Joint Probability Density Function upper bounds

Question

Let X and Y be random variables with a joint probability density function given by: f(x,y) = (x+y)/1000 for 0 <= x < 10; 0 <= y < 10.

Find P{X+Y<4}.

So to do this I've got my double integral set up with the outer integral being taken from 0 - 10 in terms of dy and the inner my lower bound is 0 however I'm not sure whether my upper bound should be 4-X or 10-X, and I'm taking this integral in terms of dx. I think it should be 10-X however I'm not 100% sure.

Any help is appreciated!

Do you know if the random variables $X$ and $Y$ are independent? It's much easier if they are. — Anthony P, Dec 08 '16 at 01:25
Check out: http://math.stackexchange.com/questions/829492/deriving-joint-cdf-from-joint-pdf — Anthony P, Dec 08 '16 at 01:28
They're not independent. That was another part of this question which I already solved. — quwerty, Dec 08 '16 at 02:00

score 1 · Accepted Answer · edited Dec 08 '16 at 03:16

1

Your joint PDF is given by

$$ f(x,y) = \frac{x+y}{1000}~\mathbf 1_{(x,y)\in[0;10]^2} $$

And the probability in question is

$$ P(x+y < 4) = P(x < 4-y) $$

Using what is above, you can set up the integral

$$\begin{align} P(x<4-y) & = \int_0^{10}\int_0^{4-y} f(x,y) \mathrm{d}x \mathrm{d}y \\[1ex] & = \int_0^{4}\int_0^{4-y} \left( \frac{x+y}{1000} \right) \mathrm{d}x \mathrm{d}y \end{align}$$

edited Dec 08 '16 at 03:16

Graham Kemp

129,094

answered Dec 08 '16 at 02:56

Anthony P

609

Oh, my outer integral is also 0-4, I thought that was 0-10. Thanks for catching that! – quwerty Dec 08 '16 at 10:22

Joint Probability Density Function upper bounds

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