Transcript Week4-3
Computing probabilities by conditioning
Let E denote some event. Define a random variable X by
1, if E occurs
X
0, if E does not occur
E[ X ] P( E )
Computing probabilities by conditioning
Let E denote some event. Define a random variable X by
1, if E occurs
X
0, if E does not occur
E[ X ] P( E )
E[ X | Y y ] xP ( X x | Y y ) P ( X 1| Y y ) P ( E | Y y )
x
Computing probabilities by conditioning
Let E denote some event. Define a random variable X by
1, if E occurs
X
0, if E does not occur
E[ X ] P ( E )
E[ X | Y y ] xP ( X x | Y y ) P ( X 1| Y y ) P ( E | Y y )
x
P ( E ) E[ X ] E[ E[ X | Y y ]]
= P ( E | Y y )P (Y y ) if Y is discrete
y
P ( E | Y y ) fY ( y )dy if Y is continuous
Example 1: Let X and Y be two independent continuous
random variables with densities fX and fY. What is P(X<Y)?
Example 1: Let X and Y be two independent continuous
random variables with densities fX and fY. What is P(X<Y)?
P ( X Y ) P ( X Y | Y y ) fY ( y )dy
Example 1: Let X and Y be two independent continuous
random variables with densities fX and fY. What is P(X<Y)?
P( X Y ) P( X Y | Y y ) fY ( y )dy
P( X y) fY ( y) dy
Example 1: Let X and Y be two independent continuous
random variables with densities fX and fY. What is P(X<Y)?
P ( X Y ) P ( X Y | Y y ) fY ( y )dy
P( X y) fY ( y) dy
FX ( y) fY ( y) dy
Example 2: Let X and Y be two independent continuous
random variables with densities fX and fY. What is the
distribution of X+Y?
Example 2: Let X and Y be two independent continuous
random variables with densities fX and fY. What is the
distribution of X+Y?
FX Y (a ) P ( X Y a )
Example 2: Let X and Y be two independent continuous
random variables with densities fX and fY. What is the
distribution of X+Y?
FX Y (a) P( X Y a) P( X Y a | Y y ) fY ( y )dy
Example 2: Let X and Y be two independent continuous
random variables with densities fX and fY. What is the
distribution of X+Y?
FX Y (a ) P ( X Y a ) P ( X Y a | Y y ) f Y ( y )dy
P( X y a ) fY ( y )dy
Example 2: Let X and Y be two independent continuous
random variables with densities fX and fY. What is the
distribution of X+Y?
FX Y (a ) P ( X Y a ) P ( X Y a | Y y ) f Y ( y )dy
P( X y a ) fY ( y )dy
P( X a y ) fY ( y )dy
Example 2: Let X and Y be two independent continuous
random variables with densities fX and fY. What is the
distribution of X+Y?
FX Y (a) P( X Y a ) P ( X Y a | Y y ) f Y ( y )dy
P( X y a ) fY ( y )dy
P( X a y ) fY ( y )dy
FX (a y ) fY ( y )dy
Example 3: (Thinning of a Poisson) Suppose X ~Poisson(l)
and {Ui} are i.i.d. Bernoulli(p) independent of X.
What is the distribution of Y i 1U i ?
X
n k
For n 0 and k n, P(Y k | X n) p (1 p ) n k
k
l n
n k
n k
nk e l
P(Y k ) n k p (1 p) P( X n) n k p (1 p)
n!
k
k
l n
l
k
n!
e
l
e
(
l
p
)
P(Y k ) n k
p k (1 p) n k
k !(n k )!
n!
k!
1
nk
l
(1
p
)
)
nk (n k )! (
e l (l p)k l (1 p ) e l p (l p) k
P(Y k )
e
Y is Poisson(l p)
k!
k!
Stochastic Processes
A stochastic process {X(t), t T} is collection of
random variables
Stochastic Processes
A stochastic process {X(t), t T} is collection of
random variables
For each value of t, there is a corresponding random
variable X(t) (state of the system at time t)
Stochastic Processes
A stochastic process {X(t), t T} is collection of
random variables
For each value of t, there is a corresponding random
variable X(t) (state of the system at time t)
When t takes on discrete values (e.g., t = 1, 2, ...)
discrete time stochastic process (the notation Xn is often
used instead, n = 1, 2, ...)
Stochastic Processes
A stochastic process {X(t), t T} is collection of
random variables
For each value of t, there is a corresponding random
variable X(t) (state of the system at time t)
When t takes on discrete values (e.g., t = 1, 2, ...)
discrete time stochastic process (the notation Xn is often
used instead, n = 1, 2, ...)
When t takes on continuous values continuous
time stochastic process
Example 1: X(t) is the number of customers waiting in
line at time t to check their luggage at an airline counter
(continuous stochastic process)
Example 1: X(t) is the number of customers waiting in
line at time t to check their luggage at an airline counter
(continuous stochastic process)
Example 2: Xn is the number of laptops a computer
store sells in week n.
Example 1: X(t) is the number of customers waiting in
line at time t to check their luggage at an airline counter
(continuous stochastic process)
Example 2: Xn is the number of laptops a computer
store sells in week n.
Example 3: Xn = 1 if it rains on the nth day of the
month and Xn = 0 otherwise.
Example 4: A gambler wins $1 with probability p, loses $1
with probability 1-p. She starts with $N and quits if she reaches
either $M or $0. Xn is the amount of money the gambler has after
playing n rounds.
Example 4: A gambler wins $1 with probability p, loses $1
with probability 1-p. She starts with $N and quits if she reaches
either $M or $0. Xn is the amount of money the gambler has after
playing n rounds.
P(Xn=i+1|Xn-1 =i, Xn-2 =i-1, ..., X0 =N}=P(Xn =i+1|Xn-1 =i}=p
(i≠0, M)
P(Xn=i-1| Xn-1 =i, Xn-2 =i-1, ..., X0 =N} = P(Xn =i-1|Xn-1 =i}=1–p
(i≠0, M)
Pi, i+1=P(Xn=i+1|Xn-1 =i}; Pi, i-1=P(Xn=i-1|Xn-1 =i}
Pi, i+1= p; Pi, i-1=1-p for i≠0, M
P0,0= 1; PM, M=1 for i≠0, M (0 and M are called
absorbing states)
Pi, j= 0, otherwise
Markov Chains
{Xn: n =0, 1, 2, ...} is a discrete time stochastic process
If Xn = i the process is said to be in state i at time n
{i: i=0, 1, 2, ...} is the state space
If P(Xn+1 =j|Xn =i, Xn-1 =in-1, ..., X0 =i0}=P(Xn+1 =j|Xn =i} = Pij,
the process is said to be a Discrete Time Markov Chain
(DTMC).
Pij is the transition probability from state i to state j
Pij 0,
P00
P
10
.
P .
Pi 0
.
.
i, j 0
P01
P11
.
.
Pi1
.
.
P02
P12
.
.
Pi 2
.
.
...
...
.
.
...
.
.
P: transition matrix
j 0
Pij 1,
i 0,1,...
Example 1: Probability it will rain tomorrow depends only
on whether it rains today or not:
P(rain tomorrow|rain today) = a
P(rain tomorrow|no rain today) = b
State 0 = rain
State 1 = no rain
a 1 a
P
b
1
b
Example 2 (random walk): A Markov chain whose state
space is 0, 1, 2, ..., and Pi,i+1= p = 1 - Pi,i-1 for i=0, 1, 2, ...,
and 0 < p < 1 is said to be a random walk.
Defining a DTMC
To define a DTMC, we need
Specify the states
Demonstrate the Markov property
Obtain the stationary probability transition matrix P