\begingroup$ Just one side note: If you want to obtain the marginal pdf only, say $f_X$ , then you may also try to take $y \to +\infty$ such that the joint CDF will goes to the marginal CDF of $X$, and then do differentiation. $\endgroup$
    – BGM
    Jan 31, 2018 at 3:34

1 Answer 1


Note that $$ \frac{\partial}{\partial x} F(x,y) = e^{-x} \left(1 - e^{-y (1 + x \alpha)} (1 + y \alpha) \right) $$ and $$ \frac{\partial}{\partial y} F(x,y) = e^{-y} \left(1 - e^{-x (1 + y \alpha)} (1 + x \alpha) \right)\, . $$

As you correctly point out, $f(x,y) = \frac{\partial}{\partial x}\frac{\partial}{\partial y} F(x,y)$. The marginal pdf $f_X(x)$ is given by: \begin{align} f_X(x) &= \int_0^{\infty} dy\; f(x,y)\\ &= \int_0^{\infty} dy\; \frac{\partial}{\partial x}\frac{\partial}{\partial y} F(x,y)\\ &= \frac{\partial}{\partial x} F(x,y)\; \Bigg\rvert_{y=0}^{y=\infty}\\ &= e^{-x} \end{align} Similarly, one finds that the marginal pdf $f_Y(y) = e^{-y}$.

  • $\begingroup$ Please replace $f_x$ and $f_y$ by $f_X$ and $f_Y$ respectively. $\endgroup$
    – Did
    Jan 30, 2018 at 19:56

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