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

 Messages: [ Previous | Next ]
 Alexander Solla Posts: 9 Registered: 12/17/10
Conditional distribution problem
Posted: Jun 22, 2013 8:14 PM

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!

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