ENGG 2040C: Probability Models and Applications

Andrej Bogdanov

Spring 2013

5. Continuous Random Variables

Delivery time

A package is to be delivered between noon and 1pm.

When will it arrive?

Delivery time

A probability model

Sample space S1 = {0, 1, …, 59}

equally likely outcomes

Random variable X: minute when package arrivesX(0) = 0, X(1) = 1, …, X(59) = 59E[X] = 0⋅1/60 + … + 59⋅1/60 = 29.5

X(w) = w

Delivery time

A more precise probability model

equally likely outcomes

X: minute when package arrives

E[X] = 29.983…

S2 = {0, , , …, 1, 1 , …, 59 }

160

260

160

5960

Taking precision to the limit

S = the (continuous) interval [0, 60)

equally likely outcomes

p = 1/60

p = 1/3600

p = 0

S1 = {0, 1, …, 59}

S2 = {0, , …, 59 }

160

5960

S = [0, 60)

Uncountable sample spaces

“The probability of an event is the sum of the probabilities of its elements”

In Lecture 2 we said:

but in S = [0, 60) all elements have probability zero!

To specify and calculate probabilites, we have to work with the axioms of probability

The uniform random variable

Sample space S = [0, 60)

Events of interest:

intervals [x, y) ⊆ [0, 60)their intersections, unions, etc.

Probabilities:

P([x, y)) = (y – x)/60

Random variable:

X(w) = w

How to do calculations

You walk out of the apartment from 12:30 to 12:45. What is the probability you missed the delivery?SolutionEvent of interest: E = [30, 45) or E = “30 ≤ X < 45”

P(E) = (45 – 30)/60 = 1/4

E0 60

How to do calculations

From 12:08 - 12:12 and 12:54 - 12:57 the doorbell wasn’t working.

Event of interest: E = “8 ≤ X < 12” ∪ “54 ≤ X < 57”

0 60

P(E) = P([8, 12)) + P([54, 57)) = 4/60 + 3/60 = 7/60

Cumulative distribution function

The probability mass function doesn’t make much sense because P(X = x) = 0 for all x.

Instead, we can describe X by its cumulative distribution function (c.d.f.) F:

F(x) = P(X ≤ x)

The c.d.f. makes sense for discrete as well as continuous random variables.

Cumulative distribution functions

f(x) = P(X = x) F(x) = P(X ≤ x)

F(x)

x

Uniform random variable

If X is uniform over [0, 60) then

P(X ≤ x) =

0 60X ≤ x

x

x/60for x ∈ [0, 60)0

1 for x > 60

for x < 0

Cumulative distribution functions

p.m.f. f(x) = P(X = x) discretec.d.f. F(x) = P(X ≤ x)

continuousc.d.f. F(x) = P(X ≤ x)?

Discrete random variables:

p.m.f. f(x) = P(X = x) c.d.f. F(x) = P(X ≤ x)

F(a) = ∑x ≤ a f(x)f(x) = F(x) – F(x – d)for small d

Continuous random variables:The probability density function (p.d.f.) of a random variable with c.d.f. F(x) is

f(x) =F(x) – F(x – d)

dlim d → 0

dF(x)dx=

Discrete random variables:

p.m.f. f(x) = P(X = x) c.d.f. F(x) = P(X ≤ x)

F(a) = ∑x ≤ a f(x)

Continuous random variables:

p.d.f. f(x) = dF(x)/dx c.d.f. F(x) = P(X ≤ x)

F(a) = ∫x ≤ a f(x)dx

Uniform random variable

F(x) = x/60if x ∈ [0, 60)0

1 if x ≥ 60

if x < 0

c.d.f. F(x)

dF(x)/dx =1/60if x ∈ (0, 60)0

0 if x > 60

if x < 0

p.d.f. f(x)

1/60

Cumulative distribution functions

discretec.d.f. F(x) = P(X ≤ x)

continuousc.d.f. F(x) = P(X ≤ x)

p.m.f. f(x) = P(X = x)

p.d.f. f(x) = dF(x)/dx

Uniform random variable

A random variable X is Uniform(0, 1) if its p.d.f. isf(x) =

0if x ∈ (0, 1)1

if x < 0 or x > 1

f(x)

A Uniform(a, b) has p.d.f.

f(x) =0

if x ∈ (a, b)1/(b - a)if x < a or x > b

a b

X

Some practice

A package is to be delivered between noon and 1pm.

When will it arrive?

It is now 12.30 and the package is not in yet.

Some practice

Arrival time X is Uniform(0, 60)

Probability model

We want the c.d.f. of X conditioned on X > 30:

P(X ≤ x | X > 30)

=P(X ≤ x and X > 30)

P(X > 30)

=P(30 < X ≤ x)

P(X > 30)

(x – 30)/30==1/2

(x – 30)/60

Some practice

G(x) = (x – 30)/30

The c.d.f. of X conditioned on X > 30 is

The p.d.f. of X conditioned on X > 30 is

for x in [30, 60)

g(x) = dG(x)/dx = 1/30

for x in [30, 60)

and 0 outside.

So X conditioned on X > 30 is Uniform(30, 60).

Waiting for a friend

Your friend said she’ll show up between 7 and 8 but probably around 7.30.

It is now 7.30.

What is the probability you have to wait past 7.45?

Waiting for a friend

Let’s assume arrival time X has following p.d.f.:

Probability model

1/30

0 6030x

f(x)

We want to calculate

P(X > 45 | X > 30)

=P(X > 45)P(X > 30)

Waiting for a friend

1/30

0 6030x

f(x)

P(X > 30) = ∫30 f(x)dx

60 = 1/2

so P(X > 45 | X > 30) = (1/8)/(1/2) = 1/4.

45

P(X > 45) = ∫45 f(x)dx

60 = 1/8

1/60

x

Interpretation of the p.d.f.

The p.d.f. value f(x) d approximates the probability that X in an interval of length daround x

ExampleIf X is uniform, then

P(x ≤ X < x + d) = d/60

f(x) = 1/60

d

P(x ≤ X < x + d) = f(x) d + o(d)

P(x – d ≤ X < x) = f(x) d + o(d)

Discrete versus continuous

p.d.f. f(x)p.m.f. f(x)

∑x ≤ a f(x) ∫x ≤ a f(x)dx

E[X] ∑x x f(x) ∫x x f(x)dx

E[X2] ∑x x2 f(x) ∫x x2 f(x)dx

Var[X] E[(X – E[X])2] = E[X2] – E[X]2

P(X ≤ a)

Uniform random variable

A random variable X is Uniform(0, 1) if its p.d.f. isf(x) =

0if x ∈ (0, 1)1

if x < 0 or x > 1

E[X] = ∫0 x f(x)dx1

∫0 f(x)dx1 = ∫0 dx = 11

= x2/2|01 = 1/2

m

E[X2] = ∫0 x2 f(x)dx1 = x3/3|01 = 1/3

Var[X] = 1/3 – (1/2)2 = 1/12

m – s m + s

f(x)

x

m = E[X]s = √Var[X]

√Var[X] = 1/√12 ≈ 0.289

Uniform random variable

A random variable X is Uniform(a, b) if its p.d.f. is

f(x) =0

if x ∈ (a, b)1/(b - a)if x < a or x > b

Then

E[X] = (b – a)/2 Var[X] = (b – a)2/12

Rain is falling on your head at an average speed of l drops/second.

Raindrops again

1 2

How long do we wait until the next drop?

Raindrops again

Probability model

Time is divided into intervals of length 1/n

Events Ei = “raindrop hits in interval i” have probability p = l/n and are independent

X = interval of first drop

1 2

P(X = x)

= P(E1c…Ex-1

cEx)= (1 – p)x-1p

X

Raindrops again

X = interval of first drop

T = time (in seconds) of first drop

X = xP((x – 1)/n ≤ T < x/n) = P(X = x)

(x – 1)/n ≤ T < x/n

means that

= (1 – p)x-1p

= (1 – l/n)x-1(l/n)

X = x

x-1n

xnT

Raindrops again

T = time (in seconds) of first drop

P((x – 1)/n ≤ T < x/n) = (1 – l/n)x-

1(l/n)If we set t = (x – 1)/n and d = 1/n we get

P(t ≤ T < t + d) = (1 – d l)t/d(dl)

f(t) =d

lim d → 0

P(t ≤ T < t + d) = l e-lt

= dl e– ( d l + o( d l))

t/d

The exponential random variable

The p.d.f. of an Exponential(l) random variable T is

f(t) =l e-lt

0if x ≥ 0if x < 0.

p.d.f. f(t)

l = 1

c.d.f. F(t) = P(T ≤ t)

l = 1

The exponential random variable

The c.d.f. of T is

F(a) = ∫0 l e-lt dt = e-lt|0 = 1 – e-la a a if a ≥ 0

What should the expected value of T be?

(Hint: Rain falls at l drops/secondHow many seconds till the first

drop?)E[T] = 1/l Var[T] = 1/l2

Poisson vs. exponential

1 2

T

N

number of events

within time unit

time until first event happens

Exponential(l)Poisson(l)

description

expectation

1/ll

std. deviation

1/ll

Memoryless property

How much time between the second and third drop? 1 2

T

Solution

We start time when the second drop falls.

Then T is Exponential(l)

What happened before is irrelevant.

T

Expected time

What is the expected time of the third drop?

1 2

T3T2T1

Solution

T = T1 + T2 + T3

T is not exponential but

E[T] = E[T1] + E[T2] + E[T3]

= 3/l