how to make random vector with a certain profile

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Well, im looking for a way to produce a random vector that behave like the profile in the picture.
It needs to start from let's say around 2, then climbs up to a max value not higher than 20, and then drops to zero.
That is the general idea, the distribution doesnt have to be exactly like in the graph.
does anyone have a good idea how to make that happened?
  2 Comments
Walter Roberson
Walter Roberson on 21 Jul 2020
Do you happen to have the data for that plot available? Rather than us having to read it off the graph and enter the values by hand.
samuel
samuel on 21 Jul 2020
the data is random, it is just an example.
the profile is what matter. do you still need the data? i will send it

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Answers (2)

Bruno Luong
Bruno Luong on 21 Jul 2020
Edited: Bruno Luong on 21 Jul 2020
r=linspace(0,1.5);
fv=(1.5-r).*(2+10*exp(-((r-1)./0.3).^2)); % whatever unnormaized pdf
% NOTE: This method assumes r is an equidistance vector
% otherwise you need to multiply fv by dr before cumulative sum
c = cumsum(fv);
c=(c-c(1))/(c(end)-c(1));
[cu,loc] = unique(c);
rs=r(loc);
x=interp1(cu, rs, rand(1,1e6)); % your random vector
% Check
subplot(2,1,1);
plot(r,fv);
subplot(2,1,2);
hist(x,100)
  5 Comments
Bruno Luong
Bruno Luong on 21 Jul 2020
If you want change for different pdf fv then change it, it in the line #1 & 2 of my code example.
samuel
samuel on 21 Jul 2020
actualy r is given and constant. that i can not change.
for fv (#2 line), if i play with numbers, wont it mess up the profile?

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Bruno Luong
Bruno Luong on 21 Jul 2020
Edited: Bruno Luong on 21 Jul 2020
Modified code in case r is not equidistance (but monotonic)
rmid = 0.5*(r(1:end-1) + r(2:end));
dr = r(2:end)-r(1:end-1);
fvfun = @(r)(1.5-r).^3.*(2+50*exp(-((r-1)./0.3).^2)); % whatever
fv = fvfun(rmid);
c = cumsum(fv.*dr);
c = [0, c]/c(end);
[cu, loc] = unique(c);
x = interp1(cu, r(loc), rand(1,1e6));

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