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May 1st, 2017, 06:32 PM   #1
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Finding a cdf

A point P in the unit square has coordinates X and Y chosen at random in
the interval [0, 1]. Let D be the distance from P to the nearest edge of the
square, and E the distance to the nearest corner.
(a) D < 1/4?
(b) E < 1/4?
find the cumulative distribution F and density f for the random variable D

attempt: I thought about using a distance formula to solve this but the forumla had too many coordinates. So then I decided to use P (Y is less than or equal to y) to get P (D < or equal to y), to eventually get P (D < or equal to y - (1/2)), so I eventually came up with the Cdf of F (d) = d - (1/2) on the interval of [1\2,1 and a half] but I got the wrong answer, how should I find the correct cdf?
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May 1st, 2017, 07:34 PM   #2
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these will be the regions of $D<c$ and $E<c$



$D$ is pretty clearly

$P[D<c] = 1-(1-2c)^2 = 4c(1-c),~0\leq c \leq \dfrac 1 2$

$E$ will take a bit more doing since the regions start to overlap.

I'll leave that to you.
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May 2nd, 2017, 12:26 AM   #3
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Quote:
Originally Posted by romsek View Post
these will be the regions of $D<c$ and $E<c$



$D$ is pretty clearly

$P[D<c] = 1-(1-2c)^2 = 4c(1-c),~0\leq c \leq \dfrac 1 2$

$E$ will take a bit more doing since the regions start to overlap.

I'll leave that to you.
oh actually if you just need to find

$P[E<1/4]$ then the regions won't overlap and it's really pretty simple.
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May 2nd, 2017, 06:48 PM   #4
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Hey Romsek, may I ask how did u get the 1-2c? Also, thank u for urnhelp!
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May 2nd, 2017, 06:48 PM   #5
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Quote:
Originally Posted by poopeyey2 View Post
Hey Romsek, may I ask how did u get the 1-2c? Also, thank u for urnhelp!
total length is 1, c is being subtracted from each side, 1-2c
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