Download Math 4250 Spring 2007 Take-Home Assignment # 5 with solutions

Math 4250
Spring 2007
Take-Home Assignment # 5
with solutions
Instructions: Your solutions must appear in an organized and legible format to be
given full consideration.
1. (ex9 pp314) The joint probability density function of X and Y is given by
6 2 xy f (x, y) =
x +
7
2
0 < x < 1, 0 < y < 2.
(a) Verify that this is indeed a joint density function.
(b) Compute the density function of X.
(c) Find P [X
> Y1 ] .
(d) Find P Y > 2 |X < 12 .
(e) Find E[X].
(f) Find E[Y ].
Solution:
(a)
6
7
Z
1
2
Z
0
x2 +
0
xy dxdy = 1.
2
(b)
6
fX (x) =
7
Z
2
0
xy 6
x +
dy = (2x2 + x).
2
7
2
(c)
6
P (X > Y ) =
7
Z
0
1
Z
0
x
x2 +
xy 15
dxdy = .
2
56
(d)
P (X > 1/2|X < 1/2) = P (X > 1/2, X < 1/2)P (X < 1/2)
R 2 R 1/2 2 xy x + 2 dxdy
1/2 0
.
=
R 1/2
2 + x)dx
(2x
0
(e)
Z
Z
6 1
6 1 3
2
E[X] =
xfX (x)dx =
x(2x + x)dx =
(2x + x2 )dx
7 0
7 0
0
6 1 4 1 3 1 5
=
x + x |0 = .
7 2
3
7
Z
1
2. (ex20 pp315) The joint density of X and Y is given by
f (x, y) =
xe−(x+y) x > 0, y > 0
0
otherwise
Are X and Y independent? What if f (x, y) were given by
f (x, y) =
2 0 < x < y, 0 < y < 1
0
otherwise
Solution:
(a)
∞
Z
∞
Z
f (x, y)dy =
fX (x) =
xe
−∞
0
∞
Z
Z
fY (y) =
−(x+y)
xe−x
x>0
0
otherwise
e−y
y>0
0
otherwise
dy =
∞
f (x, y)dx =
ye
−∞
−(x+y)
dx =
0
Hence, it is clear that
f (x, y) = fX (x) · fY (y).
and as a consequence X and Y are independent.
(b)
Z ∞
Z 1
2(1 − x) 0 < x < 1
fX (x) =
f (x, y)dy =
2dy =
0
otherwise
−∞
x
and
Z
∞
Z
−∞
2dx =
f (x, y)dx =
fY (y) =
y
0
2y 0 < y < 1
0 otherwise
We have that
f (x, y) 6= fX (x) · fY (y),
hence X and Y are not independent.
3. (ex27 pp316) If X is uniformly distributed over (0, 1) and Y is exponentially distributed with parameter λ = 1, find the distribution of
(a) Z = X + Y .
(b) Z = X
. Assume independence.
Y
Solution:
fX (x) =
fY (y) =
1 0<x<1
0 otherwise
e−y
y>0
0
otherwise
(a) If a ≤ 0 then FZ (a) = 0.
R a R a−x −y
e dydx = a − 1 + e−a
0<a<1
P (X + Y ≤ a) = R01 R0a−x −y
−a
e dydx = 1 − e (e − 1)
a>1
0 0
(b)
1
Z
∞
Z
e−y dydx = a(1 − e(−1/a) ).
P (Y ≥ X/a) =
0
x/a
4. (ex5 pp319) If X and Y are independent continuous positive random variables,
express the density function of
(a) Z = X
.
Y
(b) Z = XY in terms of the density functions of X and Y . Evaluate these expressions in the special case where X and Y are both exponential random variables.
Solution:
(a) For a > 0
∞
Z
ay
Z
FZ (a) = P (X ≤ aY ) =
Z
∞
fX (x)fY (y)dxdy =
0
FX (ay)fY (y)dy.
0
0
∞
Z
fZ (a) =
yfX (ay)fY (y)dy.
0
(b)
Z
∞
a/y
Z
FZ (a) = P (XY ≤ a) =
Z
0
∞
FX (a/y)fY (y)dy.
fX (x)fY (y)dxdy =
0
0
∞
Z
fZ (a) =
0
1
fX (a/y)fY (y)dy.
y
If X ∼ exp(λ) and X ∼ exp(µ) then (a) and (b) reduce to
(a)
Z
∞
FZ (a) =
λye−λay µe−µy dy.
0
(b)
Z
FZ (a) =
0
∞
1
λ e−λa/y µe−µ/y dy.
y
5. (ex2 pp323) The joint probability mass function of the random variables X, Y, Z is
1
p(1, 2, 3) = p(2, 1, 1) = p(2, 2, 1) = p(2, 3, 2) = .
4
Find
(a) E[XY Z].
(b) E[XY + XZ + Y Z].
Solution: Let us recall the following formula
X
E[g(X, Y, Z)] =
g(x, y, z)pX,Y,Z (x, y, z).
x,y,z
Moreover, the rv
X, Y, Z : S 7−→ {1, . . . , 27} .
where S is the sample space.
(a) Here g(x, y, z) = xyz. Hence
1
E[XY Z] = 1·2·3p(1, 2, 3)+2·1·1p(2, 1, 1)+2·2·1p(2, 2, 1)+2·3·2p(2, 3, 2) = (6+2+4+12) = 6.
4
(b) Here g(x, y, z) = xy + xz + yz. Hence
E[XY + XZ + Y Z] = (1 · 2 + 1 · 3 + 2 · 3)p(1, 2, 3)
+ (2 · 1 + 2 · 1 + 1 · 1)p(2, 1, 1)
+ (2 · 2 + 2 · 1 + 2 · 1)p(2, 2, 1)
1
+ (2 · 3 + 2 · 2 + 3 · 2)p(2, 3, 2) = (11 + 5 + 8 + 16) = 10.
4