Variance=lambda in poisson distribution

1. Expected value and variance of Poisson random variables. We said thatis the expected
value of a Poisson() random variable, but did not prove it. We did not (yet) say what
the variance was. For the expected value, we calculate, forXthat is a Poisson() random
variable:
E(X) =
1
X
x=0
x
e


x
x!
=
1
X
x=1
x
e


x
x!
since thex= 0 term is itself 0
=
1
X
x=1
e


x
(x1)!
divided on top and bottom byx
=e

1
X
x=1

x1
(x1)!
factor oute

andtoo
=e



0
0!
+

1
1!
+

2
2!
+: : :

=e

1
X
x=0

x
x!
=e

e

=
So in summaryE(X) =. For Var(X) =E(X
2
)(E(X))
2
=E((X)(X1) +X)
(E(X))
2
=E((X)(X1)) +E(X)(E(X))
2
=E((X)(X1)) +
2
. Now we calculate
E((X)(X1)) =
1
X
x=0
(x)(x1)
e


x
x!
=
1
X
x=2
(x)(x1)
e


x
x!
becausex= 0 andx= 1 terms are themselves 0
=
1
X
x=2
e


x
(x2)!
divide out byxandx1
=
2
e

1
X
x=2

x2
(x2)!
factor oute

and
2
=
2
e



0
0!
+

1
1!
+

2
2!
+: : :

(I had extrae

in the video on this line)
=
2
e

e

=
2
In summary, Var(X) =
2
+
2
=.
So both the expected value and the variance ofXare equal to.
1