Download Further Topics on Random Variables: Sum of a Random Number of

Further Topics on Random Variables:
Sum of a Random Number of Independent
Random Variables
Berlin Chen
Department of Computer Science & Information Engineering
National Taiwan Normal University
Reference:
- D. P. Bertsekas, J. N. Tsitsiklis, Introduction to Probability , Sections 4.2-3
Sum of a Random Number of Independent
Random Variables (1/4)
X 1 , X 2 , X 3 ,K , X
Y = X1 + X
2
+L + X
N
,K K K
N
• If we know that
– N is a random variable taking positive integer values N = 1 , 2 , K
– X 1 , X 2 , K are independent, identically distributed (i.i.d.)
random variables (with common mean μ and variance σ 2 )
• A subset of X i ' s ( X 1 , X 2 , L , X N ) are independent as well
• What are the formulas for the mean, variance, and the
transform of Y ?
Y = X1 + X
2
+L + X
N
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Sum of a Random Number of Independent
Random Variables (2/4)
• If we fix some number n , the random variable
X 1 + X 2 + L + X n is independent of random variable N
E [Y N = n ]
= E [X 1 + X
= E [X 1 + X
= E [X 1 + X
= n E [X
i
]=
2
+L + X
N
2
+L + X
n
2
+L + X
n
N = n]
N = n]
]
nμ
– E [Y N ] can be viewed as a function of random variable N
• E [Y N ] is a random variable
• The mean of E [Y N ] (i.e. E [Y ]) can be calculated by using
the law of iterated expectations
E [Y ] = E [E [Y N
]] = E [N μ ] =
μ E [N
]
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Sum of a Random Number of Independent
Random Variables (3/4)
• Similarly, var (Y N = n ) can be expressed as
(
var Y N = n
(
var (X 1 +
)
= var X 1 + X
2
+L + X
N
=
2
+L + X
n
2
+L + X
n
X
= var ( X 1 + X
= nσ
)
= n)
N = n
N
)
2
– var (Y N ) can be viewed as a function of random variable N
• var (Y N ) is a random variable
– The variance of Y can be calculated using the law of total
variance
var (Y
)=
E [var (Y N
[
= E Nσ
=σ
2
2
)] +
var (E [Y N
]+ var (N μ )
E [N ] + μ
2
var ( N
])
)
Probability-Berlin Chen 4
Sum of a Random Number of Independent
Random Variables (4/4)
• Similarly, E [e sY
[
N = n
]
E e sY N = n
= E e s (X 1 + X 2 +L + X
[
= E [e
= (M
–
[
] can be expressed as
] [
) N = n = E e s (X 1 + X 2 +L + X n ) N = n
s (X 1 + X 2 +L + X n )
= E e sX 1 e sX 2 L e sX n
N
] [
]
]
(s ))n
X
] can be viewed as a function of random variable N
E [e
N ] is a random variable
The mean of E [e N ] (i.e. the transform of Y , E [e ]
E e sY N
•
•
sY
sY
sY
)
can be calculated by using the law of iterated expectations
M
Y
(s ) =
[ ] = E [E [e
E e sY
sY
N
]] = E [(M
X
(s ))N
]=
∞
∑ (M
n =1
X
(s ))n p N (n )
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Properties of the Sum of a Random Number of
Independent Random Variables
⇒ E [Y ] = E [N ]E [ X
i
]
⇒ var (Y ) = E[N ] var ( X i ) + (E[X i ])2 var ( N )
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Illustrative Examples (1/5)
• Example 4.21. A remote village has three gas stations,
and each one of them is open on any given day with
probability 1/2, independently of the others. The amount
of gas available in each gas station is unknown and is
uniformly distributed between 0 and 1000 gallons.
– We wish to characterize the distribution ( Y ) of the total amount
of gas available at the gas stations that are open
Y = X1 +L
Total amount of gas available
2
The amount of gas provided by
one gas station, out of three
( X i is uniformly distributed)
The transform of X i (uniformll y
distribute d) is :
e1000 s − 1
M X (s ) =
1000 s
1
The number N of gas stations open at a day
is a binomial distributi on with parameter (3, p )
⇒ the transform of random variable N is
3
3
1
M N (s ) = 1 − p + pe s = 1 + e s
8
(
)
(
)
3
Using the property introduced in
the previous slide, we have
1 ⎛⎜ ⎛⎜ e1000 s − 1 ⎞⎟ ⎞⎟
M Y (s ) = 1 +
8 ⎜⎝ ⎜⎝ 1000 s ⎟⎠ ⎟⎠
cf. 8
3
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Illustrative Examples (2/5)
• Example 4.22. Sum of a Geometric Number of
Independent Exponential Random Variables.
– Jane visits a number of bookstores, looking for Great
Expectations. Any given bookstore carries the book with
probability p , independently of the others. In a typical
bookstore visited, Jane spends a random amount of time,
exponentially distributed with parameter λ , until she either finds
the book or she decides that the bookstore does not carry it.
Assuming that Jane will keep visiting bookstores until she buys
the book and that the time spent in each is independent of
everything else
– We wish to determine the mean, variance, and PDF of the total
time spent in bookstores.
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Illustrative Examples (3/5)
X
X1
1− p
p
X
2
1− p
n
p
p
found
Y = X1
Y = X1 + X 2
Total amount of time spent
The amount of time spent
in a given bookstore
1
⇒ E [Y ] = E [N ]E [ X
i ]=
found
found
1 1
⋅
p λ
Y = X1 + X 2 + L + X n
2
var (Y ) = E [N ] var ( X i ) + (E [ X i ])2 var ( N )
2
1 1 ⎛ 1 ⎞ 1− p
= ⋅ 2 +⎜ ⎟ ⋅ 2
p λ
p
⎝λ ⎠
1
= 2 2
λ p
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Illustrative Examples (4/5)
M X (s ) =
3
M N (s ) =
⇒ M Y (s ) =
λ
λ−s
pe s
1 − (1 − p )e s
p
λ
λ−s
1 − (1 − p )
λ
λ−s
=
pλ
pλ
=
λ − s − λ + pλ pλ − s
∴ Y is an exponentia lly distribute d random variable with parameter p λ
f Y ( y ) = p λ e − pλ y , y ≥ 0
Recall that if Y is the sum of a fixed number of independent
random variables (e.g., Y = X 1 + X 2 ),
its associated transform M Y (s ) is
(Assume that X 1, X 2 are identical exponential distributi ons
with parameter λ )
2
⎛ λ ⎞
M Y (s ) = ⎜
⎟
⎝λ−s⎠
⇒ Y is not an exponentia l random variable
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Illustrative Examples (5/5)
• Example 4.23. Sum of a Geometric Number of
Independent Geometric Random Variables.
– This example is a discrete counterpart of the preceding one.
– We let N be geometrically distributed with parameter p . We
also let each random variable X i be geometrically distributed
with parameter q . We assume that all of these random
variables are independent.
M X (s ) =
M N (s ) =
qe s
1 − (1 − q )e s
pe s
1 − (1 − p )e s
qe s
p
pqe s
pqe s
1 − (1 − q )e s
⇒ M Y (s ) =
=
=
s
s
s
qe
1 − (1 − q )e − (1 − p )qe
1 − (1 − pq )e s
1 − (1 − p )
1 − (1 − q )e s
∴ Y is a geometric distribute d random variable with parameter pq
Probability-Berlin Chen 11