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Re: Conditional distribution problem
Posted:
Jun 23, 2013 11:36 AM


On Saturday, June 22, 2013 5:14:44 PM UTC7, 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(yx) = \frac{2(x + y)}{3x^2} > > > > and, fixing the condition, > > > > f(yx = 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 <= 1/2X = 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.



