5. Continuous Random Variables
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Transcript of 5. Continuous Random Variables
ENGG 2040C: Probability Models and Applications
Andrej Bogdanov
Spring 2014
5. Continuous Random Variables
Delivery time
A package is to be delivered between noon and 1pm.
When will it arrive?
Delivery time
A probability modelSample 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 arrivesE[X] = 29.983…
S2 = {0, , , …, 1, 1 , …, 59 }
160
260 1
6059
60
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)
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)
dlimd → 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) = 0
if x ∈ (0, 1)1if 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 d around x
ExampleIf X is uniform, then
P(x ≤ X < x + d) = d/60
f(x) = 1/60d
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) = 0
if x ∈ (0, 1)1if 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) = 0if x ∈ (a, b)1/(b - a)if x < a or x > b
Then
E[X] = (a + b)/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 modelTime is divided into intervals of length 1/n
Events Ei = “raindrop hits in interval i” have probability p = l/n and are independentX = 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 getP(t ≤ T < t + d) = (1 – d l)t/d(dl)
f(t) =d
limd → 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)
descriptionexpectation
1/ll
std. deviation
1/ll
Memoryless property
How much time between the second and third drop? 1 2
T
SolutionWe 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
SolutionT = T1 + T2 + T3
T is not exponential butE[T] = E[T1] + E[T2] + E[T3]
= 3/l