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Let X and Y have the joint probability density function

$f(x,y) = \frac{3}{2}(x^2 +y^2)$, $0<x<1,0<y<1$

a.) Find $P(Y<.5|X>.5)$. My answer is $\frac{5}{11}$.

b.) Find $P(Y>2X)$. I got $\frac{1}{4}$ but I'm not sure if I'm supposed to integrate the joint pdf or the marginal function $f_Y$.

c.) Find $P(.5<X+Y<1.5)$. I set up the integral as $\int_{.5+y}^{1.5-y}$ but again, not sure. This is the one I'd prefer to be answered most. My book has one example but it's only for $X+Y<a$, and the integrand is $f_Y(y)dyf_X(x)dx$.

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For parts (b) and (c), it is most helpful to draw a diagram of the $x$-$y$ plane, showing the unit square where the joint pdf is nonzero, and marking on it the regions where $(X,Y)$ must lie in order to satisfy the stated conditions. Then, setting up the integrals becomes very straightforward. One can even think about whether it is better to have the inner integral be with respect to $x$ or $y$, or whether calculating the complementary probability and subtracting from $1$ might give an easier way of getting at the answer. – Dilip Sarwate Apr 24 '13 at 2:13

For part b, I got .203125.

I set it up as follows: $$ \frac{3}{2} \int_0^\frac{1}{2} \int_{2x}^1 x^2 + y^2 dy dx $$

My explanation is, set y = 2x. Find the area bound by the space of (0,1) for x and y, that is above y = 2x. It needs to be a double integral (thus the joint probability) to get an actual probability. (The marginal would result in a function, not a probability value.)

For part c, I am pretty sure this is a convolution which requires the convolution formula, which is a lot of tedious integration with the joint pdf you have.

You'd have to break it up into the intervals $.5<z<1$ and $1<z<1.5$

I am learning this too, so I post my answer with limited confidence, arithmetic errors or otherwise. I will watch to see what the more experienced have to say.

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No, you don't want to do a convolution for part (c) because (i) $X$ and $Y$ are not independent random variables and so convolution will not give you the density of $X+Y$, and (ii) if $X$ and $Y$ were independent so that convolution could be used, convolution would be a much longer way of arriving at the answer than a straightforward double integral. – Dilip Sarwate Apr 24 '13 at 2:18
In that case, is the double integral, int 0 to .5, int (.5-x) to 1 dy dx + int .5 to 1, int 0 to (1.5-x) dy dx? I tried that but second guessed myself. (I got .7266) – Katarzyna Apr 24 '13 at 3:50
No, your limits are incorrect. Draw a diagram as I suggested in a comment on the question and it will help you figure out what the correct limits are. – Dilip Sarwate Apr 24 '13 at 17:02
@Dilip - I actually did have it right, you probably just couldn't translate my lazy shorthand. (I didn't know you could use latex in the comments.) Anyway, for posterity: $\int_0^.5 \int_{.5-x}^1 f(x,y)~dydx$ + $\int_.5^1 \int_0^{1.5-x} f(x,y)~dydx$ checks out with wolfies 's answer below. Also, I did the complement as you suggested, and got the same answer. – Katarzyna Apr 26 '13 at 0:07
... But my arithmetic above was wrong... (added instead of subtracted one of the dozen or so terms.) But I re-did it today with a satisfactory result. – Katarzyna Apr 26 '13 at 0:15

This looks a bit like homework, but I thought it fun to check some of your solutions using the mathStatica add-on to Mathematica. Presumably, you will need to show workings anyway :).

Given: random variables X and Y have joint pdf f(x,y):

f = (3/2)*(x^2 + y^2);     domain[f] = {{x, 0, 1}, {y, 0, 1}}; 

Then, the solution to part b is:

Prob[y > 2 x, f]


and the solution to part c is:

Prob[1/2 < x + y < 3/2, f]


Hope this helps

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