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October 19th, 2015, 09:59 AM  #1 
Member Joined: Jan 2012 Posts: 57 Thanks: 0  function of random variable
suppose X is uniformly distributed over [1,3] and $\displaystyle Y=X^2$ Find the CDF $\displaystyle F_{Y}(y)$? since X is unoformly distrbuted $\displaystyle f_{X}(x)=\frac{1}{4} $ for [1,3] otherwise 0. $\displaystyle F_{Y}(y)=P[X^{2}\leqslant y]=P[\sqrt{y} \leqslant X\leqslant \sqrt{y} ]=\int_{\sqrt{y}}^{\sqrt{y}}f_{X}(x)dx $ the calculation of the integral depends on the value of y I don't know how to deal the rest of the problem for example what is the value of $\displaystyle F_{Y}(y)$ for $\displaystyle 0\leq y\leq 1$ or $\displaystyle 1\leq y\leq 9$ thanks Last edited by mhhojati; October 19th, 2015 at 10:11 AM. 
October 19th, 2015, 02:58 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,395 Thanks: 545 
You need to separate the integral into 2 cases. For y < 1, use the integral as is. For y > 1, the lower limit should be 1.

October 19th, 2015, 08:56 PM  #3 
Member Joined: Jan 2012 Posts: 57 Thanks: 0 
can you please tell me the general ideas for solving such problems or introduce some good books I'm stuck with such problems

October 20th, 2015, 06:46 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,395 Thanks: 545 
I don't have any particular recommendation. However, your approach was correct, except for not taking care of the lower limit of x (1) when applicable.


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