Solutions to Quiz ECE 534 Fall 2005
Problem 1 (6 points) Let Xhave the pdf fX(x) = sin(x)
2x∈[0, π]
0 else
(a) Find the cumulative distribution function FX. In particular, what is FX(2π)?
(b) Compute E[sin(X)].
(a) The support of fXis the interval [0, π]. Thus, FX(x) = 8
<
:
0x < 0
Rx
0
sin(u)
2du =−cos(u)
2|x
0=1−cos(x)
20≤x < π
1x≥π
In particular, FX(2π) = 1.
(b) E[sin(X)] = Rπ
0sin(x)fX(x)dx =Rπ
0
sin2(x)
2dx =π
4.
Problem 2 (12 points) Suppose Xand Yare jointly continuous random variables distributed over the
unit square with the joint pdf given by
fX,Y (x, y) = 3x2
2+ 2xy x, y ∈[0,1]
0 else
(a) Calculate E[X].
(b) Are Xand Yindependent? Briefly justify your answer.
(c) Calculate the pdf, fY(y), of Y. Be sure to specify it for −∞ < y < ∞.
(d) Calculate the conditional density fX|Y(x|y). Be sure to indicate what values of yit is well-defined
for, and for such y, specify it for −∞ < x < ∞.
(a) E[X] = R1
0R1
0x(3x2
2+ 2xy)dxdy =R1
0
3x3
2dx R1
01dy +R1
02x2dx R1
0ydy =3
8+2
3
1
2=17
24
(b) No, because the joint density cannot be expressed as a function of xtimes a function of y.
(c) fY(y) = R1
0(3x2
2+ 2xy)dx =1
2+y y ∈[0,1]
0 else
(d) fX|Y(x|y) is not defined unless y∈[0,1].For y∈[0,1], fX|Y(x|y) = 8
<
:
3x2
2+2xy
1
2+yx∈[0,1]
0 else
Problem 3 (8 points) Suppose a N(0,9) random variable is passed through the quantizer function f
shown. The output is Y=f(X).
!1
!2 1 2
1
0.5
!0.5
!1
Y=f(x)
x
(a) Express the pmf of Yin terms of the Qfunction. (Check your answer. Make sure your answer is
positive for the right values of y.)
(b) Express the variance of Yin terms of the Qfunction. Give as simple an answer as possible.
(a) pY(y) = 8
>
>
<
>
>
:
1−Q(−2
3) = Q(2
3)y=±1
Q(1
3)−Q(2
3)y=±0.5
1−2Q(1
3)y= 0
0 else
(b) Var(Y) = E[Y2] = P[Y∈ {−1,1}] + 1
4P[Y∈ {−0.5,0.5}] = 2Q(2
3) + 1
42(Q(1
3)−Q(2
3)) = 1.5Q(2
3) + 0.5Q(1
3).
1