Convolution CDF formula?

Question

In reference to this post, the pdf of dependent random variables $A+B$ is given by:

$$f_{A+B}(z) = \int_{-\infty}^{\infty} f_{A,B}(a,z-a) \mathrm da = \int_{-\infty}^{\infty} f_{A,B}(z-b,b) \mathrm db.$$

Is there also an expression for the convolution CDF $F_{A+B}(z)$ (except as double integral)?

Answer 1

If $A^-$ and $B^-$ are integrable, then $$F_{A+B}(z)=\left.\frac{\mathrm d}{\mathrm ds}\left(\int_\mathbb RF_{A,B}(x,s-x)\mathrm dx\right)\right|_{s=z}$$

Answer 2

There are different solutions for you question, however the following equation is suggested:

$$ F_{A+B}(z) = F_{A} * F_{B} = \int_0^z \! F_{A}(z-a) \, \mathrm{d}G_{B}(a). $$

where * denotes convolution. If you use MATLAB and you need parametric solution, you can use following code:

function [f, F] = Param_Conv(F1,F2,a,z)
    f1 = diff(F1, a);
    f2 = diff(F2, a);
    f2_n = subs(f2,a,z - a);
    f = int(f1*f2_n,a,0,z);
    F = int(f,Tau,0,a);
end

Note that when you have Weibull CDF, parametric solution is hard and time consuming. You can use numerical solution. For example, in MATLAB, you can use following function:

function F = Numerical_Conv(F1,F2)
m = max(size(F1)); F = zeros(m,1);
for i=2:m
    for j=2:i
        F(i) = F(i) + 0.5 * [F1(j) - F1(j-1)]*[F2(i-j+1) + F2(i-j+2)];
    end
end
end

To find out numerical convolution, reference [1] is suggested.

[1]: Cleroux, R., & McConalogue, D. J. (1976). A numerical algorithm for recursively-defined convolution integrals involving distribution functions. Management Science, 22(10), 1138-1146.