Hi there hazeleyes,
Here's a kick off...
for the mean
hey guys, I'm just wondering if you guys can help me with this question, thank you in advance.
A random variable, X, has a probability density function (p.d.f.) given by
f(x)=2xe^(-x^2)
where x>0
Y is a random variable such that Y = X^2.
Show that Y has an exponential distribution and state its mean.
thanks.
Sorry, but that's not correct.
There are several approaches that can be taken.
The OP has carelessly neglected to mention that the support of X is (so that f(x) = 0 for x < 0). The support of Y will be .
The most basic approach to finding the pdf of Y is to calculate the cdf of Y and then recall that the pdf is the derivative.
.
Differentiating this is simple - either integrate (use a substitution) directly and then differentiate, or use the chain rule and the Fundamental Theorem of Calculus.
You should get for and zero elsewhere. Calculating the mean of Y is trivial.
Some trivia: The random variable X follows a Weibull distribution (http://en.wikipedia.org/wiki/Weibull_distribution) with parameters and . Since it is well known that Y follows an exponential distribution.
Apologies in advance as I am not yet familiar with the Weibull distribution, and also I haven't yet learned to use LaTex...
I follow mr. fantastic's logic up to the point of G(y) = P(-y^{1/2} <= X <= y^{1/2})
But shouldn't this mean that G(y) = the integral from 0 to y^{1/2} of 2e^{-2x} dx = 1 - e^{-2y1/2}?
and therefore, g(y) = dG/dy = y^{-1/2} e^{-2y1/2}?
Sorry, regarding my last post I realized I did not read the original post carefully enough - I thought the original pdf was f(x) = 2e^{-2x}. In the process however, I think I learned that where X~exp(lambda), Y= X^{2} is also exp(lambda). Is that correct?