Convolution of a CDF and a PDF
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Jeremie Schutz
am 19 Jun. 2018
Kommentiert: Jeremie Schutz
am 20 Jun. 2018
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
2 Kommentare
Jeff Miller
am 20 Jun. 2018
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?
Akzeptierte Antwort
Torsten
am 20 Jun. 2018
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.
2 Kommentare
Jeff Miller
am 20 Jun. 2018
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.
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