Problem Set#4 Solution

    1. A company owns two online social networking sites, Lookbook and Quickgram. On average, 7.5 users sign up for Lookbook each minute, while on average 5.5 users sign up for Quickgram each minute. The number of users signing up for Lookbook and for Quickgram each minute are independent. A new user is defined as a new account, i.e., the same person signing up for both social networking sites will count as two new users.

        a. What is the probability that more than 10 new users will sign up for the Lookbook social networking site in the next minute?

        b. What is the probability that more than 13 new users will sign up for the Quickgram social networking site in the next 2 minutes?

        c. What is the probability that the company will get a combined total of more than 40 new users across both websites in the next 2 minutes?

    2. The median of a continuous random variable having cumulative distribution function F is the value m such that F (m) = 0:5. That is, a random variable is just as likely to be larger than its median as it is to be smaller. Find the median of X (in terms of the respective distribution parameters) in each case below.

        a. X   Uni(a; b)

        b. X  N(;2)

        c. X   Exp( )



    b) The normal distribution is symmetric about the mean, i.e. , the median (an mean and mode) would occurs at . Thus, m = . For a more rigorous/formal approach, we can use the standard normal, since the integral doesn’t give an analytical solution. To use the standard normal RV,

we would substitute Z = X    N (  = 0;  = 1). Thus, we have:

=

    c) For the exponential, we can use the integral in a similar process to what we’ve done in the previous parts. Thus, we get the following but with the lower bound of 0 since the f(x) = 0 outside the range of x 0:

Natasha Ong    Problem Set #4    February 21, 2020



    3. Let X, Y , and Z be independent random variables, where X N ( 1; 12), Y N ( 2; 22), and Z N ( 3; 32).

        a. Let A = X + Y . What is the distribution (along with parameter values) for A?

        b. Let B = 4X + 3. What is the distribution (along with parameter values) for B?

c. Let C = aX b2Y + cZ, where a, b, and c are real-valued constants. What is the distribution (along with parameter values) for C? Show how you derived your answer.



    4. Say the lifetimes of computer chips produced by a certain manufacturer are normally distributed with parameters = 1:5 106 hours and = 9 105 hours. The lifetime of each chip is independent of the other chips produced. What is the approximate probability that a batch of 100 chips will contain at least 65 whose lifetimes are less than 1:9 106 hours?

    5. You roll 6 six-sided dice. How much more likely is a roll with [1 one, 1 two, 1 three, 1 four, 1 five, 1 six] than a roll with 6 sixes? Think of your dice roll as a multinomial.

    6. You are testing software and discover that your program has a non-deterministic bug that causes catastrophic failure (aka a “hindenbug”). Your program was tested for 400 hours and the bug occurred twice.

        a. Each user uses your program to complete a three hour long task. If the hindenbug manifests they will immediately stop their work. What is the probability that the bug manifests for a given user?

        b. Your program is used by one million users. Use a normal approximation to estimate the probability that more than 10,000 users experience the bug. Use your answer from part
(a).




    7. Say that of all the students who will attend Stanford, each will buy at most one laptop computer when they first arrive at school. 40% of students will purchase a PC, 30% will purchase a Mac, 10% will purchase a Linux machine and the remaining 20% will not buy any laptop at all. If 15 students are asked which, if any, laptop they purchased, what is the probability that exactly 6 students will have purchased a PC, 4 will have purchased a Mac, 2 will have purchased a Linux machine, and the remaining 3 students will have not purchased any laptop?







8. The joint probability density function of continuous random variables X and Y is given by:

fX;Y (x; y) = c
y
where 0 < y < x < 1



x


    a. What is the value of c in order for fX;Y (x; y) to be a valid probability density function?

    b. Are X and Y independent? Explain why or why not.

    c. What is the marginal density function of X?

    d. What is the marginal density function of Y ?

    e. What is E[X]?




d) We can repeat a similar process as in part c but summing over xs instead of y:
Z 1    y
fY (y) =    y    4xdx


= 4y(ln(1)    ln(y)) =    4y ln(y)

    e) E[X] = R01 x f(x)dx for a continuous function with the bounds of 0 and 1. Plugging in our marginal density function from part c and solving, we have the following:


    • 1

E[X] =    x f(x)dx

0
        ◦ 1

    • 2x2dx
0

    • 23x3 evaluated at 0,1

    • 23



    9. A robot is located at the center of a square world that is 10 kilometers on each side. A package is dropped off in the robot’s world at a point (x; y) that is uniformly (continuously) distributed in the square. If the robot’s starting location is designated to be (0; 0) and the robot can only move up/down/left/right parallel to the sides of the square, the distance the robot must travel to get to the package at point (x; y) is jxj + jyj. Let D = the distance the robot travels to get to the package. Compute E[D].



    10. Choose a number X at random from the set of numbers f1; 2; 3; 4; 5g. Now choose a number at random from the subset no larger than X, that is from f1; : : : ; Xg. Let Y denote the second number chosen.

        a. Determine the joint probability mass function of X and Y .

        b. Determine the conditional mass function of X given Y = i. Do this for i = 1; 2; 3; 4; 5.

        c. Are X and Y independent? Justify your answer.



    b) Notice that the values that X can take on given any Y is Y to 5 inclusive. For example, given Y = 4, X can take on the values 4; 5 based on the definition of Y .


    c) No these are not independent as the value of X influences the sample space of Y since the upper value of Y is capped at X.

For a more rigorous example, notice that if X and Y were independent, then P (X = 1; Y = 1) = P(X = 1)P(Y = 1).

We see that P (X = 1; Y = 1) = 15 from part (a). However, P (X = 1)P (Y = 1) = 15 137300 = 1500137 6= 15 from part b’s graph. Clearly, these variables are not independent.

    11. Let X1; X2; : : : be a series of independent random variables which all have the same mean and the same variance 2. Let Yn = Xn + Xn+1. For j = 0, 1, and 2, determine Cov(Yn; Yn+j). Note that you may have different cases for your answer depending on the value of j.


    12. Consider a series of strings that independently get hashed into a hash table. Each such string can be sent to any one of k + 1 buckets (numbered from 0 to k). Let index i denote the i-th bucket. A string will independently get hashed to bucket i with probability pi, where Pki=0 pi = 1. Let N denote the number of strings that are hashed until one is hashed to any bucket other than bucket

        0. Let X be the number of that bucket (i.e. the bucket not numbered 0 that receives a string).

            a. Find P (N = n); n   1.

            b. Find P (X = j), j = 1; 2; : : : ; k.

            c. Show that N and X are independent.



    13. Consider the following function, which simulates repeatedly rolling a 6-sided die (where each integer value from 1 to 6 is equally likely to be “rolled”) until a value 3 is “rolled”.

def roll(): total = 0 while(True):

    • randomInteger is equally likely to return 1, ..., 6 roll = randomInteger(1, 6)

total += roll

    • exit condition:

if (roll >= 3):

break

return total

    a. Let X = the value returned by the function roll(). What is E[X]?

    b. Let Y = the number of times that the die is "rolled" (i.e., the number of times that randomInteger(1, 6) is called) in the function roll(). What is E[Y ]?





b) We can use a similar process, where we realize that if the roll is 3, then it is only rolled once. If one rolls a one or two for the first roll, we add one to the E [y] since it is essentially a recursive case like earlier. Moreover, each value for the roll is equally likely, like in the previous case. Therefore, we have the following:

E[Y ] = 16(1) + 16(1) + 16(1) + 16(1) + 16(1 + E[Y ]) + 16(1 + E[Y ])

E[Y ]
E[Y]=1+


E[Y]= 32



    14. Our ability to fight contagious diseases depends on our ability to model them. One person is exposed to llama-flu. The method below models the number of individuals who will get infected.

from scipy import stats

"""

Return number of people infected by one individual.

"""

def num_infected():

        ◦ most people are immune to llama flu.

        ◦ stats.bernoulli(p).rvs() returns 1 w.p. p (0 otherwise)

immune = stats.bernoulli(p = 0.99).rvs()

if immune: return 0

    • people who are not immune spread the disease far by

    • making contact with k people (up to 100).

spread = 0

    • returns random # of successes in n trials w.p. p of success k = stats.binom(n = 100, p = 0.25).rvs()

for i in range(k): spread += num_infected()

    • total infections will include this individual

return spread + 1

What is the expected return value of numInfected()?


    15. Did you know that computers can identify you not only by what you write, but also by how you write? Coursera uses Biometric Keystroke signatures for plagiarism detection. If you cannot write a sentence with the same statistical distribution of key press timings as in your previous work, they assume that you are not the person sitting behind the computer. In this problem we provide you with three files:

        ◦ personKeyTimingA.txt has keystroke timing information for a user A writing a passage. The first column is the time in milliseconds (since the start of writing) when the user hit each key. The second column is the key that the user hit.

        ◦ personKeyTimingB.txt has keystroke timing information for a second user (user B) writing the same passage as the user A. Even though the content of the passage is the same the timing of how the second user wrote the passage is different.

        ◦ email.txt has keystroke timing information for an unknown user. We would like to know if the author of the email was user A or user B.

Let X and Y be random variables for the duration of time, in milliseconds, for users A and B (respectively) to type a key. Assume that each keystroke from a user has a duration that is an independent random variable with the same distribution.

    a. [Coding] Complete the function part_a provided in the starter code. This function takes in a filename (which is either personKeyTimingA.txt or personKeyTimingB.txt) and should return E[X] or E[Y ], respectively.

    b. [Coding] Complete the function part_b provided in the starter code. This function should return E[X2] or E[Y 2], depending on which file is supplied as filename.
    c. [Written] Use your answers to part (a) and (b) and approximate X and Y as Normal random variables with mean and variance that match their biometric data. Report both distributions.

    d. [Written] Calculate the ratio of the probability that user A wrote the email over the probability that user B wrote the email. You do not need to submit code, but you should include the formula that you attempted to calculate and a short description (a few sentences) of how your code works.




For Y:

= E[Y ]   8:0308

V ar(Y ) =    2 = E[Y 2]    (E[Y ])2 = 68:17685    (8:0308)2    3:68358


        ◦ N (  = 8:0308;  2 = 3:68358)

    d) We can use the normal distributions calculated in part c and combine it with the expected/mean value for the email.txt file. Using some code, I extracted the mean value for the email.txt file as 7:9442.


We can then use relative probabilities of continuous random variables to find the probability for each and then divide. In other words, we are trying to calculate P (emailjA) and P (emailjB) and would use
P (emailjA) > 1 : : : writer A

P (emailjB)

Thus, we have the following:

P (emailjA)
=
P (X = 7:9442)
=
f(X = 7:9442)

P (emailjB)


P (Y = 7:9442)


f(Y = 7:9442)










=
0:92838







B is slightly more likely to have written the email than A.