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




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(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!



