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ENGG 2040C: Probability Models and Applications
Spring 2014
5. Continuous Random
Variables
Andrej Bogdanov
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 arrives
X(0) = 0, X(1) = 1, …, X(59) = 59
E[X] = 0⋅1/60 + … + 59⋅1/60 = 29.5
X(w) = w
Delivery time
A more precise probability model
S2 = {0, 1 60 ,2 60 , …, 1, 11 60 , …, 595960}
equally likely outcomes
X: minute when package arrives
E[X] = 29.983…
Taking precision to the limit
S = the (continuous) interval [0, 60)
equally likely outcomes
S1 = {0, 1, …, 59}
p = 1/60
S2 = {0, 1 60 , …, 595960 }
p = 1/3600
S = [0, 60)
p=0
Uncountable sample spaces
In Lecture 2 we said:
“The probability of an event is the sum of the
probabilities of its elements”
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?
Solution
Event of interest: E = [30, 45) or E = “30 ≤ X < 45”
E
0
P(E) = (45 – 30)/60 = 1/4
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)
Cumulative distribution functions
f(x) = P(X = x)
F(x) = P(X ≤ x)
Uniform random variable
If X is uniform over [0, 60) then
X≤x
0
P(X ≤ x) =
x
60
F(x)
for x < 0
0
x/60 for x ∈ [0, 60)
for x > 60
1
x
Cumulative distribution functions
p.m.f. f(x) = P(X = x)
discrete
c.d.f. F(x) = P(X ≤ x)
?
continuous
c.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(x) = F(x) – F(x – d)
F(a) = ∑x ≤ a f(x)
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) = lim
d→0
F(x) – F(x – d)
d
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) =
if x < 0
0
x/60 if x ∈ [0, 60)
if x ≥ 60
1
c.d.f. F(x)
1/60
if x < 0
0
dF(x)/dx = 1/60 if x ∈ (0, 60)
if x > 60
0
p.d.f. f(x)
Cumulative distribution functions
p.m.f. f(x) = P(X = x)
discrete
c.d.f. F(x) = P(X ≤ x)
p.d.f. f(x) = dF(x)/dx
continuous
c.d.f. F(x) = P(X ≤ x)
Uniform random variable
A random variable X is Uniform(0, 1) if its p.d.f. is
1 if x ∈ (0, 1)
0 if x < 0 or x > 1
f(x) =
f(x)
A Uniform(a, b) has p.d.f.
1/(b - a) if x ∈ (a, b)
0
if x < a or x > b
f(x) =
X
a
b
Some practice
A package is to be delivered
between noon and 1pm.
It is now 12.30 and the
package is not in yet.
When will it arrive?
Some practice
Probability model
Arrival time X is Uniform(0, 60)
We want the c.d.f. of X conditioned on X > 30:
P(X ≤ x and X > 30)
P(X ≤ x | X > 30) =
P(X > 30)
P(30 < X ≤ x)
=
P(X > 30)
(x – 30)/60
=
= (x – 30)/30
1/2
Some practice
The c.d.f. of X conditioned on X > 30 is
G(x) = (x – 30)/30
for x in [30, 60)
The p.d.f. of X conditioned on X > 30 is
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
Probability model
Let’s assume arrival time X has following p.d.f.:
f(x)
1/30
0
30
60
x
We want to calculate
P(X > 45)
P(X > 45 | X > 30) =
P(X > 30)
Waiting for a friend
f(x)
1/30
0
P(X > 30) =
60
∫30
f(x)dx = 1/2
P(X > 45) =
60
∫45
f(x)dx = 1/8
30
45
60
x
so P(X > 45 | X > 30) = (1/8)/(1/2) = 1/4.
Interpretation of the p.d.f.
The p.d.f. value f(x) d approximates the probability
that X in an interval of length d around x
P(x – d ≤ X < x) = f(x) d + o(d)
P(x ≤ X < x + d) = f(x) d + o(d)
Example
1/60
If X is uniform, then
f(x) = 1/60
P(x ≤ X < x + d) = d/60
d
x
Discrete versus continuous
p.m.f. f(x)
p.d.f. f(x)
P(X ≤ a)
∑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
Uniform random variable
A random variable X is Uniform(0, 1) if its p.d.f. is
f(x) =
1
1 if x  (0, 1)
0 if x < 0 or x > 1
m
f(x)
1
∫0 f(x)dx = ∫0 dx = 1
m–s m+s
1
1
E[X] = ∫0 x f(x)dx = x2/2|0 = 1/2
E[X2]
=
1
∫0 x2
1
f(x)dx = x3/3|0 = 1/3
Var[X] = 1/3 – (1/2)2 = 1/12
√Var[X] = 1/√12 ≈ 0.289
m = E[X]
s = √Var[X]
x
Uniform random variable
A random variable X is Uniform(a, b) if its p.d.f. is
f(x) =
1/(b - a) if x  (a, b)
0
if x < a or x > b
Then
E[X] = (a + b)/2
Var[X] = (b – a)2/12
Raindrops again
Rain is falling on your head at an average speed of l
drops/second.
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
X
1
2
P(X = x) = P(E1c…Ex-1cEx) = (1 – p)x-1p
Raindrops again
X = interval of first drop
X=x
T
x-1
n
x
n
T = time (in seconds) of first drop
X = x means that
(x – 1)/n ≤ T < x/n
P((x – 1)/n ≤ T < x/n) = P(X = x)
= (1 – p)x-1p
= (1 – l/n)x-1(l/n)
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) = dl e– (d l + o(d l)) t/d
f(t) = lim
d→0
P(t ≤ T < t + d)
d
= l e-lt
The exponential random variable
The p.d.f. of an Exponential(l) random variable T is
f(t) =
l e-lt if x ≥ 0
0
l=1
p.d.f. f(t)
if x < 0.
l=1
c.d.f. F(t) = P(T ≤ t)
The exponential random variable
The c.d.f. of T is
F(a) =
a
∫0 l e-lt dt
a
= e-lt|0 = 1 – e-la
if a ≥ 0
What should the expected value of T be?
(Hint: Rain falls at l drops/second
How many seconds till the first drop?)
E[T] = 1/l
Var[T] = 1/l2
Poisson vs. exponential
N
2
1
T
Poisson(l)
Exponential(l)
description
number of events
within time unit
time until first
event happens
expectation
l
1/l
std. deviation
l
1/l
Memoryless property
How much time between the second and third drop?
2
1
T
Solution
We start time when the second drop falls.
What happened before is irrelevant.
Then T is Exponential(l)
Expected time
What is the expected time of the third drop?
2
1
T
T1
T2
T3
Solution
T = T1 + T2 + T3
T is not exponential but
E[T] = E[T1] + E[T2] + E[T3] = 3/l