Midterm 3 Solutions
Math 5010–1, Spring 2005
(a) Compute C.
Solution: We have
Z ∞ Z 2x
1. Let λ > 0 be a fixed number. Recall that the pmf for a
Poisson(λ) random variable is: p(k) = e−λ λk /k! for k =
0, 1, 2, . . . .
Now suppose X and Y are independent, each is Poisson(2).
Find the conditional mass function of X given that X + Y =
10. [Show all of your work; remember, ‘no detail = no
credit’.]
Solution: Recall that X +Y is Poisson(4). Therefore,
e−4 410
.
P{X +Y = 10} =
10!
Here is why:
10
P{X +Y = 10} =
∑ P{X = a}P{Y = 10 − a}
a=0
e−2 2a e−2 210−a
(10 − a)!
a=0 a!
10
=
∑
= e−4 210
10
1
∑ a!(10 − a)!
1=C
Therefore, C = 2 .
0
C
.
2
e−y dy dx =
x
(b) Let (R, Θ) denote the polar coordinates of (X,Y ). Compute the joint pdf of (R, Θ). Are R and Θ independent?
Solution: We have seen in lecture that J(x, y) = 1/r.
Therefore,
fR,Θ (r, θ) = r fX,Y (x, y)
= 2re−r sin θ ,
r > 0,
π
4
<θ<
3π
8
This cannot
be written as fR(r) × fΘ (θ). Therefore, R
and Θ are not independent .
3. Let R be a fixed positive number. A point (X,Y ) is selected
uniformly at random from the circle of radius R centered at
the origin, (0, 0).
a=0
=
e−4 210
10!
10
∑
a=0
10
e−4 410
=
.
a
10!
The last identity follows from the binomial theorem.
(a) Compute P{X > R2 }.
Solution:
√
Z
R
P X > 2 =2
This yields,
pX|Y (x|10) =
e−2 2x
Z
R2 −x2
1
dy dx.
πR2
Compute directly to finish.
e−2 210−x
10!
×
×
−4
e 410
x!
(10 − x)!
x
10−x
10
1
1
= x 2
1− 2
,
for x = 0, . . . , 10. [The probab. is zero for other values of
x.] So, conditional on {X + Y = 10}, X is Binomial with
parameters n = 10 and p = 1/2.
=
R
R/2 0
pX (x)pY (10 − x)
pX+Y (10)
2. Let (X,Y ) be a random vector with joint pdf
(
Ce−y , if 0 < x < y < 2x,
f (x, y) =
0,
otherwise.
Here, C is a positive, unknown constant that does not depend
on (x, y).
.
(b) Compute P{X >
R
2
|Y = R2 }.
Solution: We are after
R R
P X > Y =
2
2
Z √ 2
2
R −(R/2)
=
R/2
fX|Y (x|R/2) dx.
You find the latter conditional pdf as follows:
fX|Y (x|R/2) =
f (x, R/2)
fY (R/2)
=
1/πR2
,
fY (R/2)
p
√
√
if x is between − R2 − (R/2)2 = − 12 R 3 and 12 R 3;
else fX|Y (x|R/2) = 0. To compute fY (R/2), we merely
compute
√
Z 1 R √3
2
1
3
dx =
fY (R/2) = 1 √
.
2
πR
πR
−2R 3
Therefore,
1
fX|Y (x|R/2) = √ ,
R 3
√
√
1
1
if x is between − 2 R 3 and 2 R 3; else, fX|Y (x|R/2) =
0. Therefore,
Z 1 R √3
2
R
1
R √ dx
P X > Y =
=
2
2
R/2
R 3
!
√
3 1
1
−
=√
2
2
3
=
1
1
− √ .
2 2 3