probability, random processes and inference

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INSTITUTO POLITÉCNICO NACIONAL CENTRO DE INVESTIGACION EN COMPUTACION Probability, Random Processes and Inference Dr. Ponciano Jorge Escamilla Ambrosio [email protected] http://www.cic.ipn.mx/~pescamilla/ Laboratorio de Ciberseguridad

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Page 1: Probability, Random Processes and Inference

INSTITUTO POLITÉCNICO NACIONAL CENTRO DE INVESTIGACION EN COMPUTACION

Probability, Random Processes and Inference

Dr. Ponciano Jorge Escamilla Ambrosio [email protected]

http://www.cic.ipn.mx/~pescamilla/

Laboratorio de

Ciberseguridad

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CIC

1. Choose two numbers X1, X2 without

replacement and with equal probabilities from

the set {1, 2, 3}, and let X = max{X1, X2} and

Y = min{X1, X2}. a) Find the joint PMF of X

and Y; b) find the marginals of X = max(X1,

X2) and Y = min(X1, X2) from the joint PMF.

2

Problem 1

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Solution Problem 1

X1 1 1 2 2 3 3

X2 2 3 1 3 1 2

X 2 3 2 3 3 3

Y 1 1 1 2 1 2

Y\X 2 3 X Marginal

1 1/3 1/3 2/3

2 0 1/3 1/3

Y Marginal 1/3 1/3 1

The possible combinations are:

a) PMF y b) son:

Page 4: Probability, Random Processes and Inference

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2. You just rented a large house and the realtor

gave you 6 keys, one for each of the 6 doors of

the house. Unfortunately, all keys look identical,

so to open the front door, you try them at

random. Find the PMF of the number of trials

you will need to open the door under the

assumption that after an unsuccessful trial, you

mark the corresponding key, so you never try it

again.

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Problem 2

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Solution Problem 2

𝑃 𝑋 = 1 =1

6

𝑃 𝑋 = 2 =5

6∙1

5=1

6

𝑃 𝑋 = 3 =5

6∙4

5∙1

4=1

6

𝑃 𝑋 = 4 =5

6∙4

5∙3

4∙1

3=1

6

𝑃 𝑋 = 5 =5

6∙4

5∙3

4∙2

3∙1

2=1

6

𝑃 𝑋 = 6 =5

6∙4

5∙3

4∙2

3∙1

2∙1

1=1

6

𝑃𝑋 𝑥 =

1

6para 𝑋 = 1, 2, 3, 4, 5 , 6

0 cualquier otro caso

Page 6: Probability, Random Processes and Inference

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3. Accidentally, two depleted batteries got into a

set of five batteries. To remove the two depleted

batteries, the batteries are tested one by one in a

random order. Let the random variable X denote

the number of batteries that must be tested to

find the two depleted batteries. What is the

probability mass function of X?

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Problem 3

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Solution Problem 3

To find the two depleted batteries, you need at least two tests but no more than

four tests. Label the batteries as 1, 2, . . . , 5. Think of the order in which the

batteries are placed for testing as a random permutation of the numbers 1, 2, . . . ,

5. The sample space has 5! equally likely outcomes. You need two tests if the first

two batteries tested are depleted. The number of outcomes for which the two

depleted batteries are on the positions 1 and 2 is 2 × 1 × 3!. Hence:

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Solution Problem 3

You need three tests if the first three batteries tested are not depleted or if a

second depleted battery is found at the third test. This leads to:

The probability P (X = 4) follows from P (X = 4) = 1 − P (X = 2) − P (X = 3) =

6/10.

Alternatively, the probability mass function of X can be obtained by using

conditional probabilities. This gives

P (X = 0) = 2/5 × 1/4 = 1/10 and P (X = 2) = 3/5 × 2/4 × 1/3 + 2/5 × 3/4 × 1/3 +

3/5 × 2/4 × 1/3 = 3/10.

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4. You roll a fair dice twice. Let the random variable X be the

product of the outcomes of the two rolls. What is the probability

mass function of X? What are the expected value and the

standard deviation of X?

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Problem 4

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Solution Problem 4

The random variable X is defined on a sample space consisting of the 36

equiprobable elements (i, j), where i, j = 1, 2, . . . , 6. The random variable X

takes on the value i × j for the realization (i, j). The random variable X takes

on the 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 30, and 36 with

respective probabilities 1/36, 2/36, 2/36, 4/36, 2/36, 4/36, 2/36, 1/36, 2/36,

4/36, 2/36, 1/36, 2/36,‘ 2/36, 2/36, 1/36, and 1/36 . The expected value of X is

easiest computed as:

Also,

Hence E(X) = 12.25 and σ(X) =

Page 11: Probability, Random Processes and Inference

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5. Bobo, the amoeba, currently lives alone in a

pond. After one minute Bobo will either die,

split into two amoebas, or stay the same, with

equal probability. Find the expectation and

variance for the number of amoebas in the pond

after one minute.

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Problem 5

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Problem 5

Let X denote the number of amoebas after one minute, then by definition of

expected value:

E[X] = 0 · P (X = 0) + 1 · P (X = 1) + 2 · P (X = 2)

= 0 + 1/3 + 2/3 = 1

To compute the variance first evaluate the second moment E[X2] (or use the

definition directly):

E[X2 ] = E[X] = 0 · P (X = 0) + 1 · P (X = 1) + 4 · P (X = 2)

= 0 + 1/3 + 4/3 = 5/3

Thus,

Var[X] = E[X2 ] - (E[X])2 = 5/3 - 1 = 2/3

Page 13: Probability, Random Processes and Inference

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6. In a digital communication system a "0" or "1" is

transmitted to a receiver. Typically, either bit is equally

likely to occur so that a prior probability of 1/2 is

assumed. At the receiver a decoding error can be made

due to channel noise, so that a 0 may be mistaken for a

1 and vice versa. Defining the probability of decoding a

1 when a 0 is transmitted as ϵ and a 0 when a 1 is

transmitted also as ϵ, what is the overall probability of

an error?

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Problem 6

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Solution Problem 6

The effect of the channel is to introduce an error so that even if we know

which bit was transmitted, we do not know the received bit.

The probability of error is:

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7. A radioactive source emits particles toward a Geiger

counter. The number of particles that are emitted in a

given time interval is Poisson distributed with expected

value λ. An emitted particle is recorded by the counter

with probability p, independently of the other particles.

Let the random variable X be the number of recorded

particles in the given time interval and Y be the number

of unrecorded particles in the time interval. What are

the probability mass functions of X and Y ? Are X and

Y independent?

15

Problem 7

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Solution Problem 7

Let the random variable N be the number of particles emitted in the given

time interval. Noting that P (X = j, Y = k) = P (X = j, Y = k, N = j + k) and

using the formula P (AB) = P (A | B) P (B), it follows that:

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Solution Problem 7

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8. Let X be a random variable with PMF:

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Problem 8

(a) Find a and E[X].

(b) What is the PMF of the random variable Z = (X - E[X])2 ?

(c) Using the result from part (b), find the variance of X.

Find the variance of X using the formula var(X) = 𝑥 − 𝐸 𝑋 2𝑥 𝑝𝑋 𝑥 .

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Solution Problem 8

(a) The scalar a must satisfy:

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Solution Problem 8

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9. Alvin throws darts at a circular target of

radius r and is equally likely to hit any point in

the target. Let X be the distance of Alvin’s hit

from the center.

(a) Find the PDF, the mean and the variance of

X.

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Problem 9

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Solution Problem 9

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10. A person playing darts finds that the

probability at which the dart falls between r and r

+ dr is:

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Problem 10

where, R is the distance of the impact from the center of the dartboard, c is a

constant and a is the radius of the dartboard. Find the probability of hitting a

target of radius b concentric with the dartboard (see figure below). Assume also

that impact on the dartboard is always done.

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Solution Problem 10