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




Re: Conditional distribution problem
Posted:
Jun 23, 2013 6:27 PM


On Sunday, June 23, 2013 8:36:18 AM UTC7, Ray Vickson wrote: > On Saturday, June 22, 2013 5:14:44 PM UTC7, 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 <= 1/2X = 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.



