Bertsekas Probability Notes

Ravi Mohan

July 12, 2012

1 Chapter 1. Sample Space and Probability

(1) Definiton:

2 Chapter 2. Discrete Random Variables

(1) Definiton: Random Variable: A Random Variable is a mapping from anoutcome of an experiment to a real number.

e.g: Experiment: Throw 2 six sided dice. Outcome: { (1,1) ... (6,6) }Random Variable X maps outcome to the sum of the numbers on the dice.

Thus X( (1,1) ) 2The domain of the RV is the Sample Space. The range is Real Number.

(2) Notation: X(o) = r, o Sample Space is written as X = r;

(3) An RV is discrete if its range is discrete. Discrete = Finite or CountablyInfinite.

(4) A Random Variable X has a Probablity Mass Function (PMF) denoted byX which maps each value of its range to the probability of its occurence.

If x is an element of Xs range then probablity mass of x is X(x) =P (outcome {o | X(o) = x}).e.g: Experiment: Throw 2 six sided dice. Outcome: { (1,1) ... (6,6) }Random Variable X maps outcome to the sum of the numbers on the dice.

then

X(2) = P (outcome {o | X(o) = 2})=

X(2) = P (outcome {(1, 1)})=

X(2) = P (outcome = (1, 1))

1

=

X(2) =136

(5) A Random Variable can be discrete even if its domain is infinite. Only therange needs to be finite or countably infinite.

e.g: sgn(P )

1 if p > 00 if p = 01 if p < 0is a discrete random variable, though the domain is infinite.

(6) A Discrete Random Variable has an associated Probability Mass Functionor PMF that maps each value of the *range* of the RV to the probabilityof its occurrence.

Thus

RV: outcome RPMF: range(RV ) probabilityIf x is a variable over the range of discrete Random Variable X, the prob-ablitiy mass of x, denoted X(x) = the probability of the event P({ X =x}).To calculate the PMF of a discrete RV X, for each possible value x of therange of the RV,

(1) Collect all experiment outcomes, oi that give rise to X having a value x

(2) Collect all these outcomes oi into a single event A

(3) Calculate P(A)

Check for PMF (X) isx

X(x) = 1

Example:

Experiment = 2 consecutive rolls of a fair die

Sample Space = { (1,1),(2,2)...(6,6)}Random Variable X = number of sixes in outcome = { 0,1,2 }X(2) = P(outcome = (6,6)) =

136

X(1) = P(outcome {(1,6),(2,6)..(5,6),(6,1),(6,2)..(6,5) } = 1036X(0) = 1 -

1136 =

2536

Thus,

X(x) =

2536 if x = 01036 if x = 1136 if x = 00 otherwise

2

3 Chapter 3. General Random Variables

(1) Brief Review of Integration

(a) Integration by linearity (corresponding to linearity of dervatives)

(1) sum rule[v(x) + w(x)] dx =

v(x)dx+

w(x)dx

(2) constant rulecv(x)dx = c

v(x)

(3) linearity[av(x) + bw(x)] dx = a

v(x)dx+ b

w(x)dx

(b) Integration as Anti Derivative (corresponding to chain rule of deriva-tives)

(1) Chain Rule of Derivatives: dfdx =dfdu

dudx

Example: d(sin2x)

dxd(sin2x)

dx= Let u = sinx, then

d(u2)dx

= Chain Rule of Derivatives: dfdx =dfdu

dudx

2ududx= Substitute back u = sinx

2sin(x)d sin(x)dx= Differentiating sin(x)

2sin(x)cos(x)

(2) Integration by Anti Derivatives reverses this process.

The key is to find a function u within the given function f so that f= u du (ignoring constants) Thus given

sin(x)cos(x)= Identify u to be sin(x) ,

u du

=undu = u

n+1

n + Cu2

2 + C= Substituting back u = sin(x),

sin2x2 + C

(c) Integration by Parts (corresponding to product rule of derivatives)

(1) Product Rule of Derivatives: ddxu(x)v(x) = v(x)ddxu(x)+u(x)

ddxv(x)

Example

ddxx

2sin(x)

= Product Rule of Derivatives: ddxu(x)v(x) = v(x)ddxu(x) +

u(x) ddxv(x)

3

x2 ddxsin(x) + sin(x)ddxx

2

= Differentiating each partx2cos(x) + 2 x sin(x)

(2) Integration by parts reverses Product Rule.

The Product Rule for differentiation isddxu v = v

ddxu+ u

ddxv

=u ddxv =

ddxu v v

ddxu

Integrating both sides,u(x) ddxv(x)dx = u(x)v(x)

v(x) ddxu(x)dx

This is the equation for Integration by Parts.

The problem of integrating u dv is converted into the problem ofintegrating v du.

Very Important: Note the presence of dx in the integral forms.

The key is to find functions u and v such that the given integralfdx is equivalent to

u ddxv dx -note explicit presence of dx s

Exampleln x dx

= Let u(x) = ln x, v(x) = x.Then ddxv(x) = 1 and

ddxv(x)dx = 1dx = dx

u(x) ddxv(x)dx= Integration by Partsu(x)v(x)

v(x) ddxu(x)dx

= Substituting for u(x) and v(x)x ln(x)

x ddx ln(x)dx

= ddx ln(x) =1x

x ln(x)x 1xdx

=x 1xdx

x ln(x)

1dx=x ln(x) x

(d) Definite Integrals

ifv(x)dx = f(x) + C then

bav(x) = f(b) f(a)

(e) Partial Derivatives

(f) Multiple Integrals

4

4 Chapter 4. Further Topics on Random Vari-ables

5 Chapter 5. Limit Theorems

6 Chapter 6. Bernoulli and Poisson Processes

7 Chapter 7. Markov Chains

8 Chapter 8. Bayesian Statistical Inference

9 Chapter 9. Classical Statistical Inference

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