Let X have the uniform distribution U(-1,3). How do you find the p.d.f. of $\displaystyle Y=X^{2}$?

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- Apr 22nd 2009, 01:24 PMlaseruniform distribution / pdf
Let X have the uniform distribution U(-1,3). How do you find the p.d.f. of $\displaystyle Y=X^{2}$?

- Apr 22nd 2009, 01:39 PMmr fantastic
- Apr 23rd 2009, 10:16 PMmatheagle
You need to be careful.

This one to one on 1<x<3 and two to one -1<x<1.

USE $\displaystyle f_Y(y)= \biggl(f_X(-\sqrt y) +f_X(\sqrt y)\biggr)\biggl({1\over 2\sqrt y}\biggr)$.

Now $\displaystyle f_X(x)={1\over 4}$ on -1<x<3.

Next notice that y's range is 0 to 9.

WHEN 0<y<1, we have a 2-1 mapping and both

$\displaystyle f_X(\sqrt y)=f_X(-\sqrt y)=1/4$.

So, for 0<y<1, we have

$\displaystyle f_Y(y)= \biggl( {1\over 4} +{1\over 4} \biggr)\biggl({1\over 2\sqrt y}\biggr)= {1\over 4\sqrt y}$.

But when 1<y<9, we have a 1-1 mapping (with probability one)

So, $\displaystyle f_X(\sqrt y)=1/4$ while $\displaystyle f_X(-\sqrt y)=0$.

Thus for 1<y<9, we have

$\displaystyle f_Y(y)= \biggl( 0 +{1\over 4} \biggr)\biggl({1\over 2\sqrt y}\biggr)= {1\over 8\sqrt y}$.

AND zero everwhere else.