You can find E(X) just from f(x), or from f(x,y). You
can
find E(XY) only from f(x,y) (unless X and Y are indep.)
You can find E(X+Y) from f(x,y), or by finding E(X) +E(Y).
These things work because of the rules of iterated integration, where
the
variable not being integrated acts like a constant for the
moment.
Reread Ch. 7-1, the parts with integration.
Find E(XY) for f(x,y) = y/18,
0<x<4,
0<y<3. oS3oS4
xyy/18 dxdy = (1/18)
oS
3 [(1/2) x2y2 x=o]4dy
= (1/18) oS 3 [8y2
- 0]dy = (1/18) [8y3/3
y=o]3 = (1/18) [8·27/3
] = 4
Note E(XY) = oS3oS4xyy/18
dxdy = oS32y2/9 [oS4
x/4 dx]dy =[oS4 x/4
dx][ oS32y2/9 dy]
=
EX EY = 2·2
Find E (X/Y) for f(x,y)
= y/18, 0<x<4,
0<y<3. oS3oS4
(x/y)(y/18) dxdy = oS3oS4
(x/18) dxdy
= (1/18)
oS
3 [(1/2) x2 x=o]4dy =
(1/18)
oS
3 (8 - 0)dy = (1/18) [8y
y=o]3 = (1/18) [8·3
] = 4/3
--- --- --- --- --- --- --- --- ---
HW Multivariate Continuous, a few more
on Independence:
A. On the Handout "MultCont", are the variables in
problem 17 independent? How about in 18?
Ash p. 191, #5. For each part: If they are independent, what are
the marginals?. Sketch the regions where f is positive, before
bothering to look for the marginals.
#9
HW: (Re)read Ash Ch. 7,
now focusing on Continuous random variables in Joint situations
Expected values: Ash p. 222 #4. (One-variable, good
review.)
On Handout "MultCont":
For the density of problem #17, find E(XY), and E(Y/X).
#18 c (You found the marginals before. Use
them to find the E's)
#19 c (do it by noting the form of the densities
of X, and Y, and appealing to your list of means for known
distributions)
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