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$f(x,y) = c(x^3+ \frac {xy}4), 0<x<1, 0<y<2$

a) For what value of c is this a joint density function:

edit:

$c*[\int_0^1\int_0^2 x^3 + \frac {xy}4 = 1 $

$c=\frac 43$

If i have done this all correct, then $c=\frac{16}5$

b) Using the value of c, computer the density function of Y

c) find P{X>Y}

$f(x,y) = \frac{4}3(x^3+ \frac {xy}4)$

i dont know what to do.

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The $c$ is incorrect. You need to "double integrate" the $x^3$ part too. –  André Nicolas Feb 26 '13 at 22:19
    
can you tell me what i did wrong? edit: i dont get it? –  Special--k Feb 26 '13 at 22:21
    
We are integrating $x^3+\frac{xy}{4}$. Integrate first with respect to $x$, we get $\frac{x^4}{4}+\frac{x^2y}{8}$. Plug in endpoints, We get $\frac{1}{4}+\frac{y}{8}$. Integrate with respect to $y$. We get $\frac{y}{4}+\frac{y^2}{16}$. Plug in endpoints. We end up with $\frac{3}[4}$. Your mistake was finding the double integral of $x^3$ incorrectly. Since you didn't see a $y$, you didn't integrate w.r.t. $y$. –  André Nicolas Feb 26 '13 at 22:27
    
changed. thanks –  Special--k Feb 26 '13 at 22:34
    
For the last part, make sure you draw a picture. If you don't, there is a very good chance of getting the limits of integration wrong. –  André Nicolas Feb 26 '13 at 22:35

1 Answer 1

up vote 1 down vote accepted

For b) You need to compute the marginal density function of Y and to do that you integrate over $x$

$f_{Y}(y)=\int_0^1 f(x,y)dx$

For c) $P(X>Y)=\int_0^1 \int_0^x f(x,y)dxdy$

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