STAT 302
Intro to Probability
Alexandre Bouchard-Côté
www.stat.ubc.ca/~bouchard/courses/stat302-fa2019-20/
Announcement
• Extra office hour: Thursday, 3pm, ESB 4182
Variance
Ex. 40
Compare these 3 ‘games’
• X = ...
Same expectation
lose
$1,000
with
pr
1/2
•
(zero), but very
• win $1,000 with pr 1/2 different risks!
• Y = ...
• lose $1 with pr 1/2
• win $1 with pr 1/2
• Z = No amount is exchanged with pr 1
Def 13
Variance
• A measure of how ‘risky’ a r.v. is
• In other words how much does it deviate
from its mean, in expectation?
• New notation:
• Var[X] = E[ ( X - E[X] )
• Examples on the board...
2
]
Prop 11
Useful properties (*)
• Var(X) =
•
2
E[X ]
-
2
(E[X])
Var(aX+b) = a2 Var(X)
Ex. 40b
Example
• Value X of a volatile stock after 1 year:
• $1 with probability p = 1/3
• $0 with probability p = 2/3
• X ~ ???
A.
1/3
• Var[X] ? B. 2/3
C. 0
D. 2/9
Variance of Bern(p)
Linearity
Recall: Ex 35d
Where is ‘linearity’ useful?
Find:
E[S] = E[X1 + X2]
‘Brute force way’
B1 ~ Unif({0, 1/2, 1})
B2 ~ Unif({0, 1/2})
1/2
X1 | B1 ~ Bern(B1)
X2 | B2 ~ Bern(B2)
(B1 = 0)
(.., B2 = 1/2)
1/2
1/2
1/2
S = X1 + X2
1/2
1/2
1/3
1/3
1
(.., B2 = 0)
(B1 = 1/2)
1/2
(.., B2 = 0)
1/2
1/4
(.., B2 = 1/2)
1/2
* 1/6 +
(..., S = 0) * 1/12 +
(..., S = 0)
(..., S = 1)
* 1/12 +
(..., S = 0)
* 1/12 +
(..., S = 1)
* 1/12 +
(..., S = 0)
* 1/24 +
(..., S = 1)
* 1/12 +
(..., S = 2)
* 1/24 +
1/4
1/3
1/2
(B1 = 1)
(.., B2 = 0)
1/2
1
* 1/6 +
(..., S = 1) * 1/12 +
(..., S = 1)
1/2
(.., B2 = 1/2)
1/2
(..., S = 2) * 1/12
Recall: Ex 35d
Where is ‘linearity’ useful?
Find:
E[S] = E[X1 + X2]
B1 ~ Unif({0, 1/2, 1})
B2 ~ Unif({0, 1/2})
X1 | B1 ~ Bern(B1)
X2 | B2 ~ Bern(B2)
S = X1 + X2
Shortcut:
1) Compute E[X1]
2) Compute E[X2]
3) When can we write
E[X1 + X2] = E[X1] + E[X2] ??
Always!!
Independence NOT needed
Ex 35d - Shortcut way
Where is ‘linearity’ useful?
B1 ~ Unif({0, 1/2, 1})
B2 ~ Unif({0, 1/2})
1) Compute E[X1] = 1/2
X1 | B1 ~ Bern(B1)
X2 | B2 ~ Bern(B2)
S = X1 + X2
1
(B1 = 0)
(..., X1 = 0) 1/3
1/3
1/2
1/3
(B1 = 1/2)
(.., X1 = 0) 1/6
1/2
(.., X1 = 1) 1/6
1/3
1
(B1 = 1)
(.., X1 = 1) 1/3
Ex 35d - Shortcut way
Where is ‘linearity’ useful?
B1 ~ Unif({0, 1/2, 1})
B2 ~ Unif({0, 1/2})
X1 | B1 ~ Bern(B1)
X2 | B2 ~ Bern(B2)
1) Compute E[X1] = 1/2
2) Compute E[X2] = 1/4
S = X1 + X2
1/2
(B2 = 0)
1
1/2
1/2
(B2 = 1/2)
(..., X2 = 0) 1/2
(.., X2 = 0) 1/4
1/2
(.., X2 = 1) 1/4
Ex 35d - Shortcut way
Where is ‘linearity’ useful?
B1 ~ Unif({0, 1/2, 1})
B2 ~ Unif({0, 1/2})
1) Compute E[X1] = 1/2
X1 | B1 ~ Bern(B1)
X2 | B2 ~ Bern(B2)
2) Compute E[X2] = 1/4
S = X1 + X2
3) Use linearity!
E[S] = E[X1 + X2]
= E[X1] + E[X2]
= 3/4
Linearity for expectation
E[X1 + X2] = E[X1] + E[X2]
For all random variables X1 and X2
Linearity for variance?
Prop 10
Linearity for expectation
E[X1 + X2] = E[X1] + E[X2]
For all random variables X1 and X2
Linearity for variance
Var[X1 + X2] = Var[X1] + Var[X2]
If the random variables X1 and X2 are independent
Computing expectation
with ‘joint PMFs’
Computing E[Y] when Y = g(X1, X2)
Slower
Faster
g(x1, x2) = x1 + x2 ?
YES
E[X1+X2] = E[X1]+E[X2]
NO
A few other shortcuts that we will cover later
(ignore for now)
‘Joint PMF’
Two ‘default’
methods
(always work)
Covering this one
today
or
Ex 35d
(B1 = 0)
Method of Ex53d: find PMF of Y
using a decision tree
1/2
1/2
(.., B2 = 1/2)
1/3
1/2
1/2
1/2
1/3
1
(.., B2 = 0)
1/2
(B1 = 1/2)
1/2
(.., B2 = 0)
1/2
1/4
(.., B2 = 1/2)
1/2
(..., S = 0)
* 1/6 +
* 1/12 +
(..., S = 1)
* 1/12 +
(..., S = 0)
* 1/12 +
(..., S = 1)
* 1/12 +
(..., S = 0)
(..., S = 1)
* 1/24 +
* 1/12 +
(..., S = 2)
* 1/24 +
(..., S = 1)
* 1/6 +
* 1/12 +
(..., S = 0)
1/4
1/3
1/2
(B1 = 1)
(.., B2 = 0)
1/2
1
(..., S = 1)
1/2
(.., B2 = 1/2)
1/2
(..., S = 2) * 1/12
Prop 9 (version 2)
Expectation when Y=g(X1, X2)
• You know:
• P(X = x , X
• Y = g(X , X )
1
1
1
This is
called the
joint PMF,
p(x1, x2)
2
Definition
= x2)
E[Y] =
2
y ∈ Image(Y)
Second method:
E[Y] =
∑ ∑ g(x1,x2) * P(X1 = x1, X2 = x2)
Im
)
X2
e(
ag
1)
Im
(X
∈
x 2 age
∈
x1
Def 14
∑ y * P(Y = y)
One more shortcut
• If:
• g(x , x ) = x
, ie. we want the
expectation of the product, E[X1X2]
1
•X
1
E[X1X2] =
and X2 are independent
∑ ∑ x1 x2
x1
=
1 x2
2
x2
* P(X1 = x1, X2 = x2)
∑ x1 P(X1 = x1) ∑ x2
x1
= E[X1]E[X2]
x2
P(X2 = x2)
Geometric PMF
(‘distribution’)
Ex 45
Motivating example
• You need to hire 1 person
• You can only interview one random person
per business day
• This person will be qualified for the job and
hired with independent probability 2/3
• What is the probability you succeed in 3
business days or less?
A.2/3
B.2/27
C.26/27
D.26/31
Decision tree
1/3
1/3
2/3
Do not find
employee on
day 1
2/3
(X1 = 0)
Find employee
on day 1
(X1 = 1) 2/3
(T = 1)
1/3
Do not find
employee on
2/3
day 2
(.., X2 = 0)
Find employee
on day 2
(..., X2 = 1) 2/9
(T = 2)
Do not find
employee on
day 3
(.., X3 = 0)
1/3
2/3
Find employee
on day 3
(..., X3 = 1) 2/27
(T = 3)
Geometric PMF
This gives us a new PMF!
P (X = k) = (1
0.4
0.3
0.2
0.1
0.0
Approximation of P(Y = k)
0.5
0.6
Probability Mass Function (PMF)
4
6
8
k
1
p
k in 1, 2, 3, ...
What do we need to check?
2
p)k
10
12
14
Polya distribution
Also known as negative binomial
Ex 46
Motivating example
• You need to hire 3 people
• You can only interview one random person
per business day
• This person will be qualified for the job and
hired with independent probability 2/3
• What is the probability you succeed in
exactly 6 business days
If T’ = # of days,
compute P(T’ = 6)
Note: there are several ways (paths) to hire
3 people in exactly 6 days
#
people
#
people
days
days
P(T’=6) = P(X1=0, X2=0, X3=0, ..., X6=1) + P (X1=1, X2=0, X3=1 ..., X6=1) + ...
If T’ = # of days,
compute P(T’ = 6)
Is this a valid path in (T’ = 6) ?
#
people
days
NO: here T’ = 5 !
Probability of one path
#
people
• Chain rule
• Recall: Events A and B are independent if:
P(A ∩ B) = P(A) * P(B)
• Equivalent definition (when P(B) > 0): Events A
and B are independent if:
P(A | B) = P(A)
0)
3=
(X
* P 0)
2=
(X )
* P =0
X1
P(
P (X1=0, X2=0, X3=0 ..., X6=1) =
3 (1/3)3
(2/3)
=
...
days
If T’ = # of days,
compute P(T’ = 6)
P(T’=6) = P (X1=0, X2=0, X3=0 ..., X6=1) + P (X1=1, X2=0, X3=1 ..., X6=1) + ..
= number of paths * probability of each path
???
(2/3)3 (1/3)3
Counting
• How many scenarios lead to a 6 days
hiring process?
• Notes:
• Each scenario ends in X = 1
• Assign the remaining two 1’s among the
6
remaining five Xi’s
If T’ = # of days,
compute P(T’ = 6)
Note: there are several ways (paths) to hire
3 people in exactly 6 days
4 5
#
people
#
people
1
3
days
days
{4, 5}
{1, 3}
Polya PMF
This gives us a new PMF!
1 r
p (1
1
p)
i r
i in r, r+1, r+2, ...
0.05
0.10
0.15
0.20
0.25
0.30
Probability Mass Function (PMF)
0.00
i
r
◆
Approximation of P(Y = k)
P (X = i) =
✓
5
10
15
k
20