# Convolution of two log normal distributions

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Joshua Woodard on 17 Sep 2019
Edited: Bruno Luong on 18 Sep 2019
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)

Jeff Miller on 18 Sep 2019
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.

Image Analyst on 18 Sep 2019
For convolution, use conv() on your numerical vectors.

Bruno Luong on 18 Sep 2019
Edited: Bruno Luong on 18 Sep 2019
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