Random Variable with exponential distribution of Probablity Density Function

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Hydro
Hydro 2014-9-16
编辑： ABHILASH SINGH 2018-8-17
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 .

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回答（3 个）

ABHILASH SINGH
ABHILASH SINGH 2018-8-14
编辑：ABHILASH SINGH 2018-8-17
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)')
Roger Stafford
Roger Stafford 2014-9-17
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.
Youssef Khmou
Youssef Khmou 2014-9-16
编辑：Youssef Khmou 2014-9-16
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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