Random Variable with exponential distribution of Probablity Density Function

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Hydro
Hydro on 16 Sep 2014
Edited: ABHILASH SINGH on 17 Aug 2018
Any thought/hint for solving this question
Assume that the random variable X has an Exponential distribution with PDF given by:
* f(x) = 1/α exp(-x/α); x => 0*
Using the theory of transformed random variables, determine an expression for the PDF of Y , where Y = X^2.
Plot the PDFs of X and Y in the same plot. The answer to the question should be the analytical derivation of the PDF of Y , as well as the plots of the PDFs of X and Y .

Answers (3)

Roger Stafford
Roger Stafford on 17 Sep 2014
If we call the pdf of y, 'pdfy', then
pdfy(t) = 1/(2*a*sqrt(t))*exp(-1/a*sqrt(t))
Note the interesting fact that as t approaches the lower limit of zero, the probability density approaches infinity, but that's all right because, of necessity, it is still integrable, and in fact has an integral of one as the upper limit for t approaches infinity. You can see that trend to infinity at t = 0 in Youssef's plot.

ABHILASH SINGH
ABHILASH SINGH on 14 Aug 2018
Edited: ABHILASH SINGH on 17 Aug 2018
alpha=2;
X=0:0.1:10;
fx=(1/alpha)*exp(-X./alpha);
plot(X,fx)
fy=fx./(2*X);
hold on
plot(X,fy)
legend('f(X)','f(Y)')
xlabel('X')
ylabel('F(X),F(Y)')

Youssef  Khmou
Youssef Khmou on 16 Sep 2014
Edited: Youssef Khmou on 16 Sep 2014
The beginning of the answer can be as the following :
% first part
alpha=2; % parameter
N=400; % size of a sample
r=random('exp',alpha,1,N);
figure; plot(r);
x=linspace(min(r),max(r),40);
figure;hist(r,x);
title(' Exponential distribution') ; % hold on..........

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