Convolution of a CDF and a PDF

Good afternoon,

I come from Mathematica and I would like to compute a convolution (the function ANOF, below) thanks to MATLAB :

Unfortunately, I can't do it despite reading this post: https://fr.mathworks.com/matlabcentral/answers/230010-performing-a-convolution-of-two-exponential-functions-in-matlab?s_tid=srchtitle.

Can you help me?

The expected results are:

• ANOF[50]=0.0772848
• ANOF[100]=0.956483

Thanks, Jérémie

[Jeff Miller]

Is the expression inside the square brackets [] the upper tail probability of a random variable which is the sum of independent Weibull and normal RVs?

[Jeremie Schutz]

Dear Jeff,

For a better readability, I isolated the part that causes me problems.

• The expression CDF[WeibullDistribution[2, 100] can be formulated as 'wblcdf(t-y,100,2)'
• The expression 'PDF[NormalDistribution[2.2, 0.1], y]' can be formulated as '* The expression 'PDF[NormalDistribution[2.2, 0.1], y]' can be formulated as 'normpdf(y,2.2,0.1)'.

I tried to execute the code below but it returns many errors.

assume(y > 0)
y=sym('y');
t=sym('t');
% I = int(wblcdf(t-y,100,2)*normpdf(y,2.2,0.1),y,0,100)
I = int(wblcdf(t-y,100,2)*normpdf(y,2.2,0.1),y,0,t)

Answer 1

[Torsten]

ANOF=@(t)-log(1-integral(@(y)wblcdf(t-y,100,2).*normpdf(y,2.2,0.1),0,t,'ArrayValued',true));
ANOF(50)
ANOF(100)

Best wishes

Torsten.

[Jeff Miller]

FWIW, Torsten's answer looks right to me, but it gives

ANOF(50) =      0.22848
ANOF(100) =     0.95648

which are not the "expected values" stated in the original question.

[Jeremie Schutz]

Thank you Torsten and Jeff Miller!

It work fine. There was a mistake in my first post... It was ANOF[30]= 0.0772848