NORTHEASTERN UNIVERSITY

Department of Electrical and Computer Engineering

EECE-7204
DS5020 Appl. Prob. & Stoch. Proc. FALL-2016
Homework-310Solutions
Solutions

P3.1 Stark/Woods Text Problem 3.18 (Page 207) and 3.25 Page (210).

(a) Text Problem 3.18 (Page 207).

Solution: Random variables X and Y are iid with with f X (x) = f Y (x) = Æe°Æx u(x). A new
random variable is Z = X ° Y . Let V , °Y . Then Z = X +V . The CDF and the pdf of V are
given by

dF v
FV (v) = Pr [V ∑ v] = Pr [Y ∏ °v] = 1 ° F Y (°v) ) f Y (y) = = f Y (°v)
dv

since Y is continuous. Therefore, f V (v) = ÆeÆv u(°v). Now since X and V are indepen-
dent, the pdf of Z = X + V is given by
£ § £ §
f z (z) = f X (z) § f V (z) = Æe°Æz u(z) § ÆeÆz u(°z)
Z1 Z1
2 °Æx Æ(z°x) 2 Æz
= Æ e e u(x)u(x ° z) dx = Æ e e°2Æx dx
x=°1 max(0,z)
(R1 (
°2Æx Æz
e dx z < 0, Æ e z < 0, Æ °Æ|z|
= Æ2 eÆz R01 °2Æx = °Æz
= e , 8z
z e dx z ∏ 0. 2 e z ∏ 0. 2

(b) Text Problem 3.25 (Page 210).

Solution: In Text Problem 3.18 we showed that the random variable Z = X ° Y has the pdf
Æ °Æ|z|
f Z (z) = e .
2

Now let (
Z, Z ∏ 0,
W , |Z | =
°Z , Z < 0.

Hence there are two monotone functions over w ∏ 0 and f W (w) = 0 for w < 0. Consider

w = g 1 (z) , Z , z ∏ 0, w ∏ 0 ) |g 10 (z)| = 1
w = g 2 (z) , °Z , z < 0, w ∏ 0 ) |g 20 (z)| = 1

Then (
f Z (w)/1 + f Z (°w)/1, for w ∏ 0 ,
f W (w) =
0, for w < 0.

Hence if f Z (w) = f Z (°w), as it is here, then finally

f W (w) = 2 f Z (w)u(w) = Æe°Æw u(w).

EECE-7204 : H OMEWORK -3 S OLUTIONS : FALL-2016 5

P3.6 Text Problem 4.27 (Page 287).

Solution: The random variables X and Y are jointly Gaussian with mean 0 and variance æ2 . A
new random variable is defined as Z = 12 (X + Y ).

(a) The mean and variance of Z when X and Y are independent: The mean is given by
£1 § 1
E [Z ] = E 2
(X + Y ) = 2 (0) = 0.

Using independence, the variance is given by

æ2Z = æ2X /2 + æ2Y /2 = 14 æ2 + 14 æ2 = 12 æ2.

(b) The mean and variance of Z when X and Y are correlated with coefficient Ω: The mean
of Z is still zero. Hence the variance is also the second moment and the covariance is also
the correlation. Using this the variance of Z is given by
£ § £ § £ §
æ2Z = E Z 2 = 14 E (X + Y )2 = 14 E X 2 + 2X Y + Y 2
° ¢
= 14 æ2 + 2Ωæ2 + æ2 = 12 (1 + Ω)æ2 .

(c) For Ω = °1, 0, and 1, the values of æ2Z are 0, æ2 /2, and æ2 , respectively.

P3.7 Text Problem 4.54 (Page 290).

Solution: If random variables X and Y are independent, then the moment generating function
of Z = X + Y is given by the products of the moments generating functions of X and Y , that is,

™ Z (s) = ™ X (s)™Y (s).

a k e°a
Now, for a Poisson random variable X with parameter a or pmf Pr [X = k] = , the moment
k!
generating function is

1
X a k e°a X1 (sa)k
™ X (s) = sk = e°a = e°a esa = ea(s°1) .
k=°1 k! k=°1 k!

Thus for the given Poisson random variables X : Poisson(2) and Y : Poisson(3), we have

™ Z (s) = e2(s°1) e3(s°1) = e5(s°1) .

Therefore, the random variable Z is Poisson with parameter 5, or the pmf is

5k e°5
Pr [Z = k] = .
k!
6 FALL-2016: H OMEWORK -3 S OLUTIONS : EECE-7204

P3.8 A professor is studying the performance of various groups in his class. Let the random variable
X represent a student’s final score in the class and define the following conditioning events:

• Fr = {Student is a freshman}
• So = {Student is a sophomore}
• Jr = {Student is a junior}
• Sr = {Student is a senior}
• M = {Student is a male}
• F = {Student is a female}

(a) Suppose the average score among males is 73.2 and the average score among females is
75.8. If the overall class average score is 74.6, what percentage of the class is female?
Solution: Let the percentage of females be 100p, then the percentage of males is 100(1°p).
Using E [X ] = E [X |F] p + E [X |M] (1 ° p), we have

74.6 = 75.8p + 73.2(1 ° p) = 73.2 + 2.6p ) p = 0.5385

or the female percentage is 53.85%.

(b) Suppose the conditional average scores by class are:

E [X |Fr] = 65.8, E [X |So] = 71.2, E [X |Jr] = 75.4, and E [X |Sr] = 79.1.

If the class average score is 72.4, what can you say about the percentage of freshmen,
sophomores, juniors, and seniors in the class?
Solution: Let the percentages of freshmen, sophomores, juniors, and seniors be, 100p Fr ,
100p So , 100p Jr , and 100p Se , respectively. Then we have

1 = p Fr + p So + p Jr + p Se (1)
and E [X ] = p Fr E [X |Fr] + p So E [X |So] + p Jr E [X |Jr] + p Se E [X |Se]
or 0.724 = 0.658p Fr + 0.712p So + 0.754p Jr + 0.791p Se (2)

which is not enough to determine percentages.

(c) Given the class statistics as in part (b), if it is known that there are 10 freshmen, 12 sopho-
mores, and 9 juniors in the class, how many seniors are in the class? Note that this number
must be an integer (round to the next integer).
Solution: Let the number of seniors be equal to N . Then the total number of students in
the class is equal to M , 31 + N . Hence to satisfy (1) we must have

10 12 9 N
p Fr = , p So = , p Jr = , and p Se = . (3)
M M M M
Substituting in (2), we obtain

0.724M = 0.658(10) + 0.712(12) + 0.754(9) + 0.791(N )

or 0.724(31 + N ) = 21.91 + 0.791N (4)

Solving for N in (4), we obtain N = 7.98 or N = 8 since it must be an integer.

EECE-7204 : H OMEWORK -3 S OLUTIONS : FALL-2016 7

P3.9 A random variable X is distributed according to the pdf

The above pdf f X (x) is given as (Note that A = 47 for unit area under the pdf)
8
8 8
7 + 7 x, °1 ∑ x ∑ °0.5;
>
>
>
< 4
7
, °0.5 ∑ x ∑ 0;
f X (x) = 4 2
> 7 ° 7 x, 0 ∑ x ∑ 2;
>
>
: 0, otherwise.

In this problem, it is helpful to obtain the following result:

Z1 Z°0.5 Ω æ Z0 µ ∂ Z2 Ω æ
n 8 8 n 4 n 4 2
£ n§ n
E X = x f X (x) dx = x + x dx + x dx + x ° x dx
°1 °1 7 7 °0.5 7 0 7 7
µ ∂
2 4(°1)n + 4(2n ) ° (°2)°n
= ; n∏0 (using M APLE)
7 n 2 + 3n + 2
Hence we have
µ ∂
2 °4 + 4(2) ° (°2)°1
E [X ] = = 0.2143
7 1+3+2
µ ∂
£ 2 § 2 4 + 4(4) ° (°2)°2
E X = = 0.4702
7 4+6+2
µ ∂
£ 3 § 2 °4 + 4(8) ° (°2)°3
E X = = 0.4018
7 9+9+2
µ ∂
£ 4 § 2 4 + 4(16) ° (°2)2°4
E X = = 0.6470
7 16 + 12 + 2

(a) Determine the mean µ X = E [X ] of X .

Solution: The mean value from above is

µ X = E [X ] = 0.2143
h° ¢2 i
(b) Determine the variance æ2X = E X ° µX of X .
Solution: The variance is
h° ¢2 i £ §
æ2X = E X ° µx = E X 2 ° (E [X ])2
= 0.4702 ° (0.2143)2 = 0.4243

Hence the standard deviation is

p
æX = 0.4243 = 0.6514.
8 FALL-2016: H OMEWORK -3 S OLUTIONS : EECE-7204

(c) Determine the skewness of X given by

∑µ ∂3 ∏
X ° µX
SX , E .
æX

Solution: The skewness is

∑µ ∂ ∏
X ° µX 3 1 £ §
SX = E = 3 E X 3 ° 3X 2 µ X + 3X µ2X ° µ3X
æX æX
1 ° £ § £ § ¢
= 3 E X 3 ° 3µ X E X 2 + 2µ3X = 0.4311
æX

(d) Determine the kurtosis of X given by

∑µ ∂4 ∏
X ° µX
KX , E ° 3.
æX

Solution: Consider
∑µ ∂ ∏
X ° µX 4 1 £ §
E = 4 E X 4 ° 4X 3 µ X + 6X 2 µ2X ° 4X µ3X + µ4X
æX æX
1 ° £ § £ § £ § ¢
= 4 E X 4 ° 4µ X E X 3 + 6µ2X E X 2 ° 3µ4X = 2.3653
æX

or kurtosis K X = 2.3653 ° 3 = °0.6347.

P3.10 Random variable X is specified by the pdf f X (x) = 12 |x|e°|x| for all x.

(a) Determine the even-ordered moments ªn = E [X n ], n ª even, of random variable X .

Solution: Since the given pdf f X (x) is an even function in x, all its odd-ordered moments
are zero. For n-even, the nth moment is given by
Z1
£ n§
ªn = E X = x n 12 |x|e°|x| dx
x=°1
Z1
1 n+1 °x
=2 2x e dx = (n + 1)!, n ª even. (5)
x=0

Since the mean µ X = 0, the variance from (5) is given by the second moment

ª2 = (2 + 1)! = 6 = æ2X . (6)

(b) Determine the Kurtosis K X of random variable X .

Solution: Again due to zero mean the Kurtosis is given by
∑µ ∂4 ∏
X ° µX 1 £ § 5! 120 1
KX = E °3 = E X4 °3 = °3 = ° 3 = = KX . (7)
æX æ4X 6 2 36 3

Since K X > 0, the pdf has more area in the tails than near the mean.
EECE-7204 : H OMEWORK -3 S OLUTIONS : FALL-2016 9

£ §
(c) Determine the moment generating function (MGF) © X (s) = E es X and express it as a
rational function in variable s to get the maximum credit.
Solution: The MGF is given by
Z1 ΩZ0 Z1 æ
£ sX
§
© X (s) = E e = esx 12 |x|e°|x| dx
= 1
2 (°x)e (s+1)x
dx + (s°1)x
(x)e dx
x=°1 x=°1 x=0
ΩZ1 Z1 æ
1 °x sx
© £ § £ §Ø ™
=2 °x °sx
xe e dx + x e e dx = 12 L x e°x + L x e°x Øs!°s
0 0
Ω∑ ∏ ∑ ∏æ
1 1 1 s2 + 1
=2 , s > °1 + , s < 1 = ° ¢2 , °1 < s < 1 . (8)
(s + 1)2 (°s + 1)2 s2 ° 1

(d) Determine the mean µ X using the MGF © X (s).

Solution: The mean µ X is given by
Ø ° ¢2 ° ¢ ° ¢Ø
d Ø 2s s 2 ° 1 ° s 2 + 1 4s s 2 ° 1 ØØ
µX = © X (s)ØØ = ° ¢4 Ø
ds s=0 s2 ° 1 Ø
s=0
= 0 (as expected) . (9)