1. 5. Continuous Random Variables

2. Delivery time
A package is to be delivered between noon and 1pm.
When will it arrive?

3. 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
X(w) = w
E[X] = 0⋅1/60 + … + 59⋅1/60 = 29.5

4. A more precise probability model
Delivery time
A more precise probability model
S2 = {0, , , …, 1, 1 , …, } 1 60 2 59
equally likely outcomes
X: minute when package arrives
E[X] = …

5. 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 59
p = 1/3600
S = [0, 60)
p = 0

6. 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

7. 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

8. How to do calculations Solution
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 60
P(E) = (45 – 30)/60 = 1/4

9. How to do calculations
From 12: :12 and 12: :57 the doorbell wasn’t working.
Event of interest: E = “8 ≤ X < 12” ∪ “54 ≤ X < 57” 60
P(E) = P([8, 12)) + P([54, 57)) = 4/60 + 3/60 = 7/60

10. 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.

11. Cumulative distribution functions
f(x) = P(X = x)
F(x) = P(X ≤ x)

12. Uniform random variable
If X is uniform over [0, 60) then 60
X ≤ x x
F(x) x
for x < 0
P(X ≤ x) =
x/60
for x ∈ [0, 60) 1
for x > 60

13. 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) ?

14. 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)
for small d
F(a) = ∑x ≤ a f(x)
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) d
lim
d → 0
dF(x) dx =

15. 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

16. Uniform random variable
if x < 0
F(x) =
x/60
if x ∈ [0, 60) 1
if x ≥ 60
c.d.f. F(x)
p.d.f. f(x)
1/60
if x < 0
dF(x)/dx =
1/60
if x ∈ (0, 60)
if x > 60

17. Cumulative distribution functions
discrete c.d.f. F(x) = P(X ≤ x)
p.m.f. f(x) = P(X = x)
continuous c.d.f. F(x) = P(X ≤ x)
p.d.f. f(x) = dF(x)/dx

18. Uniform random variable
A random variable X is Uniform(0, 1) if its p.d.f. is
f(x) =
if x ∈ (0, 1) 1
if x < 0 or x > 1
f(x)
A Uniform(a, b) has p.d.f.
f(x) =
if x ∈ (a, b)
1/(b - a)
if x < a or x > b X a b

19. Some practice
A package is to be delivered between noon and 1pm.
It is now and the package is not in yet.
When will it arrive?

20. 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 > 30)
P(X ≤ x | X > 30) =
P(30 < X ≤ x)
P(X > 30) =
1/2
(x – 30)/60
(x – 30)/30 =

21. 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).

22. 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?

23. Waiting for a friend Probability model
Let’s assume arrival time X has following p.d.f.: 60 30 x
f(x)
1/30
We want to calculate =
P(X > 45)
P(X > 30)
P(X > 45 | X > 30)

24. Waiting for a friend P(X > 30) = ∫30 f(x)dx 60 = 1/2
1/30 x 30 45 60
P(X > 30) = ∫30 f(x)dx 60
= 1/2
P(X > 45) = ∫45 f(x)dx 60
= 1/8
so P(X > 45 | X > 30) = (1/8)/(1/2) = 1/4.

25. 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 x d
If X is uniform, then
f(x) = 1/60
P(x ≤ X < x + d) = d/60

26. 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

27. Uniform random variable
A random variable X is Uniform(0, 1) if its p.d.f. is 1
if x ∈ (0, 1)
f(x) =
if x < 0 or x > 1
f(x) m
m – s
m + s
∫0 f(x)dx 1
= ∫0 dx = 1 1
E[X] = ∫0 x f(x)dx 1
= x2/2|0 1
= 1/2
E[X2] = ∫0 x2 f(x)dx 1
= x3/3|0 1
= 1/3
m = E[X]
s = √Var[X]
Var[X] = 1/3 – (1/2)2 = 1/12 x
√Var[X] = 1/√12 ≈ 0.289

28. Uniform random variable
A random variable X is Uniform(a, b) if its p.d.f. is
1/(b - a)
if x ∈ (a, b)
f(x) =
if x < a or x > b
Then
E[X] = (b – a)/2
Var[X] = (b – a)2/12

29. 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?

30. 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

31. Raindrops again X = interval of first drop
X = x
x-1 n x n T
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)

32. 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) = d
lim
d → 0
P(t ≤ T < t + d)
= l e-lt

33. The exponential random variable
The p.d.f. of an Exponential(l) random variable T is
l e-lt
if x ≥ 0
f(t) =
if x < 0.
p.d.f. f(t)
l = 1
c.d.f. F(t) = P(T ≤ t)
l = 1

34. 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
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

35. Poisson vs. exponential1 2 T
Poisson(l)
Exponential(l)
description
number of events within time unit
time until first event happens l
1/l
expectation l
1/l
std. deviation

36. Memoryless property Solution
How much time between the second and third drop? 1 2 T
Solution
We start time when the second drop falls.
What happened before is irrelevant.
Then T is Exponential(l)

37. Expected time Solution What is the expected time of the third drop?1 2 T T3 T2 T1
Solution
T = T1 + T2 + T3
T is not exponential but
E[T] = E[T1] + E[T2] + E[T3]
= 3/l