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October 19th, 2015, 09:59 AM   #1
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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.
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October 19th, 2015, 02:58 PM   #2
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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.
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October 19th, 2015, 08:56 PM   #3
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can you please tell me the general ideas for solving such problems or introduce some good books I'm stuck with such problems
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October 20th, 2015, 06:46 PM   #4
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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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