Random Variable with exponential distribution of Probablity Density Function

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 .

Answer 1

[ABHILASH SINGH]

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)')

Answer 2

[Roger Stafford]

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.

Answer 3

[Youssef Khmou]

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..........