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so
E
X
2
=
E
(
X
2

X
) +
E
X
=
λ
2
+
λ
, and hence Var
X
=
λ
.
An example: Suppose on average there are 5 homicides per month in a
given city. What is the probability there will be at most 1 in a certain month?
Answer: If
X
is the number of homicides, we are given that
E
X
= 5.
Since the expectation for a Poisson is
λ
, then
λ
= 5. Therefore
P
(
X
=
0) +
P
(
X
= 1) =
e

5
+ 5
e

5
.
An example. Suppose on average there is one large earthquake per year
in California. What’s the probability that next year there will be exactly 2
large earthquakes?
Answer:
λ
=
E
X
= 1, so
P
(
X
= 2) =
e

1
(
1
2
).
We have the following proposition.
Proposition 6.1
If
X
n
is binomial with parameters
n
and
p
n
and
np
n
→
λ
,
then
P
(
X
n
=
i
)
→
P
(
Y
=
i
)
, where
Y
is Poisson with parameter
λ
.
Proof.
For simplicity, let us suppose
λ
=
np
n
. In the general case we use
λ
n
=
np
n
. We write
P
(
X
n
=
i
) =
n
!
i
!(
n

i
)!
p
i
n
(1

p
n
)
n

i
=
n
(
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This note was uploaded on 12/29/2011 for the course MATH 316 taught by Professor Ansan during the Spring '10 term at SUNY Stony Brook.
 Spring '10
 ansan
 Probability

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