How can I Calculate the PDF and CDF of a product of two i.i.d exponentially distributed random variables with mean a and b respectively

22 Ansichten (letzte 30 Tage)
Hi
  3 Kommentare
Mayank Agrawal
Mayank Agrawal am 6 Mär. 2020
hello sir
I have a similar question but a little advanced one where I want to plot pdf and cdf of Z = (x*y)/(y+b) where 'b' is a positive real constant and x & y have same exponential distributions.
Can you please help me out how to plot the pdf and cdf of 'z' in MATLAB?
CDF of Z: P(Z<=z) = 1 - (e^(-x/lambda1)/lambda2)*(sqrt(4*b*z*lambda2/lambda1))*K1(sqrt(4*b*z/(lambda1*lambda2)))
so we differentiated this cdf w.r.t. z using chain rule and found the pdf which had 3 terms in addition.
we are not able to plot the pdf and cdf of z in MATLAB. please do help us out.
Thanks

Melden Sie sich an, um zu kommentieren.

Akzeptierte Antwort

Torsten
Torsten am 26 Nov. 2015
Bearbeitet: Torsten am 26 Nov. 2015
You get
F(x)=1-2*sqrt((lambda1)*(lambda2)*x)*K1(2*sqrt((lambda1)*(lambda2)*x))
f(x)=2*(lambda1)*(lambda2)*K0(2*sqrt((lambda1)*(lambda2)*x))
with
lambda1 = 1/m
lambda2 = 1/n
K0, K1 : Modified Bessel functions of the second kind
Best wishes
Torsten.
  2 Kommentare
kader
kader am 27 Nov. 2015
Dear Torsten, I really grateful to you. Thanks a lot for this answer. In MatLab, the function is like as follows. For K1, it is besselK(1,x). Right? Though I got the expected result from the mentioned changes in MatLab. So, for K0 it will be besselK(0,x)? F(x)=1-2*sqrt((lambda1)*(lambda2)*x)*besselk(1,2*sqrt((lambda1)*(lambda2)*x))
Torsten
Torsten am 27 Nov. 2015
Yes, in MATLAB notation it's
f=@(x)2*(lambda1)*(lambda2)*besselk(0,2*sqrt((lambda1)*(lambda2)*x))
F=@(x)1-2*sqrt((lambda1)*(lambda2)*x)*besselk(1,2*sqrt((lambda1)*(lambda2)*x))
Best wishes
Torsten.

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (1)

Torsten
Torsten am 26 Nov. 2015
Bearbeitet: Torsten am 26 Nov. 2015
  2 Kommentare

Melden Sie sich an, um zu kommentieren.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by