ECE 340: PROBABILISTIC METHODS IN ENGINEERING
SOLUTIONS TO HOMEWORK #6
Y is the difference between the number of heads and the number of tails in the 3 tosses
of a fair coin. Let m be the number of tails 0 ≤ m ≤ 3. Then 3-m is the number of heads
and the difference is
Y=3-m-m = 3-2m
with 0 ≤ m ≤ 3.
Thus, SY={-3,-1,1,3} and the probabilities are:
4.5
P{Y=-3} = P{(T,T,T)} =
=
P{Y=-1} = P{(T,T,H), (T,H,T), (H,T,T)} =
P{Y=1} = P{(H,H,T), (H,T,H), (T,H,H)} =
P{Y=3} = P{(H,H,H)} =
=
a) The cdf of Y is:
0
1
8
1
8
1
8
7
8
1 3
3
3
8
3
8
1
8
1
2
3
8
1
7
8
1
1
1
1
3
3
ECE 340
Homework #6 Solutions
b) To express P{|Y| < y} in terms of the cdf of Y, first notice that S|Y|={1,3}.
| |
3
8
0
3
8
1
1
6
8
1
3
3
clearly, P{|Y| < y }=FY(y) - FY(-y)
4.6
Solution:
a) The sample space is given by all the points with coordinates x,y that are at a distance
less than or equal to 2 from the origin. In other words, all the points within a circle of
radius equal to 2:
S={(x,y): x2 + y2 ≤ 4}
R is the distance of the landing point to the origin, so:
The sample space of R is given by all the real numbers between 0 and 2:
SR = {r: 0≤ r ≤ 2}
b) The mapping from S to SR:
c) The event A (dart hits the bull’s eye):
2 ECE 340
Homework #6 Solutions
1
4
2
1
4
1
64
d) The cdf of R:
For 0 ≤ r ≤ 2
2
2
4.11 Solution:
a) X is a continuous random variable whose cdf has a linear increase between -1 and 2
(because it is uniformly distributed).We know that the value of the cdf at -1 has to be 0
and 1 at 2. The line that connects these 2 points has the equation:
1
3
1
3
1
3
1
The plot is:
|
3 0
0
b)
0.5|
1
1
0.5
1
1
2
3
2
ECE 340
Homework #6 Solutions
1.5
1 3
1
3 2
2
3
1
2
4.12
0.5
1
2
1
3
1
1
2
1
1
1
3
1
2
1
5
6
0
0.5
1
1
2
0.5
Solution:
a) X is a random variable of mixed type
1
1
b)
0.5
0
1
1
1 0.5
2
0.5
0.5
1
2
3
1
0.75
0.5
1
1
2
1
0.5
1
3
0.5
1
0.5
0.25
0
4.13 Solution:
a) X is a random variable of mixed type, so we have to watch out for values where the cdf
is not continuous.
4 ECE 340
Homework #6 Solutions
2
1
0.9954
0
1
0.75 using property (vii) of the cdf
0
0 where we took the limit from the left to exclude the point X=0.
b)
Since the cdf is continuous at x=6
2
6
2
6
6
1
1
1
0.0046
1
4
4
10
1
4.14
1
10
1
4
1
2
5.15 10
1
is right continuous,
Solution: First of all, note that since
;
1/2
0
1
1.
;
;
a) X is a random variable of mixed type
1
b)
1
1
1
1
1
0
0
1
1
0.75
0.5
0
0.5
0.5
0.75
0
0.5
0.5
0.5
1
0
4
2
10 10
0.8
2
10
2
10
1
5
0.6
because the equation of the line between 0 and 1 has a slope equal to
. So
5 0.5
0.5
0.8
ECE 340
Homework #6 Solutions
|
0.5|
0.5
0.5
0.5
0.4 [recall that we write
0.5
0
0.5
0.5
1
1
0
0.5
0.5
1
0.5
0.5 ]
4.21
Solution:
1
a)
Also, by the definition of a cdf:
Furthermore,
1
1
1
1
So,
1
ζ ranges from 0 to 1, so X ranges from 1 to ∞.
If x2>1 then 0< 1
≤1
To get the cdf of X, notice that
1
1
1
1
1
1 because ζ is uniformly distributed between 0
Notice that we used the fact that
and 1. So, the plot of FX(x) is:
1
0.9
0.8
0.7
FX(x)
0.6
0.5
0.4
0.3
0.2
0.1
0
6 0
1
2
3
4
5
x
6
7
8
9
10
ECE 340
Homework #6 Solutions
The pdf of X is:
2
0
b)
2
4.44
1
2
1
2
1
1
1
1
4
Mean and variance of X in Problem 4.12:
We can plot the pdf of X by observing the cdf found in problem 4.12:
Notice that the delta function for x=-1 corresponds to the discontinuity of the cdf at that
point. A jump always leads to a delta function. For the interval [0,1] the cdf increases
linearly, which corresponds to a constant value in the pdf.
So, the expected and the variance are:
0.5
1
1
2
0.5
1
2
7 1
1 1
1
2 2
0
1
2
0.25
0.25
29/48.
ECE 340
Homework #6 Solutions
4.59
Solution
| |
| |
1
1
| |
Since X is uniform between [-2,2], its pdf is:
Consequently, the pdf ox |X| is:
From that, we can get F|X|(x)
8 ECE 340
Homework #6 Solutions
So, P{(|X|>x)}=1-F|X|(x) is:
9 ECE 340
Homework #6 Solutions