1. ## Transformation

X denotes a random variable with pdf$\displaystyle f(x)=\left\{\begin{array}{cc}\frac{2}{x^2} & 1 < x < 2\\ 0 &elsewhere \end{array}\right.$
let $\displaystyle Y=X^2$, find the cdf of Y and hence the pdf of Y and cov(X,Y).

I know that to find the pdf from the cdf we differentiate, but I'm not sure how to handle to transformation to $\displaystyle Y = X^2$

2. How can you get the covariance?
You don't have a joint density

I guess you get $\displaystyle E(XY)-E(X)E(Y)=E(X^3)-E(X)E(X^2)$

This is a one to one mapping with probability one since x>0.

$\displaystyle f_Y(y)=f_X(x)\Biggl|{dx\over dy}\Biggr|={2\over y}\Biggl|{1\over 2\sqrt y}\Biggr|=y^{-3/2}$

where 1<y<4.

3. I prefer the method of distribution functions.

$\displaystyle G(y) = P(Y \le y) = P(X^{2}\le y) = P(-\sqrt{y} \le X \le \sqrt{y}) = P(X \le \sqrt{y})$ (since X is not defined for any negative numbers)

$\displaystyle = \int^{\sqrt{y}}_{0}\ \frac{2}{x^{2}} \ dx = -\frac{2}{x} \Big|^{\sqrt{y}}_{0}= -\frac{2}{\sqrt{y}}$

then $\displaystyle g(y) = G'(y) = \frac{1}{y^{3/2}}$ for $\displaystyle \ 1 \le y \le 4$

4. Originally Posted by Random Variable
I prefer the method of distribution functions.

$\displaystyle G(y) = P(Y \le y) = P(X^{2}\le y) = P(-\sqrt{y} \le X \le \sqrt{y}) = P(X \le \sqrt{y})$ (since X is not defined for any negative numbers)

$\displaystyle = \int^{\sqrt{y}}_{0}\ \frac{2}{x^{2}} \ dx = -\frac{2}{x} \Big|^{\sqrt{y}}_{0}= -\frac{2}{\sqrt{y}}$

then $\displaystyle g(y) = G'(y) = \frac{1}{y^{3/2}}$ for $\displaystyle \ 1 \le y \le 4$
An antiderivative of $\displaystyle \frac{1}{x^2}$ is $\displaystyle \frac{-2}{x^3}$

5. Originally Posted by Moo
An antiderivative of $\displaystyle \frac{1}{x^2}$ is $\displaystyle \frac{-2}{x^3}$
On what planet?

It should be $\displaystyle P(-\sqrt{y} \le X \le \sqrt{y}) = P(1 \le X \le \sqrt{y}) = \int^{\sqrt{y}}_{1}\frac{2}{x^{2}} dx = -\frac{2}{\sqrt{y}} +2$

7. $\displaystyle P(-\sqrt{y} \le X \le \sqrt{y}) = P(1 \le X \le \sqrt{y}) = \int^{\sqrt{y}}_{1}\frac{2}{x^{2}} dx = -\frac{-2}{\sqrt{y}} +2$
$\displaystyle f(y) = \frac{-2}{\sqrt{y}}+2$
this is the resulting pdf of Y, with range 1<y<4?

Also, if my pdf is
$\displaystyle P(-\sqrt{y} \le X \le \sqrt{y}) = P(1 \le X \le \sqrt{y}) = \int^{\sqrt{y}}_{1}\frac{2}{x^{2}} dx = -\frac{-2}{\sqrt{y}} +2$
the cdf is
$\displaystyle F(x)=\frac{-2}{x}, 1<x<2$ and then if $\displaystyle Y=X^2$ how would I then find the cdf of Y?

8. and I thought I solved this, this morning

A negative cdf?

9. Originally Posted by matheagle
and I thought I solved this, this morning
I can understand the method of distribution functions better....

I've just noticed the negative cdf...sometimes I get caught up in the calculating and don't use common sense!

Perhaps I should give the question in full....
the pdf is given above, I then have to find the cdf(by integrating the pdf given) then the transformation $\displaystyle Y=X^2$ is given and I have to then find the cdf of Y and the pdf of Y, and finally the covariance of (X,Y)....

10. Originally Posted by matheagle
and I thought I solved this, this morning

A negative cdf?
I fixed my mistake in a subsequent post.

11. Let me do the problem over again from scratch to clear up the confusion.

$\displaystyle G(y) = P(Y \le y) = P( X^{2} \le y) = P( -\sqrt{y} \le X \le \sqrt{y})$

But X is not defined for values less than 1

so $\displaystyle P( -\sqrt{y} \le X \le \sqrt{y}) = P(1 < X \le \sqrt{y})$

$\displaystyle = \int^{\sqrt{y}}_{1} \frac{2}{x^{2}} \ dx = -\frac{2}{x}\Big|^{\sqrt{y}}_{1}= -\frac{2}{\sqrt{y}} +2$ for 1 < y < 4

then $\displaystyle g(y) = G'(y) = \frac{1}{y^{3/2}}$ for 1 < y < 4

12. These TWO techniques are the same. IF you differentiate...

$\displaystyle G(y) = P(Y \le y) = P( X^{2} \le y) = P( -\sqrt{y} \le X \le \sqrt{y})$

wrt the variable y (this is not a random variable, this is calc 1) you get

$\displaystyle f_Y(y) = f_X(\sqrt y ){d\over dy}(\sqrt y)={2\over y}\Biggl({1\over 2\sqrt y}\Biggr)=y^{-3/2}$

The general formula, which is in most books is

$\displaystyle f_Y(y) = f_X(\sqrt y )\cdot {d\over dy}(\sqrt y)-f_X(-\sqrt y )\cdot {d\over dy}(-\sqrt y)$

$\displaystyle =f_X(\sqrt y ){1\over 2\sqrt y}+f_X(-\sqrt y ){1\over 2\sqrt y}$

but since x>0 we have $\displaystyle f_X(-\sqrt y )=0$

13. I prefer to use the distribution function technique over the change of variable technique because I find the former more intuitive (especially when dealing with distributions of two random variables).

14. Thanks.
So given the pdf $\displaystyle f(x)=\left\{\begin{array}{cc}\frac{2}{x^2} & 1 < x < 2\\ 0 &elsewhere \end{array}\right.$....

The replies have helped alot, however I'm trying to find the cdf from this pdf then the transformation $\displaystyle Y=X^2$ and then find the cdf of Y and so then the pdf of Y....
1)What would the cdf be as I get a negative answer as I shown a couple posts ago..??
2)What is the process to obtaining the cdf of Y?
3)and finally then the pdf of Y?

15. Originally Posted by bigdoggy
Thanks.
So given the pdf $\displaystyle f(x)=\left\{\begin{array}{cc}\frac{2}{x^2} & 1 < x < 2\\ 0 &elsewhere \end{array}\right.$....

The replies have helped alot, however I'm trying to find the cdf from this pdf then the transformation $\displaystyle Y=X^2$ and then find the cdf of Y and so then the pdf of Y....
1)What would the cdf be as I get a negative answer as I shown a couple posts ago..??
2)What is the process to obtaining the cdf of Y?
3)and finally then the pdf of Y?
G(y) is the cdf of Y and g(y) = G'(y) is the pdf of Y

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