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Statistics

Notes The second cross-moment of X is computed by taking the second cross-partial derivative of the
joint moment generation function

 2 M X 1 ,X 2 (t ,t )     1 [1 exp(t  2t ) exp(2t  t )]   
1
2



 t t 2 = t  1   t  2   3 1 2 1 2   
1
  1 
=  [2 exp(t  2t ) exp(2t  t )]

2 
1
2
1
t  1  3 
1

= [2 exp(t  2t ) exp(2t  t )]
2
1
1
2
3
and evaluating it at (t , t ) = (0, 0):
1 2
 2 M , (t , t )
E[X X ] = X 1 X 2 1 2
1 2
 t t

1
2
1 t  0, 2 t  0
1
= [2 exp(0) 2exp(0)]
3
4
=
3
Therefore
Cov[X , X ] = E[X X ] – E[X ]E[X ]
1 2 1 2 1 2
4
=  1.1
3
1
=
3
13.2 Properties of Moment Generating Function
(a) The most significant propertyof moment generating function is that “the moment
generating function uniquely determines the distribution.”
(b) Let a and b be constants, and let MX(t) be the mgf of a random variable X. Then the mgf of
the random variable Y = a + bX can be given as follows
M (t) = E[e ] = E[e t(a + bX) ] = e E[e(bt)X] = eatMX(bt)
at
tY
Y
(c) Let X and Y be independent random variables having the respective mgf’s M (t) and M (t).
X Y
Recall that E[g (X)g (Y)] = E[g (X)]E[g (Y)] for functions g and g . We can obtain the mgf
1 2 1 2 1 2
Mz(t) of the sum Z = X + Y of random variables as follows.
(d) When t = 0, it clearly follows that M(0) = 1. Now by differentiating M(t)r times, we obtain
d r  d t tX 
r tX
tX
(r)
M (t) = E[e ] E  e   E[X e ]

dt r  dt r 
(r)
In particular when t = 0, M (0) generates the r-th moment of X as follows.
M (r)(0) = E[X ], r = 1, 2, 3,...
r
Example 4: Find the mgf M(t) for a uniform random variable X on [a, b]. And derive the
derivative M’(0) at t = 0 by using the definition of derivative and L’Hospital’s rule.
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