Date: Jun 23, 2013 11:36 AM
Author: RGVickson@shaw.ca
Subject: Re: Conditional distribution problem

On Saturday, June 22, 2013 5:14:44 PM UTC-7, Alexander Solla wrote:> Hi everybody!> > > > 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)> > > > To that end, I computed the marginal density> > > > f_X(x) = int_0^x 2(x + y) \d{y}> >        = 3x^2> > > > We now have enough information to compute the conditional distribution> > > > f(y|x) = \frac{2(x + y)}{3x^2}> > > > and, fixing the condition,> > > > f(y|x = 0.1) = \frac{0.2 + y}{3(0.1)^2}> > > > Finally, to compute the probability, we find the conditional distribution> > > > F(y | X=0.1)    = \frac{1}{3(0.1)^2} \int_0^y 0.2  + 2y \d{y}> >                 = \frac{1}{3(0.1)^2}          0.2y + y^2> > > > and substitute> > > > F(0.05 | X=0.1) = \frac{1}{3(0.1)^2}          0.1 + (0.05)^2> > > > 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?> > > > 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.