# Marginal distribution

• April 12th 2008, 04:58 AM
matty888
Marginal distribution
For the following joint distribution

---Y
X- 1-- 2
0 .02 .08
1 .11 .44
2 .07 .28

(i) Calculate the marginal distribution for each of X and Y
(ii) Find
a) P(X >= 1)
b) P(X = 1 | Y = 2)
• April 12th 2008, 11:39 AM
lllll
for i)
to find $X$ you just have to sum across the rows, so you get:

$X_0 = 0.02+0.08 = 0.1$
$X_1 = 0.11+0.44 = 0.55$
$X_2 = 0.07+0.28 = 0.35$

and for $Y$ you do the same thing across the columns, so you get:
$Y_1 = 0.02+0.11+0.07= 0.2$
$Y_2 = 0.08 +0.44 +0.28 = 0.8$

for ii)
a) so $P(X\geq 1) = P(X=1) \ \cup \ P(X=2) = 0.55+0.35 = 0.9$
b) $P(X=1|Y=2) = \frac{P(X=1 \ \cap \ Y=2)}{P(Y=2)} = \frac{0.44}{0.8} = \frac{11}{20}$

for iii)
they're dependent, but I'm still trying to figure out why.