# uniform distribution / pdf

• April 22nd 2009, 01:24 PM
laser
uniform distribution / pdf
Let X have the uniform distribution U(-1,3). How do you find the p.d.f. of $Y=X^{2}$?
• April 22nd 2009, 01:39 PM
mr fantastic
Quote:

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

Calculate the cdf of Y and then differentiate it to get the pdf.
• April 23rd 2009, 10:16 PM
matheagle
You need to be careful.
This one to one on 1<x<3 and two to one -1<x<1.

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

Now $f_X(x)={1\over 4}$ 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
$f_X(\sqrt y)=f_X(-\sqrt y)=1/4$.
So, for 0<y<1, we have

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

But when 1<y<9, we have a 1-1 mapping (with probability one)
So, $f_X(\sqrt y)=1/4$ while $f_X(-\sqrt y)=0$.

Thus for 1<y<9, we have

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

AND zero everwhere else.