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Let X and Y be two jointly continuous random variables with the given joint PDF; $\begin{equation} \nonumber f_{XY}(x,y) = \left\{ \begin{array}{l l} \frac{1}{3}x^3+\frac{1}{5}y^2 & \quad -2 \leq x \leq 1, 0 \leq y \leq 3 \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation}$

Could someone check my work here?

The marginal pdf of X; $f_{X}(x) = \int_{0}^{3}f_{XY}(x,y)dy$

And the marginal pdf of X is $ f_{Y}(y) = \int_{-2}^{1}f_{XY}(x,y)dx$

$P(X>0,y<2) = \int_{0}^{2} \int_{0}^{1} f_{XY}(x,y) dx dy$

$P(X<-1 or Y<2) = P(X<-1)+P(Y<2)$

Having trouble with last part of this problem which is;

$P(X+Y>0) = ?$

All I know is that $X+Y>0$ when for $X = a, Y>-a$ or when X = a & Y > -a also when X and Y are both greater than 0 i.e. $X>0,Y>0$.

Unsure where to go from here.

pyler
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  • You can use & to display the '&' symbol in math mode. Additionally, $P(X<-1\text{ or }Y<2)=P(X<-1)+P(Y<2)-P(X<-1,Y<2)$ which you can calculate as you have done in the previous question. – Ian May 03 '14 at 02:43
  • It could be of somehow helpful.

    http://math.stackexchange.com/questions/261073/finding-probability-pxy/578458#578458

     –  May 03 '14 at 02:45
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    The function $f_{XY}$ is not a joint PDF. Have you copied the question correctly? –  May 03 '14 at 04:08

1 Answers1

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The answer to the "or" question is not set up correctly. Let $A$ be the event $X\lt -1$ and $B$ the event $Y\lt 2$. Then $\Pr(A\cup B)=\Pr(A)+\Pr(B) -\Pr(A\cap B)$. Each of the probabilities on the right is an easily set up integral.

For the last question, draw the line $x+y=0$. We want to find the probability that the pair $(X,Y)$ ends up in the part of our basic rectangle above the line $x+y=0$. The picture should make it not too hard to set up an appropriate integration. But it will be substantially easier to compute the probability that $X+Y\le 0$, since the region of integration is nicer.

André Nicolas
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