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April 20th, 2017, 12:50 PM  #1 
Member Joined: Apr 2017 From: PA Posts: 45 Thanks: 0  How to find the cdf and density?
Choose a number U from the interval [0, 1] with uniform distribution. Find the cumulative distribution and density for the random variables (a) Y = U − 1/2. (b) Y = 1/(U − 1/2)^2 . 
April 20th, 2017, 01:32 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 1,977 Thanks: 1026 
this is pretty straightforward stuff do you have any particular question about it? 
April 20th, 2017, 05:50 PM  #3 
Member Joined: Apr 2017 From: PA Posts: 45 Thanks: 0 
For a, I tried to plug in 0 and 1 into the Y function and got 1/2 twice so, I got an interval (1/2,1/2), so I got confused since it was a straight line and the function needed area. So, I was wondering if there is another way to pick better intervals and find the cdf? 
April 20th, 2017, 05:58 PM  #4 
Member Joined: Apr 2017 From: PA Posts: 45 Thanks: 0 
(b) Y = (U −1/2)^2 sorry I this was actually the question for b not 1/(U1/2)^2. As for B, I pluged in 0 and 1. So, I got 1/4 twice. Then, I got the interval (1/4,1/4) again. Then I thought the cdf would be 4x, and the pdf would be 4. But in the book, it said the interval was (0,1/4) And the cdf was 2x^(1/2). I was wondering how they got the interval and the cdf? 
April 20th, 2017, 06:32 PM  #5  
Senior Member Joined: Sep 2015 From: USA Posts: 1,977 Thanks: 1026  Quote:
$F_Y(y)=P[Y < y] = 2 P[U<y],~y\in \left[0,\dfrac 1 2\right]$ $F_Y(y)= 2y,~y\in\left[0,\dfrac 1 2\right]$ $f_Y(y) = \dfrac{d}{dy} F_Y(y) = 2,~y\in \left[0,\dfrac 1 2 \right]$ I.e. $Y \sim U\left[0,\dfrac 1 2\right]$  
April 25th, 2017, 11:01 AM  #6 
Member Joined: Apr 2017 From: PA Posts: 45 Thanks: 0 
Hey Romsek, why is the CDF 2y, but not any other function?


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cdf, deb, density, find 
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