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.
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.