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Old 06-14-2018, 12:09 PM
Mitsu96 Mitsu96 is offline
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Default Joint pdf of random variables! Help!

Hello,

So, I understand that in order to get the probability of Y, you will kind of need to break it down into three probabilities.

1) P[0 <= X <= 0.5] = F(0.5) - F(0) = 1/20

2) P[9.5 <= X <= 10] = F[10] - F[9.5] = 1/20

And, these two are correct. However, I am having trouble finding out the probability of the remaining numbers, such as, 1,2,3,4,5,6,7,8 and 9...I believe that it is 9/20. However, the actual answer is 1/20. And I can't really figure out how.
Thank you for all the help!

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Old 06-14-2018, 12:50 PM
DKH21 DKH21 is offline
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P(Y = 1) = P(0.5<= X <1.5), P(Y=2) = P(1.5<= X < 2.5) etc. This is using the definition of Y directly where Y is rounded to the nearest integer
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Old 06-15-2018, 09:30 AM
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Vorian Atreides Vorian Atreides is offline
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Don't forget P(Y = 0).
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Old 06-17-2018, 02:24 AM
Mitsu96 Mitsu96 is offline
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Quote:
Originally Posted by DKH21 View Post
P(Y = 1) = P(0.5<= X <1.5), P(Y=2) = P(1.5<= X < 2.5) etc. This is using the definition of Y directly where Y is rounded to the nearest integer
Oh...I see what you mean...thank you so much for your help!
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