Download EE 302: Probabilistic Methods in Electrical Engineering Test II

1
EE 302: Probabilistic Methods in Electrical Engineering
Solution
Print Name:
(2/23/06 --sk)
Test II : Chapters 3.5 – 4
11/12/98, 1:00 PM
Write down your name on each paper. Read every question carefully and solve each problem in a
legible and ordered manner. Make sure you write down all your answers without skipping details.
However, you could use short cuts provided you have a valid argument and you clearly state it on
paper. I don’t give credit for wrong answers and partial credit will be given only when sufficient
details have been provided.
1. Consider the following probability density function which is also shown in Figure 1:


kx


 2k
0≤x<2
2≤x<4
fX (x) =
.

k(6
−
x)
4
≤x≤6



0
otherwise
Compute the following (20 points):
(a) Find the value of k which makes fX (x) a valid probability density function. (10 points).
(b) Find the cumulative distribution function FX (x). Use the space indicated for that in
Figure 2. (10 points).
(a)
1 =
=
Z
∞
−∞
Z 2
fX (x)dx
kxdx +
Z
(1)
4
2kdx +
2
0
= 2k + 4k + 2k = 8k
Z
4
6
k(6 − x)dx
(2)
(3)
so k = 1/8.
(b) First we check for discontinuities at the ends of the intervals:
0
k2
2k
k(6 − 6)
=
=
=
=
k0
2k
k(6 − 4)
0.
(4)
(5)
(6)
(7)
Finding no discontinuities we need only integrate fX (x) to determine
FX (x).
FX (x) =
 Rx

0 kydy


 2k + R x 2kdy
R2x

6k
+

4 k(6 − y)dy


1
y
y
y
y
∈ [0, 2)
∈ [2, 4)
∈ [4, 6]
> 6,
(8)
EE 302: Probabilistic Methods in Electrical Engineering
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which yields







FX (x) = 





0
kx2 /2
2k(x − 1)
−kx2 /2 + 6kx − 10k
1
x<0
x ∈ [0, 2)
x ∈ [2, 4)
x ∈ [4, 6]
x > 6.
(9)
EE 302: Probabilistic Methods in Electrical Engineering
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2. Assume that X1 and X2 are coded scores on two intelligence tests, and the probability density
function of [X1 , X2 ] is given by
(
fX1 ,X2 (x1 , x2 ) =
6x21 x2 0 ≤ x1 ≤ 1, 0 ≤ x2 ≤ 1
0
otherwise
Compute the following (20 points):
(a) Find the E(X2 |x1 ). (5 points).
2
(b) Find the variance of the score on test No. 2 given the score on test No. 1, σX
=
2 |x1
2
2
E(X2 |x1 ) − µX2 |x1 . (5 points).
(c) Find the correlation coefficient between the two coded scores. (10 points).
(a) We will need the conditional PDF fX2 |X1 (x2 |x1 ) and to find that
we’ll need the marginal fX1 (x1 ), so let’s calculate that first.
fX1 (x1 ) =
Z
∞
−∞
fX1 ,X2 (x1 , x2 )dx2 =
Z
1
0
1
6x21 x22 = 3x21 .
6x21 x2 dx2 =
2 0
Thus the conditional PDF is
(
fX1 ,X2 (x1 , x2 )
2x2 x1 ∈ [0, 1] and x2 ∈ [0, 1]
fX2 |X1 (x2 |x1 ) =
=
0
otherwise
fX1 (x1 )
(10)
(11)
and
E[X2 |X1 ] =
Z
∞
−∞
x2 fX2 |X1 (x2 |x1 )dx2 = 2/3.
(12)
(b) To find σX2 |X1 we will need E[X22 |X1 ] so we calculate that first.
E[X22 |X1 ] =
Thus
σX2 |X1 =
Z
∞
−∞
x22 fX2 |X1 (x2 |x1 )dx2 = 1/2.
q
E[X22 |X1 ] − (E[X2 |X1 ])2 =
q
1/2 − 4/9 =
(13)
√
18.
(14)
(c) We will need fX1 (x1 ) and fX2 (x2 ) in order to determine the
correlation coefficient. We already determined fX1 (x1 ), and
fX2 (x2 ) =
Z
∞
−∞
fX1 ,X2 (x1 , x2 )dx1 =
Z
0
1
1
6x21 x2 dx1 = 2x31 x2 = 2x2 .
0
(15)
At this point, we providentially notice that the joint PDF is the
product of the marginal PDF’s, hence the random variables X1 and
X2 are independent, hence their covariance is zero, hence the
correlation coefficient is zero, thereby saving ourselves a lot of
computation.
EE 302: Probabilistic Methods in Electrical Engineering
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3. The Bernoulli distribution function is given as
PX (x) =
(
px q 1−x 0 < p < 1, x = 0 or 1
0
otherwise
Compute the following (20 points):
(a) Find the moment generating function ΨX (u). (10 points).
(b) Find the mean and variance using the results from part (a). (10 points).
(a) The moment generating function is
φX (s)esX = es0 PX (0) + es1 PX (1)
= p0 (1 − p)1 + es p1 (1 − p)0
= (1 − p) + pes .
(16)
(17)
(18)
(b)
d
= pe0 = p.
((1 − p) + pes )
E[X] =
ds
s=0
(19)
Similarly
d2
E[X ] = 2 ((1 − p) + pes )
= pe0 = p
ds
s=0
2
so V ar[X] = p − p2 = p(1 − p).
(20)
EE 302: Probabilistic Methods in Electrical Engineering
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4. Two discrete random variables X1 and X2 have joint probability distribution function as
given in the following Table:
.
x2j . . x1i
0
1
2
3
4
0
1
30
1
30
2
30
3
30
1
30
1
1
30
1
30
3
30
4
30
0
2
1
30
2
30
3
30
0
0
3
1
30
3
30
0
0
0
4
3
30
0
0
0
0
PX1 (x1i )
PX2 (x2j )
P
Px (x) = 1
Compute the following: (20 points).
(a) The marginal distributions PX1 (x1i ) and PX2 (x2j ). (10 points).
(b) Find the correlation coefficient ρ between X1 and X2 . (10 points).
(a) Summing columns we obtain





7/30
8/30
PX1 (x1 ) = 
1/30



0
x1 ∈ {0, 1, 3}
x1 = 2
x1 = 4
otherwise.
(21)
Summing rows we obtain
PX2 (x2 ) =



















8/30
9/30
6/30
4/30
3/30
0
x2 = 0
x2 = 1
x2 = 2
x2 = 3
x2 = 4
otherwise.
(22)
(b) I don’t see any way of avoiding the calculations here. As a simple
check, showing that PX1 ,X2 (x1 , x2 ) 6= PX1 (0)PX2 (0)) will show that the random
EE 302: Probabilistic Methods in Electrical Engineering
variables X1 and X2 are not independent.
E[X1 ] =
E[X2 ] =
E[X12 ] =
E[X22 ] =
E[X1 X2 ] =
6
I found
7
8
7
1
48
7
(0) + (1) + (2) + (3) + (4) =
30
30
30
30
30
30
8
9
6
4
3
45
(0) + (1) + (2) + (3) + (4) =
30
30
30
30
30
30
7
8
7
1
118
7
(0)2 + (1)2 + (2)2 + (3)2 + (4)2 =
30
30
30
30
30
30
8
9
6
4
3
101
(0)2 + (1)2 + (2)2 + (3)2 + (4)2 =
30 30
30
30 30 30
3
4
2
1
+ 1(2)
+ 1(3)
+ 2(1)
+
1(1)
30
30
30
30
44
3
3
+ 3(1)
= ,
2(2)
30
30
30
(23)
(24)
(25)
(26)
(27)
(28)
so
Cov[X1 , X2 ] = E[X1 X2 ] − E[X1 ]E[X2 ] =
48 2 1236
118
=
−
V ar[X1 ] =
30
30
90
2
101
1005
45
V ar[X2 ] =
=
−
30
30
90
44 48(45)
840
−
=−
30 30(30)
90
(29)
(30)
(31)
and thus the correlation coefficient is
840
ρX1 ,X2 =
90√
√
1236 √
1005
√
90
90
840
=q
.
1236(1005)
(32)
We know that the correlation coefficient should be between −1 and 1,
which it is, so this answer is plausible.
EE 302: Probabilistic Methods in Electrical Engineering
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5. Given the Gaussian probability density function
fX (x) =
√
1
1
e− 2
2πσx
x−µx
σx
2
,
σx > 0, −∞ < x < ∞
and its associated standard Gaussian probability density function
fZ (z) =
1 2
1
√ e− 2 z ,
2π
−∞ < z < ∞,
where the two random variables are related by Z =
(20 points).
X−µx
σx .
Answer the following questions:
(a) Given two random variables X1 and X2 with respective mean and variance (µ1 ,σ12 ) and
(µ2 ,σ22 ). Consider the case where µ1 6= µ2 and σ12 ≫ σ22 . Is the area under the Gaussian
curve between (µ1 − σ1 , µ1 + σ1 ) greater, equal, or less than the area under the curve between
(µ2 − σ2 , µ2 + σ2 )? Explain your answer. (4 points).
The area under the curve between µ1 − σ1 and µ1 + σ1 equals the area under
the curve between µ2 −σ2 and µ2 +σ2 . This is true whether or not the means
are equal. To see this, note that
Z
µ1 +σ1
µ1 −σ1
−1
1
√
e 2
2πσ1
x−µ1
σ1
2
(µ1 + σ1 ) − µ1
= φ
σ1
= φ(1) − φ(−1)
!
(µ2 + σ2 ) − µ2
= φ
σ2
= φ(1) − φ(−1).
!
(µ1 − σ1 ) − µ1
−φ
σ1
!
(33)
and so does
Z
µ2 +σ2
µ2 −σ2
−1
1
√
e 2
2πσ2
x−µ2
σ2
2
(µ2 − σ2 ) − µ2
−φ
σ2
!
(34)
(b) Suppose you are a contestant at some TV show. You are given a pair of cards. The first card
is shown to you and contains a value of fX (x), for some unknown value of x. The second card
is not shown to you but you are told that it will contain a value X = x. You are asked to
select a value X = x, then only you will know what value of X = x the second card contains.
What value of X = x would you choose in order to determine the variance of the random
variable X using the information from both cards? Explain how you would solve this. Refer
to the above Gaussian density. (4 points).
I don’t understand the question, so don’t worry if you don’t either.
(c) Using the Gaussian Table provided, compute the area under the curve between (µ−3σ, µ+3σ),
i.e. , PX (µ − 3σ ≤ X ≤ µ + 3σ) (4 points).
No table having been provided, we’ll have to skip this one too.
disappointing. :)
How
EE 302: Probabilistic Methods in Electrical Engineering
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(d) Are any of the following four statements false? If so, briefly explain why (4 points).
(1)
(2)
R median
−∞
R µx +σx
µx
fX (x)dx = 0.5.
fX (x)dx = 0.68.
(3) Coefficient of skewness is always γs < 0 for a Gaussian probability density.
x
is to cause a shift towards the origin, thus a change
(4) The combined effect of Z = X−µ
σx
in the shape of the Gaussian distribution, but does not affect the scale.
(1) is false.
or ‘‘µx ’’.
(2) is false.
It would be true if we replaced ‘‘median’’ by ‘‘mean’’
R µx +σx
µx −σx
fX (x)dx = 0.6827.
(3) is not covered in this course. (But in case you are interested,
skewness is a measure of asymmetry, hence it is zero for symmetric
distributions. Since the Gaussian distribution is symmetric, (3) is
false.)
(4) I would say ‘‘false’’ because shifting the curve so that it is
centered at the origin does not change the shape.
(e) Towards what function (well known among electrical engineers) does the Gaussian probability
density converge to when σX → 0? Explain why. (4 points).
From geometrical considerations, I’d say the impulse function (because
the area under the curve is one and as σX decreases the curve becomes
taller and narrower). However, to prove this would require mathematical
theory well beyond the scope of this course.