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University of Illinois
ECE 313 FALL 2007
Professors Tamer Ba¸sar & Ada Poon November 12, 2007
TEST II
7:00 p.m. - 8:30 p.m.
NOTE: This is a closed-book closed-notes (and closed-neighbor) exam, with only one sheet of notes
1
2
′′
× 11
′′ ) allowed. Also, no calculators, laptops, palm pilots, and the like are allowed.
Name (Last, First) : ................................................................
Section : © Section C, 10 MWF © Section D, 11 MWF
Problem 1 : ........................ (20 points)
Problem 2 : ........................ (40 points)
Problem 3 : ........................ (20 points)
Problem 4 : ........................ (20 points)
TOTAL : ........................ (100 points)
Problem 1 (20 points)
X is a random variable with cumulative distribution function (cdf )
FX (a) =
0 a < − 1
(a + 1)/ 4 − 1 ≤ a < 1
1 1 ≤ a
(Sketching F X
(a) may help you in answering the questions below.)
Each of the following statements pertain to the random varaioble X whose cdf is given above. Read
each one carefully, and check the corresponding True box if the statement is true; otherwise check the
corresponding False box. Each correct choice counts +2 points, whereas an incorrect choice counts − 1
point; so guess at your own risk.
You can use the space provided at the bottom as well as the facing page for scratch work.
TRUE FALSE
t t P (X = 0) =
1
4
t t P (X = 1) =
1
2
t t P (|X| <
1
2
1
4
t t X admits a pdf, which is given by f X
(x) =
1
4
− 1 < x < 1
0 else
t t The mean value of X is E[X] =
1
2
t t P (|X| <
1
2
||X| ≤ 1) =
1
4
t t P (|X| <
1
2
||X| < 1) ≤
1
4
t t X is a continuous random variable.
t t If Y = 6X − 2, the mean value of Y is E[Y ] = 1
t t There is no value of c for which P (X ≤ c) =
3
4
END OF PROBLEM 1
(vi) Let X be a gamma random variable with parameter λ = 2 and order α = 4. The variance of X, var(X),
is:
a. 1 b. 0. 5 c. 2 d. 4
(vii) Let X be a uniform random variable on the interval [− 2 , 4]. Let Y = e
2 X
. The pdf of Y is :
a. fY (y) =
1 /(12y) e
− 4 ≤ y ≤ e
8
0 else
b. fY (y) =
1 /(2y) e
− 4 ≤ y ≤ e
8
0 else
c. Y has a pdf, but it is none of these two d. Y does not have a pdf
(viii) The hazard rate function of a positive random variable X is given as λ(t) = 3t , t ≥ 0. The pdf of X,
fX (x), is for all x ≥ 0:
a. 1 − e
−x
2
b. 3 xe
−x
2
c. 3 e
− 3 x d. none of these
(ix) Two discrete random variables, X and Y , take on only four values each, −1, 0, 1, and 2. The joint
probability mass function (pmf ) of the pair (X, Y ) is pX,Y (i, j) = c for all i, j = − 1 , 0 , 1 , 2. The value
of the constant c is:
a.
1
4
b.
1
16
c.
1
8
d. none of these
(x) Let X and Y be as in (ix) above. The expected value of Y , E[Y ], is:
a. 1 b.
1
2
c. 2 d. none of these
END OF PROBLEM 2
Problem 3 (4 + 4 + 5 + 7 = 20 points)
Let X and Z be two independent random variables, where Z is a Gaussian random variable with mean μ
and variance σ
2 , and X is a discrete random variable taking two values: −1 with probability p and 1 with
probability 1 − p. Let Y be another random variable related to X and Z by:
Y = X + Z
(The answers to the four questions below will be in terms of all or some of the parameters p, μ and σ.
Further, when necessary, you can express the answers in terms of the Gaussian error function (standard
Gaussian cdf) Φ, defined by Φ(x) =
1 √
2 π
x
−∞
e
−(y
2 /2) dy.)
(i) Given that X = −1 (that is, conditioned on X = −1), what is the probability that Y > θ, where θ is a
real number?
P (Y > θ | X = −1) =
(ii) Given that X = 1 (that is, conditioned on X = 1), what is the probability that Y < θ?
P (Y < θ | X = 1) =
Problem 4 (7(5 + 3) + 2 + 10(3 + 7) = 20 points)
Let X, Y and W be three independent random variables, where X and Y are exponential random variables
with the same parameter value λ = 1, and W is a Bernoulli random variable with parameter p =
1
2
. Let Z
be another random variable, expressed in terms of X, Y and W as follows:
Z = W X − (1 − W )Y
That is, Z is X when W = 1, and Z is −Y when W = 0.
(i) Obtain the expressions for the cumulative distribution function (cdf) and the probability density function
(pdf) of Z.
F
Z
(c) =
f Z
(z) =
(ii) Sketch the pdf of Z.
(iii) Let V = g(Z), where
g(z) =
z + 1, − 1 ≤ z < 0
z
3
, 0 ≤ z < 3
0 , otherwise
(a) Compute the probability of the event {V = 0}, and (b) obtain the expression for the cdf of V.
(In answering these two questions, you will find it useful to sketch the function g.)
P (V = 0) =
FV (a) =
END OF PROBLEM 4
SCRATCH SHEET II
