Probabilistic Systems Analysis Homework Help

Probabilistic Systems Analysis
Homework Help
For any help regarding Probabilistic Systems Analysis Homework Help
Visit :- https://www.excelhomeworkhelp.com/
Email :- info@excelhomeworkhelp.com
or call us at :- +1 678 648 4277

1.Professor May B. Right often has her science facts wrong, and answers each of her students’
questions incorrectly with probability 1/4, independently of other questions. In each lecture Professor
Right is asked either 1 or 2 questions with equal probability.
(a)What is the probability that Professor Right gives wrong answers to all the questions she gets
in a given lecture?
(b)Given that Professor Right gave wrong answers to all the questions she was asked in a given
lecture, what is the probability that she got two questions?
(c)Let X and Y be the number of questions asked and the number of questions answered correctly in
a lecture, respectively. What are the mean and variance of X and the mean and the variance of Y ?
(d)Give a neatly labeled sketch of the joint PMF pX,Y (x, y).
(e)Let Z = X + 2Y . What are the expectation and variance of Z?
For the remaining parts of this problem, assume that Professor Right has 20 lectures each
semester and each lecture is independent of any other lecture.
(f)The university where Professor Right works has a peculiar compensation plan. For each lecture,
she gets paid a base salary of $1, 000 plus $40 for each question she answers and an additional
$80 for each of these she answers correctly. In terms of random variable Z, she gets paid $1000
+ $40Z per lecture. What are the expected value and variance of her semesterly salary?
(g)Determined to improve her reputation, Professor Right decides to teach an additional 20-lecture
class in her specialty (math), where she answers questions incorrectly with probability 1/10 rather
than 1/4. What is the expected number of questions that she will answer wrong in a randomly
chosen lecture (math or science).
Problem
excelhomeworkhelp.com

Here, C is some constant. What is E[XY 3]?
Hint : This question admits a short answer/explanation. Don’t spend time doing calculations.
3.Joe Lucky plays the lottery on any given week with probability p, independently of whether
he played on any other week. Each time he plays, he has a probability q of winning, again
independently of everything else. During a ﬁxed time period of n weeks, let X be the number
of weeks that he played the lottery and Y the number of weeks that he won.
(a)What is the probability that he played the lottery any particular week, given that he did
not win anything that week?
(b)Find the conditional PMF pY |X (y | x).
1.The joint PMF of discrete random variables X and Y is given by
(c)Find the joint PMF pX,Y (x, y).
(d)Find the marginal PMF pY (y). Hint: One possibility is to start with the answer to part
(c), but the algebra can be messy. But if you think intuitively about the procedure that
generates Y , you may be able to guess the answer.
(e)Find the conditional PMF pX|Y (x | y). Do this algebraically using previous answers.
(f)Rederive the answer to part (e) by thinking as follows: For each one of the n −Y weeks
that he did not win, the answer to part (a) should tell you something.
In all parts of this problem, make sure to indicate the range of values for which your PMF
formula applies.
4.Let X and Y be independent random variables that take values in the set {1, 2, 3}. Let
V = 2X + 2Y and W = X − Y .
(a)Assume that P({X = k}) and P({Y = k}) are positive for any k ∈{1, 2, 3}. Can V and
W be independent? Explain. (No calculations needed.)

For the remaining parts of this problem, assume that X and Y are uniformly distributed on
{1, 2, 3}.
(b)Find and plot pV (v). Also, determine E[V ] and var(V ).
(c)Find and show in a diagram pV,W (v, w).
(d)Find E [V | W > 0].
(e)Find the conditional variance of W given the event {V = 8}. (f) Find and plot
the conditional PMF pX|V (x | v), for all values.
5.Suppose the waiting time until the next bus at a particular bus stop is exponentially dis
1
5
tributed, with parameter λ = 1 . Suppose that a bus pulls out just as you arrive at the
stop.
Find the probability that:
(a)You wait more than 15 minutes for a bus.
(b)You wait between 15 and 30 minutes for a
bus.
G1†. Clark is a news reporter who lives in Metropolis. Every morning he drives from his
apartment at point A to The Daily Planet oﬃce at point B for work. He travels only in the
north and east directions, and he chooses each path with equal probability. The heavy line in
the diagram indicates one such valid path.

(a)Determine the probability that the path in the diagram is the one he actually picked this
morning.
(b)There is a phone booth at point C, the intersection of 4th St. and 3rd Ave. Find the probability
that he drove past this phone booth on his way to work this morning.
(c)Find the probability that he drove past the phone booth, given that he was seen on the portion
of 3rd St. between 2nd and 3rd Ave. (point D).
(d)Repeat part (b) for the general case where his apartment is located at the intersection of Main
and Broadway, The Daily Planet is located at the intersection of wth St. and hth Ave., and the
phone booth is located at the intersection of xth St. and yth Ave. Assume the same numbering
scheme for the streets. (Note that this problem is of interest only if x ≤ w and y ≤ h.) Does
your answer make sense for the special cases
i.w = 0, i.e., the oﬃce is on Main St.; and
ii.h = 0, i.e., the oﬃce is on Broadway?

1. (a) Use the total probability theorem by conditioning on the number of questions that
Professor Right has to answer. Let A be the event that she gives all wrong answers in a
given lecture, let B1 be the event that she gets one question in a given lecture, and let B2
be the event that she gets two questions in a given lecture. Then
P(A) = P(A|B1)P(B1) + P(A|B2)P(B2).
From the problem statement, she is equally likely to get one or two questions in a given
2 4
lecture, so P(B1) = P(B2) = 1 . Also, from the problem statement, P(A|B1) = 1 , and,
4 16
because of independence, P(A|B2) = ( 1 )2 = 1 . Thus we have
(b) Let events A and B2 be defined as in the previous part. Using Bayes’s Rule:
2
P(B |A)
=
P (A|B2)P (B
2) P(A)
.
From the previous part, we said P(B2) = 1 , P(A|B2) = 1 , and P(A) = 5 . Thus
2 16 32
1 ·
1
5
3
2
1
5
P(B2|A) = 16 2 = .
As one would expect, given that Professor Right answers all the questions in a given
lecture, it’s more likely that she got only one question rather than two.
(c) We start by finding the PMFs for X and Y . The PMF pX (x) is given from the problem
statement:
Solution

she answers the first question correctly or she answers the second question correctly.
Thus, overall
Now the mean and variance can be calculated explicitly from the PMFs:
The PMF for Y can be found by conditioning on X for each value that Y can take on.
Because Professor Right can be asked at most two questions in any lecture, the range of Y
is from 0 to 2. Looking at each possible value of Y , we find
Note that when calculating P(Y = 1|X = 2), we got 2 ·3 ·1 because there are two ways
4 4
for Professor Right to answer one question right when she’s asked two questions: either
she answers the first question correctly or she answers the second question correctly.
(d) The joint PMF pX,Y (x,y) is plotted below. There are only five possible (x,y) pairs. For
each point, pX,Y (x,y) was calculated by pX,Y (x,y) = pX(x)pY |X(y|x)

Calculating var(Z) is a little bit more tricky because X and Y are not independent; therefore we
cannot add the variance of X to the variance of 2Y to obtain the variance of Z. (X and Y are
clearly not independent because if we are told, for example, that X = 1, then we know that Y
cannot equal 2, although normally without any information about X, Y could equal 2.)
To calculate var(Z), first calculate the PMF for Z from the joint PDF for X and Y . For each (x,
y) pair, we assign a value of Z. Then for each value z of Z, we calculate pZ (z) by summing over
the probabilities of all (x, y) pairs that map to z. Thus we get
(e) By linearity of expectations,
E [Z ] = E [X + 2Y ] = E [X ] + 2E [Y ] = 3
+ 2 ·9
= 15
.
2 8 4
In this example, each (x, y) mapped to exactly one value of Z, but this does not have to be
the case in general. Now the variance can be calculated as:
8
1
var(Z ) = 1 −
2
15
4 3
2
1
+ 2 − 8
2
15 3
4
+ 3 −
2
15
4
3 15
16 4 32
9
+ 4 − + 6 −
2
15
4
= .

(f)For each lecture i, let Zi be the random variable associated with the number of questions
Professor Right gets asked plus two times the number she gets right. Also, for each lecture i,
let Di be the random variable 1000 + 40Zi. Let S be her semesterly salary. Because she
teaches a total of 20 lectures, we have
Σ20 Σ20 Σ20
S = Di = 1000 + 40Zi = 20000 + 40 Zi.
i=1 i=1 i=1
By linearity of expectations,
Σ20
E [S] = 20000 + 40E [ Zi] = 20000 + 40(20)E [Zi ] = 23000.
i=1
Since each of the Di are independent, we have
Σ20
var(S) = var(Di) = 20var(Di) = 20var(1000 + 40Zi) = 20(402 var(Zi)) = 36000.
i=1
(g)Let Y be the number of questions she will answer wrong in a randomly chosen lecture. We
can find E[Y ] by conditioning on whether the lecture is in math or in science. Let M be
the event that the lecture is in math, and let S be the event that the lecture is in science. Then
E [Y ] = E [Y |M ]P (M ) + E [Y |S]P (S).
Since there are an equal number of math and science lectures and we are choosing

randomly among them, P(M ) = P(S) = 1 . Now we need to calculate E[Y |M ] and
2
E[Y |S] by finding the respective conditional PMFs first. The PMFs can be determined
in an manner analagous to how we calculated the PMF for the number of correct answers
in part (c).
Therefore
2. The key to the problem is the even symmetry with respect to x of pX,Y (x, y) combined with
the odd symmetry with respect to x of xy3.
By definition,
Σ5 Σ10
E [X Y 3] = xy3pX ,Y (x, y).
x=−5 y=0

All the x = 0 terms make no contribution to sum above because xy p
The term for any pair (x, y) with x =/ 0 can be paired with the term for (−x, y). These terms
cancel, so the summation above is zero. Thus, without making in more detailed computations
we can conclude E[XY 3] = 0.
3. (a) Let Li be the event that Joe played the lottery on week i, and let Wi be the event that
he won on week i. We are asked to find
(b) Conditioned on X, the random variable Y is binomial:
(c) Realizing that X has a binomial PMF, we have
(d) Using the result from (c), we could compute
Σn
pY (y) = pX ,Y (x, y) ,
x=y
but the algebra is messy. An easier method is to realize that Y is just the sum of n
independent Bernoulli random variables, each having a probability pq of being 1. Therefore
Y has a binomial PMF:

(f) Given Y = y, we know that Joe played y weeks with certainty. For each of the remaining
n − y weeks that Joe did not win there are x − y weeks where he played. Each of these
events occured with probability P(Li | Wc) (the answer from part (a) ). Using this logic
i
we see that that X conditioned on Y is binomial:
After some algebraic manipulation, the answer to (e) can be shown to be equal to the
one above.
4. (a) V and W cannot be independent. Knowledge of one random variable gives
information about the other. For instance, if V = 12 we know that W = 0.
(b) We begin by drawing the joint PMF of X and Y .

X and Y are uniformly distributed so each of the nine grid points has probability 1/9. The lines on
the graph represent areas of the sample space in which V is constant. This constant value of V is
indicated on each line. The PMF of V is calculated by adding the probability associated with each
grid point on the appropriate line.

By symmetry (or direct calculation),
E[V ] = 8 . The variance is:
Alternatively, note that V is twice the sum of two independent random variables, V = 2(X + Y ),
and hence
var(V ) = var 2(X +Y ) = 22var(X +Y ) = 4 var(X)+var(Y ) = 4·2var(X) = 8 var(X).
(Note the use of independence in the third equality; in the fourth one we use the fact that X and Y
are identically distributed, therefore they have the same variance). Now, by the distribution of X,
we can easily calculate that
so that in total var(V ) = 16 , as before.
3
(c) We start by adding lines corresponding to constant values of W to our first graph in part (b):
Again, each grid point has probability 1/9. Using the above graph, we get pV,W (v, w).

(d) The event W > 0 is shaded below:
By symmetry (or an easy calculation),
(e) The event {V = 8} is shaded below:
E[V | W > 0] = 8 .

When V = 8, W can take on values in the set {−2, 0, 2} with equal probability. By
symmetry (or an easy calculation), E[W | V = 8] = 0. The variance is:
var(W | V = 8) = (−2 − 0)2 ·1
+ (0 − 0)2 ·1
+ (2 − 0)2 ·1
=
3 3 3 8
3 .
(f) Please refer to the first graph in part (b). When V = 4, X = 1 with probability 1. When V
= 6, X can take on values in the set {1, 2} with equal probability. Continuing this reasoning
for the other values of V , we get the following conditional PMFs, which we plot on one set of
axes.
Note that each column of the graph is a separate conditional PMF and that the probability of
each column sums to 1. This part of the problem illustrates an important point. pX|V (x | v) is
actually not a single PMF but a family of PMFs.

(a) The probability that you wait more than 15 minutes is:
(b) The probability that you wait between 15 and thirty minutes is:
1
5
the heavy line is (E,E,E,E,E,N,N,N,N,N,E,E). There are a total of
G1†. (a) Clark must drive 12 block lengths to work, and each path is uniquely defined by saying
which 7 of those 12 block lengths is traveled East. For instance, the pathindicated by
79
2
The probability of picking any one path is 2
.
4
7 5
(b) Out of 792 paths, the ones of interest travel from point A to C and then from point C
to point B. Paths from A to C: = 35. Paths from C to B: = 10. Paths
from A
to C to B: 350. Thus, the probability of passing through C is 350 .
(c) If Clark was seen at point D, he must have reached the intersection of 3rd St. and 3rd
6 5
4 3
Ave. T he reasoning from this point is similar to that above. Paths from intersection
to B: = 15. Paths from intersection to C to B:= 10. Thus the conditional
3
probability of passing through C after being sighted at D is 2 .

(d) Paths from Main and Broadway to wth and hth: w+h w .
Paths from Main and Broadway to xth and yth: x+y x .
Paths from x and y to wth and hth: (w−x)+(h−y) (w−x) .
Thus the probability of passing by the phone booth is ( x+y x )((w−x)+(h−y) (w−x) ) ( w+h w ) . i.
If w = 0, then x = 0. The probability is ( y 0 )((h−y) 0 ) ( h 0 ) = 1, which is reasonable since
Clark must pass by the phone booth if there’s only one route to work. ii. If h = 0, then y = 0.
Similarly, the probability is

Probabilistic Systems Analysis Homework Help

More Related Content

What's hot (20)

Similar to Probabilistic Systems Analysis Homework Help (20)

More from Excel Homework Help (13)

Recently uploaded (20)

Probabilistic Systems Analysis Homework Help