My Math Forum function of random variable

 October 19th, 2015, 08:59 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 09:11 AM.
 October 19th, 2015, 01:58 PM #2 Global Moderator   Joined: May 2007 Posts: 6,276 Thanks: 516 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. Thanks from mhhojati
 October 19th, 2015, 07:56 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, 05:46 PM #4 Global Moderator   Joined: May 2007 Posts: 6,276 Thanks: 516 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. Thanks from mhhojati

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