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

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 Alexander Solla Posts: 9 Registered: 12/17/10
Re: Conditional distribution problem
Posted: Jun 23, 2013 6:27 PM

On Sunday, June 23, 2013 8:36:18 AM UTC-7, Ray Vickson wrote:
> On Saturday, June 22, 2013 5:14:44 PM UTC-7, Alexander Solla wrote:

> > Given random variables X and Y with joint density
> > f(x,y) = 2(x + y) for 0 < y < x < 1,
> >
> > I am trying to compute the probability that
> >
> > P(Y < 0.5 | X = 0.1)

<snip>
> > F(0.05 | X=0.1) = \frac{1}{3(0.1)^2} 0.1 + (0.05)^2

> 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.

Oops, that was a typo. I meant

P(Y=0.05 | X=0.1)

But, your answer was still very helpful, since I've been working to understand the derivation of CDFs for order statistics, and your observation lead to a minor eureka moment.

It turns out my mistake in doing this problem (on paper) was a calculator error.

Thanks.

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