University Problem Set #12: Solutions ECE 313
of Illinois Page 1 of 3 Spring 2002
1.(a),(b) The joint pdf is nonzero on the quarter-plane shown. For v > 0, the pdf of Y is given by
fY(v) = ∫
u=–v
v
c•(v2–u2)•exp(–v)du = c•[v2u–u3/3]•exp(–v)
v
–v = c•(4/3)•v3•exp(–v). We recognize this as a
gamma pdf with parameters (4,1) which gives us that c•(4/3) = 1/Γ(4) = 1/3! and thus c = 1/8.
For u > 0, fX(u) = ∫
v=u
∞
(1/8)•(v2–u2)•exp(–v)dv = (1/8)•(–v2 –2v – 2 + u2))•exp(–v)
∞
u= (1/4)•(1+u)•exp(–u)
while for u < 0, the integral is from v = -u to ∞. Thus, we get fX(u) = (1/4)•(1+|u|)•exp(–|u|).
(c) The pdf of X is symmetric and the integral of u•(1/4)•(1+u)•exp(–u) from 0 to ∞ is finite (what’s that got to
do with it??) Hence, E[X] = 0.
1
2
1
2
2.(a) The computation of the double integral is left as an exercise since the final answer is known.
(b) fX(u) = ∫
v=0
2
(6/7)•(u2 + uv/2)dv = (6/7)•[u2v + uv2/4]
2
0 = (6/7)•(2u2 + u) for 0 < u < 1, and 0 otherwise.
(c) P{X > Y} = P{(X,Y) ∈ shaded triangle shown} = ∫
u=0
1 ∫
v=0
u
(6/7)•(u2 + uv/2)dv du = 15/56.
(d) P{Y > 1
2 | X < 1
2} = P{Y > 1
2 , X < 1
2}/P{X < 1
2}
= P{(X,Y) ∈ shaded rectangle shown}/P{(X,Y) ∈ shaded rectangle+crosshatched square}
Now, P{X < 1
2} = ∫
u=0
1/2
(6/7)•(2u2 + u)du = (6/7)•[2u3/3 + u2/2]
1/2
0= 5
28 while
P{Y > 1
2 , X < 1
2} = ∫
v=1/2
2
∫
u=0
1/2
(6/7)•(u2 + uv/2)du dv = ∫
v=1/2
2
(6/7)•[u3/3 + u2v/4]
1/2
0dv
= ∫
v=1/2
2
(3/28)(1/3 + v/2) dv = 3
28•23
16, giving that P{Y > 1
2 | X < 1
2} = 69
80.
(e) E[X] = ∫
u=0
1
u•(6/7)•(2u2 + u)du = (6/7)•[u4/2 + u3/3]
1
0= 5
7
(f) fY(v) = ∫
u=0
1
(6/7)•(u2+uv/2)du = (6/7)•[u3/3+u2v/4]
1
0 = (6/7)•(1/3 + v/4) for 0 < v < 2, and 0 otherwise.
Hence, E[Y] = ∫
v=0
2
v•(6/7)•(1/3 + v/4)dv = (6/7)•[v2/6 + v3/12]
2
0= 8
7
