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Topic: Conditional distribution problem
Replies: 3   Last Post: Jun 23, 2013 6:59 PM

 Messages: [ Previous | Next ]
 RGVickson@shaw.ca Posts: 1,677 Registered: 12/1/07
Re: Conditional distribution problem
Posted: Jun 23, 2013 11:36 AM

On Saturday, June 22, 2013 5:14:44 PM UTC-7, Alexander Solla wrote:
> Hi everybody!
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> Given random variables X and Y with joint density
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> f(x,y) = 2(x + y) for 0 < y < x < 1,
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> I am trying to compute the probability that
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> P(Y < 0.5 | X = 0.1)
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> To that end, I computed the marginal density
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> f_X(x) = int_0^x 2(x + y) \d{y}
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> = 3x^2
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> We now have enough information to compute the conditional distribution
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> f(y|x) = \frac{2(x + y)}{3x^2}
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> and, fixing the condition,
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> f(y|x = 0.1) = \frac{0.2 + y}{3(0.1)^2}
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> Finally, to compute the probability, we find the conditional distribution
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> F(y | X=0.1) = \frac{1}{3(0.1)^2} \int_0^y 0.2 + 2y \d{y}
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> = \frac{1}{3(0.1)^2} 0.2y + y^2
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> and substitute
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> F(0.05 | X=0.1) = \frac{1}{3(0.1)^2} 0.1 + (0.05)^2
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> Plugging these into a calculator yields approximately 4.17, instead of 0.417, which my solution sheet says is the answer. Where am I going wrong?
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> Thanks!

Both you and the answer sheet are wrong. Remember: your random variables X and Y are restricted by 0 <= Y <= X <= 1, so if you are given the event {X = 1/10}, Y will be restricted to the interval [0, 1/10], and so will be < 1/2 with probability 1; that is, P{Y <= 1/2|X = 1/10} = 1. Your conditional cdf will vary with y only on the interval 0 <= y <= 1/10, and will be the constant 1 for y > 1/10.

Date Subject Author
6/22/13 Alexander Solla
6/23/13 RGVickson@shaw.ca
6/23/13 Alexander Solla
6/23/13 quasi