A transformation problem

iocal

Junior Member
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Jun 30, 2013
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Hi all,

I am having a bit of a problem with a transformation question. The question is:

Let f(x) = 1/3 for -1<x<2, zero elsewhere be the pdf of X. Find the cdf and the pdf of Y=X^2

Hint: Consider P(X^2<y) for two cases: 0<y<1 and 1<y<4.

My question is why do we have to split in two parts and not simply integrate the pdf of X from -sqrt(y) to sqrt(y) to get the cdf of Y?

I can see in the answer that we have to integrate the pdf of X from -sqrt(y) to sqrt(y) for the interval 0<y<1 and then from -1 to sqrt(y) for the interval 1<y<4.

Thanks in advance.
 
Hi all,

I am having a bit of a problem with a transformation question. The question is:

Let f(x) = 1/3 for -1<x<2, zero elsewhere be the pdf of X. Find the cdf and the pdf of Y=X^2

Hint: Consider P(X^2<y) for two cases: 0<y<1 and 1<y<4.

My question is why do we have to split in two parts and not simply integrate the pdf of X from -sqrt(y) to sqrt(y) to get the cdf of Y?

I can see in the answer that we have to integrate the pdf of X from -sqrt(y) to sqrt(y) for the interval 0<y<1 and then from -1 to sqrt(y) for the interval 1<y<4.

Thanks in advance.
Looks like an absolute-value problem. Between -1<x<0, the derivative dy/dx is negative.
 
Looks like an absolute-value problem. Between -1<x<0, the derivative dy/dx is negative.


Thank you for you reply. However, I have seen similar examples from the same book where that was not important. For example in the following exercise:

Let the pdf of x, be f(x) = 1/2 for -1<x<1 zero elsewhere. Define the random variable Y=X^2. We then wish to find the pdf of Y.

P(X^2<y) = P(-sqrt(y)<X<sqrt(y)) , so accordingly the cdf of Y can be found by integrating the pdf of X with those limits of integration. And differentiating that you get the pdf.

Is there something fudamentally different in the exercise of my first post? Because I cannot see it.

I am doing these exercises for fun but when I get stuck somewhere, it messes with my head and I cannot move on.
 
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Never mind DrPhil. I see your point. The problem is that the book is not good at all at this point. Your comment was short but enlightening, thanks dude.
 
Never mind DrPhil. I see your point. The problem is that the book is not good at all at this point. Your comment was short but enlightening, thanks dude.
Thank you for this comment! I actually try to give you hint to point you in a direction such that you can find the answer for yourself - so this time that worked!
 
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