# CDF of product of two *translated* exponential variables

It is known that the CDF of the product of two exponential random variables is given by this answer

If $$X$$ and $$Y$$ have rate parameters $$\mu$$ and $$\lambda$$, so their pdf's are $$f_X(x) = \mu e^{-\mu x}$$ and $$f_Y(y) = \lambda e^{-\lambda > y}$$ for $$x,y \ge 0$$, then for $$c > 0$$, $$Pr(X Y < c) = \int_0^\infty dx \int_0^{c/x} dy\; \mu \lambda e^{-\mu x - \lambda y} = 1-2\,\sqrt { {c\mu\lambda}}{{ K}_{1}\left(2\,\sqrt { c\mu\lambda}\right)} \tag{1}$$ where $$K_1$$ is a modified Bessel function of the second kind.

I need to calculate the CDF of $$Z = UV$$ where $$U=a+X$$ and $$V=b+Y$$ that $$a,b>0$$ are constant.

It is easy to show that the PDF of $$U,V$$ are $$f_U(u)=\mu e^{a\mu} e^{-\mu u}$$ and $$f_V(v)=\lambda e^{b\lambda} e^{-\lambda v}$$ for $$U \geq a > 0$$, $$V \geq b > 0$$.

$$Pr(Z < c) = \int_a^\infty du \int_b^{c/u} dv\; \mu \lambda e^{a \mu + b\lambda} e^{-\mu u - \lambda v} \tag{2}$$

I don't know how to process further.

• just like the original question, I'd say this fits much nicer on the math stackexchange site. – Marcus Müller Nov 27 '18 at 18:32
• also, not that $Z = ab + aY + bX + XY$ and hence, $f_Z$ is the convolution of $f_{ab}$ (which is a dirac, and hence, convolution with is but a shift), $f_{aY}$, $f_{bY}$ (which just scales $f_Y$ or $f_X$, respectively), and $f_{XY}$ (and you have formula for $F_{XY}$). – Marcus Müller Nov 27 '18 at 18:36
• @MarcusMüller thanks for suggestion. For the second comment, I believe that $bX$, $aY$ and $XY$ are not independent hence the convolution approach does not work. – Cath Maillon Nov 27 '18 at 21:31
• ah, true, my mistake. So, back to the Jacobi determinant, it is? – Marcus Müller Nov 27 '18 at 21:32
• @MarcusMüller It sounds promising. Do you mean the sum of dependent random variables? Could you please elaborate a little more or give me the online resources about Jacobi determinant techniques you have refered? – Cath Maillon Nov 27 '18 at 21:37