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April 20th, 2017, 12:50 PM   #1
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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
.
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April 20th, 2017, 01:32 PM   #2
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this is pretty straightforward stuff

do you have any particular question about it?
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April 20th, 2017, 05:50 PM   #3
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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?
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April 20th, 2017, 05:58 PM   #4
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(b) Y = (U −1/2)^2 sorry I this was actually the question for b not 1/(U-1/2)^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?
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April 20th, 2017, 06:32 PM   #5
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Quote:
Originally Posted by poopeyey2 View Post
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
.
(a)

$F_Y(y)=P[Y < y] = 2 P[U<y],~y\in \left[0,\dfrac 1 2\right]$

$F_Y(y)= 2y,~y\in\left[0,\dfrac 1 2\right]$

$f_Y(y) = \dfrac{d}{dy} F_Y(y) = 2,~y\in \left[0,\dfrac 1 2 \right]$

I.e. $Y \sim U\left[0,\dfrac 1 2\right]$
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