Notes on Probability, Study notes of Probability and Statistics

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Notes on Probability

A.E. Frazho

• 1 The Probability Measure
  • 1.1 The axioms of probability
  • 1.2 Conditional probability
    • 1.2.1 A lost hiker
  • 1.3 The birthday problem
  • 1.4 The gambler’s ruin problem
  • 1.5 Bayes Rule
    • 1.5.1 Drug testing
    • 1.5.2 A classical example of Bayes rule.
    • 1.5.3 The Monty Hall Problem.
  • 1.6 Independent events
    • 1.6.1 Exercise
  • 1.7 PageRank
    • 1.7.1 Exercise
• 2 Random Variables
  • 2.1 The distribution functions
  • 2.2 The density function
    • 2.2.1 Exercise
  • 2.3 The exponential density
    • 2.3.1 The memoryless property
  • 2.4 The uniform distribution
  • 2.5 The Gaussian random variable
    • 2.5.1 An affine function of a random variable
    • 2.5.2 Exercise
  • 2.6 The diffusion equation
  • 2.7 The binomial random variable
  • 2.8 The geometric random variable
  • 2.9 The Poisson random variable
    • 2.9.1 The Poisson process
    • 2.9.2 Exercise
  • 2.10 Functions of a uniform random variable
    • 2.10.1 Exercise
  • 2.11 The joint distribution function
    • 2.11.1 Independent random variables. 4 CONTENTS
  • 2.12 The joint density function
    • 2.12.1 Independent random variables.
  • 2.13 A uniformly distributed dart board.
  • 2.14 The Raleigh density
  • 2.15 The Box-Muller transform
  • 2.16 The maximum and minimum of random variables
    • 2.16.1 The minimum of random variables
  • 2.17 The Buffon needle problem
  • 2.18 The meeting problem
  • 2.19 The distribution function for n random variables
    • 2.19.1 Exercise
  • 2.20 Two examples
    • 2.20.1 An example when y = x
    • 2.20.2 An example involving independent random variables
  • 2.21 Functions of two random variables
    • 2.21.1 The Gaussian density via rotation matrices
    • 2.21.2 The Box-Muller transformation revisited
• 3 Expectation
  • 3.1 The expectation of a random variable
    • 3.1.1 The uniform distribution: mean and variance.
    • 3.1.2 The exponential distribution: mean and variance
    • 3.1.3 The Gaussian distribution: mean and variance
  • 3.2 Properties of the expectation
    • 3.2.1 A kinetic energy example
    • 3.2.2 A sinusoid example
    • 3.2.3 Exercise
  • 3.3 The median
    • 3.3.1 An optimization problem for the median
    • 3.3.2 Exercise
  • 3.4 A lawn sprinkler problem
    • 3.4.1 The density for Θ such that x = α sin(2Θ) is uniform
  • 3.5 The binomial distribution: mean and variance
    • 3.5.1 Chebyshev’s inequality
  • 3.6 The geometric distribution: mean and variance
  • 3.7 The Poisson distribution: mean and variance
  • 3.8 The expectation of several random variables
  • 3.9 The uniform dart board revisited.
    • 3.9.1 Exercise
  • 3.10 Independent random variables and expectation
    • 3.10.1 Example: The meeting problem revisited
  • 3.11 Independent and identically distributed random variables
    • 3.11.1 The binomial density revisited
• CONTENTS
  • 3.12 The probability generating function
  • 3.13 The negative binomial distribution
    • 3.13.1 Negative binomial and sum of geometric random variables
    • 3.13.2 The mean and variance from ϕx(z)
  • 3.14 The gambler’s ruin: mean and variance
    • 3.14.1 The gambler’s ruin and the binomial distribution
    • 3.14.2 A Gaussian approximation of yn
  • 3.15 Random walk and the diffusion equation
  • 3.16 The Cauchy distribution
  • 3.17 The sum of two waves
• 4 The Projection Theorem
  • 4.1 Notation
  • 4.2 The projection theorem
    • 4.2.1 Orthogonal spaces and the orthogonal projection
  • 4.3 The Gram matrix
    • 4.3.1 An approximation of et
    • 4.3.2 Orthogonal basis and the projection theorem
  • 4.4 The mean and variance revisited
  • 4.5 A least squares optimization problem
    • 4.5.1 An application to curve fitting
• 5 Least Squares
  • 5.1 Random vectors
  • 5.2 Least squares estimation of a random vector
    • 5.2.1 An example with additive noise
    • 5.2.2 Exercise
  • 5.3 Classical least squares estimation
  • 5.4 Estimating a random variable in white noise
• 6 Conditional Expectation
  • 6.1 The condition expectation E(x | y) = PGy x.
    • 6.1.1 The conditional density
    • 6.1.2 The conditional variance
  • 6.2 Applications of Ex = E(E(x | y))
    • 6.2.1 The geometric distribution revisited
    • 6.2.2 A classical maze example
  • 6.3 Two heads and Adam’s law
    • 6.3.1 Computing the mean for two heads
    • 6.3.2 Computing the variance for two heads
    • 6.3.3 Find ϕx(z) for 2 heads and the Z transform
  • 6.4 Adam’s law for a head followed by a tail
    • 6.4.1 Find ϕx(z) for a head tail combination
  • 6.5 Adam’s law for r heads in a row
    • 6.5.1 Find ϕx(z) for r heads
    • 6.5.2 Finite sum formulas 6 CONTENTS
  • 6.6 A stick breaking problem
  • 6.7 An example
  • 6.8 Another example
  • 6.9 An exponential example
  • 6.10 An area example
  • 6.11 The sum independent random variables
  • 6.12 The sum of two exponential random variables.
    • 6.12.1 The case when Ex = μ and Ev = μ
    • 6.12.2 The case when Ex 6 = Ev
    • 6.12.3 The linear estimate
  • 6.13 A uniform random variable example.
  • 6.14 A conditional expectation example E(cos(2πx) | y)
  • 6.15 Conditional expectation problems
    • 6.15.1 Solutions to Problems 9 and
• 7 Gaussian Random Vectors
  • 7.1 Linear estimation revisited
  • 7.2 Functions of a random vector
  • 7.3 Gaussian random vectors
  • 7.4 Gaussian estimation
• 8 The Central Limit Theorem
  • 8.1 The characteristic function
    • 8.1.1 Exercise
  • 8.2 Some convergence results
    • 8.2.1 A discrete time random walk
  • 8.3 The central limit theorem
  • 8.4 The law of large numbers.
• 9 Maximal likelihood estimation
  • 9.1 Maximal likelihood estimation
  • 9.2 Maximal likelihood estimation for Gaussian random vectors.
    • 9.2.1 The Gaussian vector case.
  • 9.3 The multinomial distribution
    • 9.3.1 The multinomial maximal likelihood estimate
  • 9.4 A multinomial example
• 10 Fixed points and the EM algorithm
  • 10.1 The EM algorithm and fixed points
  • 10.2 An exponential example.
  • 10.3 A Poisson example
  • 10.4 Mixing a finite set of random vectors
    • 10.4.1 A Gaussian and exponential mixture example.
    • 10.4.2 The discrete random vector case
• CONTENTS
  • 10.5 Exponential mixture.
    • 10.5.1 An exponential mixture example
  • 10.6 Gaussian mixture.
    • 10.6.1 A Gaussian mixture example
  • 10.7 Multinomial mixing
    • 10.7.1 A multinomial mixing example
    • 10.7.2 Exercise: Poisson mixing
• 11 Appendix: Markov matrices
  • 11.1 Markov matrices
  • 11.2 The Gershgorin circle theorem
  • 11.3 The continuous time Markov semigroup
• 12 Appendix: Absorbing Markov systems
  • 12.1 The absorbing Markov state space
  • 12.2 A M(Q, B, Ω) system for the geometric distribution
  • 12.3 A M(Q, B, Ω) system for the negative binomial
  • 12.4 A M(Q, B, Ω) system for 2 heads in a row
  • 12.5 A M(Q, B, Ω) system for a head tail
  • 12.6 A Markov approach to r heads
• 13 Appendix: state space and the gambler’s ruin
  • 13.1 The state space system
  • 13.2 The average number of games played
    • 13.2.1 The backward equation
  • 13.3 The variance for the number of games played
    • 13.3.1 An Example to compute the standard deviation
    • 13.3.2 The backward equation and variance

8 CONTENTS

10 CHAPTER 1. THE PROBABILITY MEASURE

Hence P (Ac) = 1 − P (A) and (1.1) holds. The axioms of probability guarantee that for any event 0 ≤ P (A) ≤ 1. By axiom (i), we have P (B) ≥ 0 for any event B. Since Ac^ is an event, P (Ac) ≥ 0. This readily implies that 0 ≤ P (Ac) = 1 − P (A). Hence 1 ≥ P (A). We claim that P (ABc) = P (A) − P (AB). (1.2)

Recall that for any sets F , G, and H, we have

F (G ∪ H) = F ∩ (G ∪ H) = F G ∪ F H.

Using this we obtain A = AΩ = A(B ∪ Bc) = AB ∪ ABc.

Since AB ∩ ABc^ = φ, axiom (iii), yields

P (A) = P (AB ∪ ABc) = P (AB) + P (ABc).

Hence P (ABc) = P (A) − P (AB) and (1.2) holds. For example consider a deck of 52 cards. Then the probability of drawing an ace but not an ace of spades is 3/52. To see this let A be the event of drawing an ace and ♠ the event of drawing a spade. Then we have

P (A♠c) = P (A) − P (A♠) =

Another method is to use axiom (iii′), that is,

P (A♠c) = P (A♣ ∪ A♦ ∪ A♥) = P (A♣) + P (A♦) + P (A♥) =

As expected, ♣ is the event of drawing a club, and ♦ is the event of drawing a diamond, while ♥ is the event of drawing a heart. The following result is useful.

P (A ∪ B) = P (A) + P (B) − P (AB). (1.3)

To see this first observe that

A ∪ B = ABc^ ∪ AB ∪ AcB. (1.4)

One can see this by drawing a Venn diagram. To mathematically verify this observe that

ABc^ ∪ AB ∪ AcB = (A ∩ (Bc^ ∪ B)) ∪ AcB = A ∪ AcB = (A ∪ Ac) ∩ (A ∪ B) = A ∪ B.

Thus (1.4) holds. Recall that P (EF c) = P (E) − P (EF ). Since ABc, AB and AcB have no common intersection, axiom (iii′) yields

P (A ∪ B) = P (ABc) + P (AB) + P (AcB) = P (A) − P (AB) + P (AB) + P (B) − P (AB) = P (A) + P (B) − P (AB).

1.1. THE AXIOMS OF PROBABILITY 11

Therefore (1.3) holds. For example, the probability of drawing an ace or a spade in a deck of 52 cards is 4/13. By consulting (1.3), we have

P (A ∪ ♠) = P (A) + P (♠) − P (A♠) =

Another useful formula is

P (A ∪ B ∪ C) = P (A) + P (B) + P (C) − P (AB) − P (AC) − P (BC) + P (ABC). (1.5)

To verify this recall that P (F ∪ G) = P (F ) + P (G) − P (F G). Using this we obtain

P (A ∪ B ∪ C) = P (A ∪ (B ∪ C)) = P (A) + P (B ∪ C) − P (A ∩ (B ∪ C)) = P (A) + P (B) + P (C) − P (BC) − P ((A ∩ B) ∪ (A ∩ C)) = P (A) + P (B) + P (C) − P (BC) − P (AB) − P (AC) + P (ABAC) = P (A) + P (B) + P (C) − P (AB) − P (AC) − P (BC) + P (ABC).

Therefore equation (1.5) holds. For example, find the probability of drawing an ace (denoted by A) or a queen (denoted by Q) or a spade (denoted by ♠) in a deck of 52 cards. By employing (1.5), we obtain

P (A ∪ Q ∪ ♠) = P (A) + P (Q) + P (♠) − P (AQ) − P (A♠) − P (Q♠) + P (AQ♠)

=

The probability space (Ω, A, P ). A mathematically precise definition of a probability space involves the notion of a σ-algebra. Since these notes are an introduction to the basic concepts of probability, we will not use measure theoretic techniques. However, for complete- ness we will define a probability space. A σ-algebra (Ω, A) is a pair consisting of a universal set Ω and a collection of sets A such that the empty set φ is A. Moreover, if A ∈ A, then its complement Ac^ ∈ A. In particular, Ω is in A. Finally, if any countable collection of sets {Aj }∞ 1 is in A, then its union ∪∞ 1 Aj is also in A. By De-Morgan’s law if any countable collection of sets {Aj }∞ 1 is in A, then its intersection ∩∞ 1 Aj is also in A. A probability measure space (Ω, A, P ) is a σ-algebra (Ω, A) along with a measure P mapping A into [0, 1] such that

(i) If A ∈ A, then P (A) ≥ 0.

(ii) P (Ω) = 1.

(iii) If the set {Aj }∞ 1 in A are pairwise distinct Aj ∩ Ak = φ when j 6 = k, then

P (

⋃^ ∞

j=

Aj ) =

∑^ ∞

j=

P (Aj ).

1.2. CONDITIONAL PROBABILITY 13

Since P (· | B) is a probability measure, all of our previous results concerning probability measures holds for P (· | B). In particular, the following results hold:

P (Ac^ | B) = 1 − P (A | B) P (ACc^ | B) = P (A | B) − P (AC | B) P (A ∪ C | B) = P (A | B) + P (C | B) − P (AC | B)

The following result is useful in applications:

P (A) = P (A | B)P (B) + P (A | Bc)P (Bc). (2.2)

To see this recall that P (F G) = P (F | G)P (G). Using this with the third axiom of proba- bility, we obtain

P (A) = P (AΩ) = P (A ∩ (B ∪ Bc)) = P (AB ∪ ABc) = P (AB) + P (ABc) = P (A | B)P (B) + P (A | Bc)P (Bc).

Therefore (2.2) holds. We say that {Bj }n 1 is a partition of Ω if the following two conditions hold

Bj ∩ Bk = φ for all j 6 = k and

⋃^ n

j=

Bj = Ω. (2.3)

If {Bj }n 1 is a partition of Ω, then we have the following useful result

P (A) =

∑^ n

j=

P (A | Bj )P (Bj ). (2.4)

To prove this simply observe that

P (A) = P (AΩ) = P (A ∩ (∪nj=1Bj )) = P (∪nj=1ABj )

=

∑^ n

j=

P (ABj ) =

∑^ n

j=

P (A | Bj )P (Bj ).

Therefore (2.4) holds.

1.2.1 A lost hiker

The following example is taken from Rozanov [48]. Consider a hiker who is lost in the woods. The hiker equally likely chooses a path to find his way home from a starting position and does not backtrack; see Figure 1.1. (GeoGebra was used to draw Figure 1.1.) The hiker chooses a path at random an either ends up at home or gets lost. Let A be the event that the hiker makes it home. The hiker must pass through one of the intermediate points {Bj }^41.

14 CHAPTER 1. THE PROBABILITY MEASURE

Figure 1.1: A lost hiker

Moreover, {Bj }^41 form a partition of Ω. By consulting equation (2.4) with Figure 1.1, we see that

P (A) =

∑^4

j=

P (A | Bj )P (Bj )

= P (A | B 1 )P (B 1 ) + P (A | B 2 )P (B 2 ) + P (A | B 3 )P (B 3 ) + P (A | B 4 )P (B 4 )

=

×

×

+ 1 ×

×

Therefore the probability that the hiker makes it home is P (A) = 2348. The probability the hiker gets lost in the woods is P (Ac) = 2548. Finally, one might be naive and think that the probability of arriving home is the number of paths leaving {Bj }^41 and arriving at home 5, divided by the total number of paths 14 leaving {Bj }^41. However, 145 is not the correct answer.

1.3 The birthday problem

Assume that there are n people in the room. The birthday problem is to find the probability that at least two of the n people have the same birthday. Here we exclude leapyear and assume that there are no twins in the room. Moreover, we assume that a birthday is equally likely to occur any day of the year, and n ≤ 365. Let Sn be the event that at least two people out of n have the same birthday. We claim that

P (Sn) = 1 −

n∏− 1

j=

j 365

To derive the formula for P (Sn) in (3.1), let Dn be the event that we have n people with a different birthday. Moreover, Dn means that we have chosen the n-th person at random with

16 CHAPTER 1. THE PROBABILITY MEASURE

Then P (Sn) = p(n). One can find an exponential approximation for P (Sn). To this end, recall that the Taylor series for ez^ is given by ez^ =

0

zn n!. In particular,^ e

−a (^) ≈ 1 − a when |a| is ”close” to zero.

(In fact, 1 − a < e−a^ for 0 < a ≤ 1.) Using this approximation, we have

P (Sn) = 1 −

n∏− 1

j=

j 365

n∏− 1

j=

e−^ 365 j = 1 − e−^ 3651 ∑n j=1−^1 j^ .

Now recall that the sum of the first n − 1 integers is given by n∑− 1

j=

j =

n(n − 1) 2

The derivation of this formula is left to the reader as a simple exercise. To prove this one simply rearranges the integers. For example,

∑^9

j=

j = 9 + (1 + 8) + (2 + 7) + (3 + 6) + (4 + 5) =

10 × 9

Using

∑n− 1 j=1 j^ =^

n(n−1) 2 , we see that

P (Sn) = 1 −

n∏− 1

j=

j 365

≈ 1 − e−^

n( 730 n−1) ≈ 1 − e−^ 730 n^2

. (3.2)

Let g and h be the functions defined by

g(n) = 1 − e−^

n( 730 n−1) for 1 ≤ n ≤ 365

h(n) = 1 − e−^ 730 n^2 for 1 ≤ n ≤ 365.

The graph of n vs p(n) = P (Sn) is given in Figure 1.2 in blue, the graph of g is in red, while the graph of h is in green. Finally, the approximations of P (Sn) by g and h are quite good. Since P (S 100 ) ≈ 1 we only plotted this graph for 1 ≤ n ≤ 100. In fact, using Matlab, we discovered, that

‖p − g‖ =

√√ ∑^365

j=

|p(j) − g(j)|^2 = 0. 0500

‖p − h‖ =

∑^365

j=

|p(j) − h(j)|^2 = 0. 0505

‖p − g‖∞ = max{|p(j) − g(j)| : 1 ≤ j ≤ 365 } = 0. 0103 |p − h‖∞ = max{|p(j) − h(j)| : 1 ≤ j ≤ 365 } = 0. 0124.

As expected, the approximation of p(n) = P (Sn) by g(n) is better, then the approximation of P (Sn) by h(n). Finally, it is noted that even in this simple problem, one sees an exponential function

of the form e−^

x γ^2 where γ is a constant. These exponential forms will be seen throughout probability theory. For some further comments on the birthday problem see Wikipedia.

1.3. THE BIRTHDAY PROBLEM 17

0 20 40 60 80 100 n

0

1

Probability of at least 2 with the same birthday in a group of n

Figure 1.2: The birthday problem

An lower bound for P (Sn)

Following some ideas of P.R. Halmos, observe that 1 − a < e−a^ for 0 < a ≤ 1. Recall that xn = P (Dn), and thus,

xn =

n∏− 1

j=

j 365

n∏− 1

j=

e−^ 365 j = e−^

n( 730 n−1) .

Hence P (Sn) = 1 − xn > 1 − e−^

n(n−1) (^730) = g(n) (for 1 < n < 365).

Therefore the red graph corresponding to g is always below the blue graph corresponding to P (Sn) in Figure 1.2. (However, h is both above and below P (Sn).)

Recall that xn < e−^

n( 730 n−1) for 1 < n ≤ 365. Let us find the minimum value of n over the interval [1, 365] such that

e−^

n( 730 n−1) <

Consider the equation,

e−^

n(n 730 −1)

or equivalently 2 = e

n( 730 n−1) .

1.4. THE GAMBLER’S RUIN PROBLEM 19

Here λ 1 and λ 2 are scalars determined by the characteristic equation, while α and β are constants determined by the initial or final conditions. To find λ 1 and λ 2 substitute γλn^ into the difference equation in (4.2), which yields

γλn^ = γλn+1p + γλn−^1 q.

Dividing by γλn−^1 , we arrive at the following quadratic equation

λ^2 −

λ p

q p

Now let set r = q/p. Then using p + q = 1, observe that

(λ − 1)(λ − r) = λ^2 −

λ p

q p

So the roots to the quadratic equation in (4.5) are given by

λ 1 = 1 and λ 2 = r =

q p

Using the fact that p + q = 1, we see that there is a repeated root if and only if p = q = 1/2. In this case, λ 1 = 1 is the repeated root. Now assume that p 6 = 1/2, and thus there is no repeated root of (4.5). In this case, the solution to the difference equation in (4.2) is of the form

yn = α + βrn.

By employing the initial condition y 0 = 0 and the final condition ym = 1, we arrive at the following matrix equation: (^) [ 1 1 1 rm

] [

α β

]

[

]

Since r 6 = 1, the determinant of the previous 2 × 2 matrix is nonzero. By inverting this matrix, we arrive at

[ α β

]

rm^ − 1

[

rm^ − 1 − 1 1

] [

]

rm^ − 1

[

]

So the solution to the difference equation in (4.2) is given by

yn =

1 − rn 1 − rm^

(if p 6 = 1/2).

For p = 1/2 an application of L’Hˆopital’s rule yields

yn =

n m

(if p = 1/2).

20 CHAPTER 1. THE PROBABILITY MEASURE

To obtain the case when p = 1/2 directly, recall that for repeated roots the solution is given by yn = αλn 1 + βnλn 1. Since λ 1 = 1, in this case yn = α + βn. By employing the initial condition y 0 = 0 and the final condition ym = 1, we arrive at the following matrix equation:

[ 1 0 1 m

] [

α β

]

[

]

Thus α = 0 and β = 1/m. In other words, yn = n/m. Summing up our previous analysis, yields the following solution to the gambler’s ruin problem

P (Wn,m) =

1 − rn 1 − rm^

if p 6 = 1/ 2 (4.7) =

n m

if p = 1/ 2.

The solution to the gambler’s ruin problem shows that the probability of doubling one’s money m = 2n in a fair game p = 1/2 is 50%. This result is not surprising. However, the solution to the gambler’s ruin problem also shows that if p < 1 /2, then it is better to bet all your n dollars in one game rather than play one game at a time. On the other hand, if p > 1 /2, then it is better to bet one dollar on each game rather than bet all your money in one game. To be more explicit, for the moment assume that p > 1 /2. In this case, r = q/p < 1. In particular, rm^ converges to zero as m tends to infinity. So according to (4.7), the probability of making an infinite amount of money m = ∞ starting with n dollars is given by P (Wn,∞) = 1 − rn^ (p > 1 /2).

For example, if p = 0.51 and n = 100, then the probability that a gambler will make an infinite amount of money is 1 − (49/51)^100 ≈ 0 .98. This is why a casino will not let the players count cards or use a computer. In this case, p > 1 /2 and a player can bankrupt the casino. This is also why casino’s make a tremendous amount of money. The p for a casino for many games is greater than or equal to 0.55. Now assume that p < 1 /2. In this case, r = q/p > 1. So if n and m are large, then

P (Wn,m) =

1 − rn 1 − rm^

rn rm^

p q

)m−n .

If p = 1/2, then there is a 50% probability of achieving m = 200 dollars starting with n = 100 dollars. Now assume that p = 0.49 and the gambler starts out with n = 100 dollars and m = 200 dollars. Then P (Wn,m) = 0.018. So in this case, it is better to bet the one hundred dollars in the first game which yields a 49% chance of achieving 200 dollars, rather than playing one game at a time which has only a 1.8% chance of doubling the original 100 dollars. In fact, there is only a 36.4% chance of achieving m = 125 dollars starting with 100 dollars playing one game at a time. The situation is even worse the as p becomes smaller. For example, if p = 0.45 and n = 100, then there is only a 37% chance of achieving m = 105 dollars, and 0.66% chance of achieving m = 125 dollars.