Convolution of two log normal distributions

Greetings. I am trying to do the convolutions of two lognormal distributions however, I am getting errors. I started to question my method but I cannot find a mistake in my script. Is there a better way? I seem to be getting unexpected numerical values. I would expect like any CDF to approach 1, but this one is not.

%% Convolution of two LogNormal Distributions.

clc
clear all
format long

muX = 9.7224; % For Ni Distribution (X-domain)

sigmaX = 0.3332; % For Ni Distribution (X-domain)

muY = 8.6878; % For Np Distribution (Y-domain)

sigmaY = 0.2454; % For Np Distribution (Y-domain)

t = inf; % Cycles input (would expect an answer of 1 with t = inf.)



fun = @(y,x) exp(-0.5.*((log(x) - muX).^2)./(sigmaX.^2))./(x.*sigmaX.*sqrt(2.*pi)) .* exp(-0.5.*((log(y) - muY).^2)./(sigmaY.^2))./(y.*sigmaY.*sqrt(2.*pi));


P = integral2(fun,-inf,t - 'x',-inf,inf,'RelTol',1e-12,'AbsTol',1e-12)

Answer 1

[Jeff Miller]

Convolutions are pretty easy to do in Cupid. For example, the following code gives the attached figure

muX = 9.7224; % For Ni Distribution (X-domain)
sigmaX = 0.3332; % For Ni Distribution (X-domain)
muY = 8.6878; % For Np Distribution (Y-domain)
sigmaY = 0.2454; % For Np Distribution (Y-domain)
conv = Convolution(Lognormal(muX,sigmaX),Lognormal(muY,sigmaY));
conv.PlotDens

This might be handy if you also want to try other distributions, try fitting data, etc.

Answer 2

[Bruno Luong]

LOGNORMAL is defined on (0,Inf) not (-Inf,Inf)

P = integral2(fun,0,t - 'x',0,inf,'RelTol',1e-12,'AbsTol',1e-12)

returns correctly

P =

    1.0000

Answer 3

[Image Analyst]

For convolution, use conv() on your numerical vectors.