Find $P(X+Y\le 0)$ given the joint probability function of $X$ and $Y$

Question

I am struggling with part c of this question. Could someone please tell me how to approach and solve this type of questions?

Answer 1

The marginal probability of Y means you sum across the columns for a given value of Y.

a) P(Y = -2) = 0.1 + 0.2 + 0.2 = 0.5
$\quad$ P(Y = 1) = 0.1 + 0.2 + 0 = 0.3
$\quad$ P(Y = 2) = 0.2 + 0 + 0 = 0.2

b) The possible values of XY are 2, 0, 4, -1, 0, 2, -2, 0, 4
E(XY) is simply the weighted average of the sum of their values with the weights being the probabilties corresponding to the XY value.

c) See Tom's answer

d) $E(X | Y=1) = (-1)P(X=-1|Y=1) + (0)P(X=0|Y=1) + (2)P(X=2|Y=1)$
$\quad$ $= (-1)\frac{0.1}{0.1 + 0.2 + 0} + (0)\frac{0.2}{0.1+0.2+0} + (2)\frac{0}{0.1+0.2+0} = -\frac{1}{3}$

Answer 2

Hint: You know that $X$ takes on the values $-1, 0,$ and $2$; you know $Y$ takes on the values of $-2, 1,$ and $2$. So, the values of $X+Y$ are ...? After you answer this, suppose you wanted find $P(X+Y=-2)$, the only way this can happen is if $Y=-2$ and $X=0$, so $P(X+Y=-2) = P(X=0,Y=-2) = .2$

Can you finish the problem now?

Answer 3

ad a)

$P(Y=y_j)=\sum_{i \in I} \ p(X=x_i,Y=y_j) $

$x_i \in \{-1,0,2 \}; \ y_j \in \{-2,1,2 \}$

$P(Y=-2)=0.1+0.2+0.2=0.5$

Sum of the first row. For the whole marginal probability function you have additional calculate $P(Y=1)$ and $P(Y=2)$

ad b)

$E(x\cdot y)=\sum_{i \in I} \sum_{j \in J} \ \ x_i\cdot y_i\cdot P(X=x_i,Y=y_i)=(-1) \cdot (-2) \cdot 0.1 + \ldots$

You have to finish the calculation.