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Two continuous random variables $X $and $Y$ have the joint density $$f(x,y) = C(x^2 + y), -1 \leqslant x \leqslant 1, 0 \leqslant y \leqslant 1.$$

  1. Compute the constant $C$.
  2. Compute the probability $P(\{Y<0.6\})$.
  3. Compute the probability $P({Y<0.6\mid X=0.5})$.

To find $C$ I'm thinking I should solve for $C$ when the indefinite integral of $f(x,y) = 1$.

As far as finding the probability in part 2 I'm thinking of finding the integral of $f(x,y)$ with respect to $C$ from $0.0$ to $0.6$.

Part three I am not sure of. Can anyone point me in the right direction, not looking for answers as much as an understandable solution.

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up vote 2 down vote accepted

You're correct on the first two, and three is already answered.

  1. Solve $\int_0^1 \int_{-1}^1 f(x,y) \, \mathrm{d}x \, \mathrm{d}y = 1$ for C. So, $$\int_0^1 \int_{-1}^1 c (x^2 + y) \, \mathrm{d}x \, \mathrm{d}y = 5c/3$$ implies $c = 3/5.$

  2. Compute $\int_0^{0.6} \int_{-1}^1 f(x,y) \, \mathrm{d}x \, \mathrm{d}y$. $$\int_0^{0.6} \int_{-1}^1 f(x,y) \, \mathrm{d}x \, \mathrm{d}y = .76 c = .456$$

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For part 2 you need to integrate $f(x,y)$ over the whole support of $x$ as well as over $y$ up to $0.6$.

For part 3 you can do $$\frac{\int_{y=0}^{0.6}f(0.5,y) \,dy}{\int_{y=0}^{1} f(0.5,y) \,dy}=\frac{\int_{y=0}^{0.6} C(0.5^2 + y) \,dy}{\int_{y=0}^{1} C(0.5^2 + y) \,dy}.$$

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Thanks, I understand part 3 now but could you help me a bit more on part 2? Would I just take the integral in respect to x from -1 to 1 and y from 0 to .6? – jrquick Nov 6 '12 at 22:14
@Jeremy: Yes, as jmbejara shows in more detail – Henry Nov 7 '12 at 8:26

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