Random variables DS GA 1002 Statistical and Mathematical Models - - PowerPoint PPT Presentation

Random variables

DS GA 1002 Statistical and Mathematical Models

http://www.cims.nyu.edu/~cfgranda/pages/DSGA1002_fall16 Carlos Fernandez-Granda

Motivation

Random variables model numerical quantities that are uncertain They allow us to structure the information we have about these quantities in a principled way

Definition

Given a probability space (Ω, F, P), a random variable X is a function from the sample space Ω to the real numbers R We use uppercase letters to denote random variables: X, Y , . . . Once the outcome ω ∈ Ω is revealed, X (ω) is the realization of X We use lowercase letters to denote numerical values: x, y, . . .

Characterization

Given a probability space (Ω, F, P), for any set S P (X ∈ S) = P ({ω | X (ω) ∈ S}) . We will almost never construct probabilistic models like this!

Discrete random variables Continuous random variables Conditioning on an event Functions of random variables

Discrete random variables

Discrete random variables take values on a finite or countably infinite subset of R such as the integers The probability mass function (pmf) of X is defined as pX (x) := P ({ω | X (ω) = x}) In words, pX (x) is the probability that X equals x The pmf completely specifies a random variable

Probability mass function

If D is the range of X, then

• D, 2D, pX
• is a valid probability space

Any pmf satisfies pX (x) ≥ 0 for any x ∈ D,

• x∈D

pX (x) = 1 P (X ∈ S) =

• x∈S

pX (x) for any S ⊆ D

Defining a discrete random variable

To define a discrete random variable X we just need

◮ A discrete range D ◮ A nonnegative function pX satisfying

• x∈D

pX (x) = 1

Bernoulli random variable

Experiment with two possible outcomes (coin flip with bias p) pX (0) = 1 − p pX (1) = p Special case: Indicator random variable of an event S 1S (ω) =

• 1,

if ω ∈ S, 0,

• therwise

Bernoulli with parameter P (S) Allows to translate events to random variables

Example: Coin flips

You flip a coin with bias p until you obtain heads (flips are independent) If you model the number of flips as a random variable X, what is pX?

Example: Coin flips

pX (k)

Example: Coin flips

pX (k) = P (k flips)

Example: Coin flips

pX (k) = P (k flips) = P (1st flip = tails, . . . , k − 1th flip = tails, kth flip = heads)

Example: Coin flips

pX (k) = P (k flips) = P (1st flip = tails, . . . , k − 1th flip = tails, kth flip = heads) = P (1st flip = tails) · · · P (k − 1th flip = tails) P (kth flip = heads)

Example: Coin flips

pX (k) = P (k flips) = P (1st flip = tails, . . . , k − 1th flip = tails, kth flip = heads) = P (1st flip = tails) · · · P (k − 1th flip = tails) P (kth flip = heads) = (1 − p)k−1 p

Geometric random variable

The pmf of a geometric random variable with parameter p is pX (k) = (1 − p)k−1 p k = 1, 2, . . .

Geometric random variable p = 0.2

2 4 6 8 10 0.2 0.4 0.6 0.8 k pX(k)

Geometric random variable p = 0.5

2 4 6 8 10 0.2 0.4 0.6 0.8 k pX(k)

Geometric random variable p = 0.8

2 4 6 8 10 0.2 0.4 0.6 0.8 k pX(k)

Example: Coin flips

You flip a coin with bias p n times (flips are independent) If you model the number of heads as a random variable X, what is pX?

Example: Coin flips

What is the probability of getting k heads and then n − k tails? P (k heads, then n − k tails)

Example: Coin flips

What is the probability of getting k heads and then n − k tails? P (k heads, then n − k tails) = P (1st = heads, . . . , kth = heads, k + 1th = tails,. . . , nth = tails)

Example: Coin flips

What is the probability of getting k heads and then n − k tails? P (k heads, then n − k tails) = P (1st = heads, . . . , kth = heads, k + 1th = tails,. . . , nth = tails) = P (1st = heads) · · · P (kth = heads) P (k + 1th = tails) · · · P (nth = tails)

Example: Coin flips

What is the probability of getting k heads and then n − k tails? P (k heads, then n − k tails) = P (1st = heads, . . . , kth = heads, k + 1th = tails,. . . , nth = tails) = P (1st = heads) · · · P (kth = heads) P (k + 1th = tails) · · · P (nth = tails) = pk (1 − p)n−k

Example: Coin flips

Any fixed order of k heads and n − k tails has the same probability

Example: Coin flips

Any fixed order of k heads and n − k tails has the same probability We are interested in the union of these events

Example: Coin flips

Any fixed order of k heads and n − k tails has the same probability We are interested in the union of these events Can we just add their probabilities?

Example: Coin flips

Any fixed order of k heads and n − k tails has the same probability We are interested in the union of these events Can we just add their probabilities? How many possible orders are there?

Example: Coin flips

Any fixed order of k heads and n − k tails has the same probability We are interested in the union of these events Can we just add their probabilities? How many possible orders are there? n k

• :=

n! k! (n − k)!

Example: Coin flips

Any fixed order of k heads and n − k tails has the same probability We are interested in the union of these events Can we just add their probabilities? How many possible orders are there? n k

• :=

n! k! (n − k)! pX (k) = n k

• pk (1 − p)n−k

Binomial random variable

The pmf of a binomial random variable with parameters n and p is pX (k) = n k

• pk (1 − p)n−k ,

k = 0, 1, 2, . . . , n

Binomial random variable n = 20, p = 0.2

5 10 15 20 5 · 10−2 0.1 0.15 0.2 0.25 k pX(k)

Binomial random variable n = 20, p = 0.5

5 10 15 20 5 · 10−2 0.1 0.15 0.2 0.25 k pX(k)

Binomial random variable n = 20, p = 0.8

5 10 15 20 5 · 10−2 0.1 0.15 0.2 0.25 k pX(k)

Example: Call center

Model the number of calls received per day Assumptions:

• 1. Each call occurs independently from every other call
• 2. A given call has the same probability of occurring at any given time of

the day

• 3. Calls occur at a rate of λ calls per day

Example: Call center

◮ Discretize day into n slots

Example: Call center

◮ Discretize day into n slots ◮ Probability of receiving m calls in one slot?

Example: Call center

◮ Discretize day into n slots ◮ Probability of receiving m calls in one slot?

(λ/n)m

Example: Call center

◮ Discretize day into n slots ◮ Probability of receiving m calls in one slot?

(λ/n)m

◮ If n is large enough λ/n >> (λ/n)m for all m > 1

Example: Call center

◮ Discretize day into n slots ◮ Probability of receiving m calls in one slot?

(λ/n)m

◮ If n is large enough λ/n >> (λ/n)m for all m > 1 ◮ Assume that in each slot we either receive one call or none at all.

What is probability of k calls in a day?

Example: Call center

◮ Discretize day into n slots ◮ Probability of receiving m calls in one slot?

(λ/n)m

◮ If n is large enough λ/n >> (λ/n)m for all m > 1 ◮ Assume that in each slot we either receive one call or none at all.

What is probability of k calls in a day? Binomial with parameters n and p := λ/n!

Example: Call center

P (k calls during the day )

Example: Call center

P (k calls during the day ) = lim

n→∞ P (k calls in n small intervals)

Example: Call center

P (k calls during the day ) = lim

n→∞ P (k calls in n small intervals)

= lim

n→∞

n k

• pk (1 − p)(n−k)

Example: Call center

P (k calls during the day ) = lim

n→∞ P (k calls in n small intervals)

= lim

n→∞

n k

• pk (1 − p)(n−k)

= lim

n→∞

n k λ n k 1 − λ n (n−k)

Example: Call center

P (k calls during the day ) = lim

n→∞ P (k calls in n small intervals)

= lim

n→∞

n k

• pk (1 − p)(n−k)

= lim

n→∞

n k λ n k 1 − λ n (n−k) = lim

n→∞

n! λk k! (n − k)! (n − λ)k

• 1 − λ

n n

Example: Call center

P (k calls during the day ) = lim

n→∞ P (k calls in n small intervals)

= lim

n→∞

n k

• pk (1 − p)(n−k)

= lim

n→∞

n k λ n k 1 − λ n (n−k) = lim

n→∞

n! λk k! (n − k)! (n − λ)k

• 1 − λ

n n = λk e−λ k! Identity proved in the notes lim

n→∞

n! (n − k)! (n − λ)k

• 1 − λ

n n = e−λ

Poisson random variable

The pmf of a Poisson random variable with parameter λ is pX (k) = λk e−λ k! k = 0, 1, 2, . . .

Poisson random variable λ = 10

10 20 30 40 50 5 · 10−2 0.1 0.15 k pX(k)

Poisson random variable λ = 20

10 20 30 40 50 5 · 10−2 0.1 0.15 k pX(k)

Poisson random variable λ = 30

10 20 30 40 50 5 · 10−2 0.1 0.15 k pX(k)

Example: Call center

Pmf of binomial with parameters n and p = λ

n converges to pmf of

Poisson with parameter λ This is an example of convergence in distribution

Binomial random variable n = 40, p = 20

40

10 20 30 40 5 · 10−2 0.1 0.15 k

Binomial random variable n = 80, p = 20

80

10 20 30 40 5 · 10−2 0.1 0.15 k

Binomial random variable n = 400, p = 20

400

10 20 30 40 5 · 10−2 0.1 0.15 k

Poisson random variable λ = 20

10 20 30 40 5 · 10−2 0.1 0.15 k

Call-center data

◮ Assumptions do not hold over the whole day (why?) ◮ They do hold (approximately) for intervals of time ◮ Example: Data from a call center in Israel ◮ We compare the histogram of the number of calls received in an

interval of 4 hours over 2 months and the pmf of a Poisson random variable fitted to the data

Call-center data

5 10 15 20 25 30 35 40 Number of calls 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 Real data Poisson distribution

Discrete random variables Continuous random variables Conditioning on an event Functions of random variables

Continuous random variables

Useful to model continuous quantities without discretizing Assigning nonzero probabilities to events of the form {X = x} for x ∈ R doesn’t work! Instead, we only consider events of the form {X ∈ S} where S is a union of intervals (formally a Borel set) We cannot consider every possible subset of R for technical reasons

Cumulative distribution function

The cumulative distribution function (cdf) of X is defined as FX (x) := P ({X (ω) ≤ x : ω ∈ Ω}) = P (X ≤ x) In words, FX (x) is the probability of X being smaller than x The cdf can be defined for both continuous and discrete random variables

Cumulative distribution function

The cdf completely specifies the distribution of the random variable The probability of any interval (a, b] is given by P (a < X ≤ b) = P (X ≤ b) − P (X ≤ a) = FX (b) − FX (a) To define a continuous random variable we just need a valid cdf! A valid underlying probability space exists, but we don’t need to worry about it

Properties of the cdf

lim

x→−∞ FX (x) = 0,

lim

x→∞ FX (x) = 1,

FX (b) ≥ FX (a) if b > a, i.e. FX is nondecreasing

Probability density function

When the cdf is differentiable, its derivative can be interpreted as a density Probability density function fX (x) := dFX (x) d x The pdf is not a probability measure! (It can be greater than 1)

Probability density function

By the fundamental theorem of calculus P (a < X ≤ b) = FX (b) − FX (a) = b

a

fX (x) dx Intuitively, lim

∆→0 P (X ∈ (x, x + ∆)) = fX (x) ∆

Properties of the pdf

For any union of intervals (any Borel set) S P (X ∈ S) =

• S

fX (x) dx In particular, ∞

−∞

fX (x) dx = 1 From the monotonicity of the cdf fX (x) ≥ 0

Uniform random variable

Pdf of a uniform random variable with domain [a, b]: fX (x) =

• 1

b−a,

if a ≤ x ≤ b, 0,

• therwise

Uniform random variable in [a, b]

a b

1 b−a

x fX(x)

Exponential random variable

Used to model waiting times (time until a certain event occurs) Examples: decay of a radioactive particle, telephone call, mechanical failure of a device Pdf of an exponential random variable with parameter λ: fX (x) =

• λe−λx,

if x ≥ 0, 0,

• therwise

Exponential random variables

2 4 6 8 0.5 1 1.5 x fX (x) λ = 0.5 λ = 1.0 λ = 1.5

Call-center data

◮ Example: Data from a call center in Israel ◮ We compare the histogram of the inter-arrival times between calls

• ccurring between 8 pm and midnight over two days and the pdf of an

exponential random variable fitted to the data

Call center

1 2 3 4 5 6 7 8 9 Interarrival times (s) 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Exponential distribution Real data

Gaussian or normal random variable

Extremely popular in probabilistic models and statistics Sums of independent random variables converge to Gaussian distributions under certain assumptions Pdf of a Gaussian random variable with mean µ and standard deviation σ fX (x) = 1 √ 2πσ e− (x−µ)2

2σ2

Gaussian random variables

−10 −5 5 10 0.1 0.2 0.3 0.4 x fX (x) µ = 2 σ = 1 µ = 0 σ = 2 µ = 0 σ = 4

Height data

◮ Example: Data from a population of 25 000 people ◮ We compare the histogram of the heights and the pdf of a Gaussian

random variable fitted to the data

Height data

60 62 64 66 68 70 72 74 76

Height (inches)

0.05 0.10 0.15 0.20 0.25

Gaussian distribution Real data

Problem

◮ The Gaussian cdf does not have a closed form solution ◮ This complicates computing the probability that a Gaussian belongs to

a set

Standard Gaussian

If X is Gaussian with mean µ and standard deviation σ, then U := X − µ σ is a standard Gaussian, with mean zero and unit standard deviation P (X ∈ [a, b]) = P X − µ σ ∈ a − µ σ , b − µ σ

• = Φ

b − µ σ

• − Φ

a − µ σ

• Φ is the cdf of a standard Gaussian, which is tabulated

χ2 random variable

Very important in hypothesis testing If U1, U2, . . . , Ud are d independent standard Gaussian random variables X :=

d

• i=1

U2

i

is a χ2 with d degrees of freedom

χ2 random variable

The pdf of a χ2 random variable with d degrees of freedom is fX (x) = x

d 2 −1 exp

• − x

2

• 2

d 2 Γ

d

2

• if x > 0 and zero otherwise

χ2 random variables

5 10 15 20 25 0.1 0.2 0.3 0.4 0.5 x fX (x) d = 1 d = 5 d = 10

Discrete random variables Continuous random variables Conditioning on an event Functions of random variables

Conditioning on an event

We usually define random variables using their pmf, cdf or pdf How can we incorporate the information that X ∈ S for some set S?

Conditional pmf

If X has pmf pX, the conditional pmf of X given X ∈ S is pX|X∈S (x) := P (X = x|X ∈ S) =

• pX (x)
• s∈S pX (s)

if x ∈ S

• therwise.

Valid pmf in the new probability space restricted to the event {X ∈ S}

Conditional cdf

If X has pdf fX, the conditional cdf of X given X ∈ S is FX|X∈S (x) := P (X ≤ x|X ∈ S) = P (X ≤ x, X ∈ S) P (X ∈ S) =

• u≤x,u∈S fX (u) du
• u∈S fX (u) du

Valid cdf in the new probability space restricted to the event {X ∈ S}

Example: Geometric random variables are memoryless

We flip a coin repeatedly until we obtain heads, but pause after k0 flips (which were tails) What is the probability of obtaining heads in k more flips?

Example: Geometric random variables are memoryless

P (k more flips)

Example: Geometric random variables are memoryless

P (k more flips) = pX|X>k0 (k)

Example: Geometric random variables are memoryless

P (k more flips) = pX|X>k0 (k) = pX (k) ∞

m=k0+1 pX (m)

Example: Geometric random variables are memoryless

P (k more flips) = pX|X>k0 (k) = pX (k) ∞

m=k0+1 pX (m)

= (1 − p)k−1 p ∞

m=k0+1 (1 − p)m−1 p

Example: Geometric random variables are memoryless

P (k more flips) = pX|X>k0 (k) = pX (k) ∞

m=k0+1 pX (m)

= (1 − p)k−1 p ∞

m=k0+1 (1 − p)m−1 p

= (1 − p)k−k0−1 p for k > k0 Geometric series:

∞

• m=k0+1

αm−1 = αk0 1 − α for any α < 1

Example: Exponential random variables are memoryless

Assume email inter-arrival times are exponential with parameter λ You get an email, then no email for t0 minutes How is the waiting time until the next email distributed now?

Example: Exponential random variables are memoryless

FT|T>t0 (t)

Example: Exponential random variables are memoryless

FT|T>t0 (t) = t

t0 fT (u) du

∞

t0 fT (u) du

Example: Exponential random variables are memoryless

FT|T>t0 (t) = t

t0 fT (u) du

∞

t0 fT (u) du

= e−λt − e−λt0 −e−λt0

Example: Exponential random variables are memoryless

FT|T>t0 (t) = t

t0 fT (u) du

∞

t0 fT (u) du

= e−λt − e−λt0 −e−λt0 = 1 − e−λ(t−t0) for t > t0

Example: Exponential random variables are memoryless

FT|T>t0 (t) = t

t0 fT (u) du

∞

t0 fT (u) du

= e−λt − e−λt0 −e−λt0 = 1 − e−λ(t−t0) for t > t0 Differentiating with respect to t fT|T>t0 (t) = λe−λ(t−t0)

for t > t0

Discrete random variables Continuous random variables Conditioning on an event Functions of random variables

Functions of random variables

◮ For any deterministic function g and r.v. X, Y := g (X) is a random

variable

◮ Formally, X maps elements of Ω to R, so Y does too since

Y (ω) = g (X (ω))

Discrete random variables

If X is discrete pY (y) = P (Y = y) = P (g (X) = y) =

• {x | g(x)=y}

pX (x)

Continuous random variables

If X is continuous FY (y) = P (Y ≤ y) = P (g (X) ≤ y) =

• {x | g(x)≤y}

fX (x) dx, Then we can differentiate to obtain the pdf fY

Gaussian random variable

If X is a Gaussian random variable with mean µ and standard deviation σ, derive the distribution of U := X − µ σ

Gaussian random variable

FU (u)

Gaussian random variable

FU (u) = P X − µ σ ≤ u

Gaussian random variable

FU (u) = P X − µ σ ≤ u

• =
• (x−µ)/σ≤u

1 √ 2πσ e− (x−µ)2

2σ2

dx

Gaussian random variable

FU (u) = P X − µ σ ≤ u

• =
• (x−µ)/σ≤u

1 √ 2πσ e− (x−µ)2

2σ2

dx = u

−∞

1 √ 2π e− w2

2 dw

by the change of variables w = x − µ σ

Gaussian random variable

FU (u) = P X − µ σ ≤ u

• =
• (x−µ)/σ≤u

1 √ 2πσ e− (x−µ)2

2σ2

dx = u

−∞

1 √ 2π e− w2

2 dw

by the change of variables w = x − µ σ To obtain the pdf we differentiate with respect to u

Gaussian random variable

FU (u) = P X − µ σ ≤ u

• =
• (x−µ)/σ≤u

1 √ 2πσ e− (x−µ)2

2σ2

dx = u

−∞

1 √ 2π e− w2

2 dw

by the change of variables w = x − µ σ To obtain the pdf we differentiate with respect to u fU (u) = 1 √ 2π e− u2

2

Gaussian random variable

FU (u) = P X − µ σ ≤ u

• =
• (x−µ)/σ≤u

1 √ 2πσ e− (x−µ)2

2σ2

dx = u

−∞

1 √ 2π e− w2

2 dw

by the change of variables w = x − µ σ To obtain the pdf we differentiate with respect to u fU (u) = 1 √ 2π e− u2

2

U is a standard Gaussian random variable

χ2 with one degree of freedom

What is the pdf of a χ2 random variable X with one degree of freedom? Recall that X := U2 where U is a standard Gaussian random variable

χ2 with one degree of freedom

FX (x)

χ2 with one degree of freedom

FX (x) =

• u2≤x

fU (u) du

χ2 with one degree of freedom

FX (x) =

• u2≤x

fU (u) du = √x

−√x

1 √ 2π e− u2

2 du

χ2 with one degree of freedom

FX (x) =

• u2≤x

fU (u) du = √x

−√x

1 √ 2π e− u2

2 du

To obtain the pdf we differentiate with respect to x

χ2 with one degree of freedom

d dt h(t)

−∞

g (u) du

• = g (h (t)) h′ (t)

χ2 with one degree of freedom

d dt h(t)

−∞

g (u) du

• = g (h (t)) h′ (t)

fX (x) = d dx √x

−∞

1 √ 2π e− u2

2 du −

−√x

−∞

1 √ 2π e− u2

2 du

χ2 with one degree of freedom

d dt h(t)

−∞

g (u) du

• = g (h (t)) h′ (t)

fX (x) = d dx √x

−∞

1 √ 2π e− u2

2 du −

−√x

−∞

1 √ 2π e− u2

2 du

= 1 √ 2π 1 2√x exp

• −x

2

• +

1 2√x exp

• −x

2

χ2 with one degree of freedom

d dt h(t)

−∞

g (u) du

• = g (h (t)) h′ (t)

fX (x) = d dx √x

−∞

1 √ 2π e− u2

2 du −

−√x

−∞

1 √ 2π e− u2

2 du

= 1 √ 2π 1 2√x exp

• −x

2

• +

1 2√x exp

• −x

2

• =

1 √ 2πx exp

• −x

2