Moment Generating Functions

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= e−
t t ( et )k = e − e e = e ( e −1) , for all t . k! k= 0


The Uniform MGF Let X ~ U[ a , b] . Then MX (t ) = =
Range X b a b 1 1 1 dx = × ex t b−a b−a t a
∞ xt xt 1 − x / e f ( t ) = e dx ∫ ∫e X 0
t − 1 < 0 so that t1/ ) 1 ∞ x ( t −1)/ e dx = dx ∫ ∫e
0 0
1
×
t−1
e x ( t −1)/
∞ 0
= 0−
( j) MX (t ) with respect to t is MX ( t) =
Range X
∫x e
Range X j xt
∫e
xt
f X ( x ) dx .
Then the j th derivative of
f X ( x ) dx , and thus
( j) MX (0) =
RangeX
.
2 t 2 /2
for all t . We then have
A special case is the standard normal distribution Z ~ N (0, 1) . MZ ( t ) = e for all t . Generating Moments
Let X be a random variable. The first moment of X is simply the expected value E[ X ] . In general for j ≥ 1 , the j th moment of X is E[ X j ] , which can be found by E[ X j ] =
n−1 Thus, E[ X ] = M ′ p=np X (0) = n ( p + q)
and E[ X 2 ] = MX ′′ (0) = n( n − 1) p 2 + np Thus, Var ( X ) = n( n − 1) p2 + np − ( np )2 = np − np2 = np (1 − p ) = n pq . Properties of MGF’s (i) MX + Y ( t ) = MX (t ) × MY ( t ) for independent random variables X and Y . (ii) MaX ( t ) = M X ( at ) for all constants a (iii) M( X + c ) ( t ) = e ct M X ( t ) for all constants c
Dr. Neal, WKU
Theorem. The moment generating function completely determines the distribution of a random variable. That is, if two random variable have the same mgf, then they have the same distribution. Proof. (For continuous distributions) Suppose that X and Y are continuous random variables with MX (t ) = MY (t ) . Then 1( −∞ , t] × M X ( t ) = 1( −∞ , t] × MY (t ) ; that is, 1( −∞ , t] × ∫ e xt f X ( x ) dx = 1(−∞ , t ] × ∫ e xt f Y ( x ) dx ,
k in RangeX
∑ e kt P ( X = k ) ,
and when X is a continuous random variable, then MX (t ) is obtained by MX (t ) = ∫ e x t f X ( x ) dx =
−∞ ∞ xt ∫ e f X ( x ) dx .
Dr. Neal, WKU
MATH 329
Moment Generating Functions
Definition . Let X be a random variable. The moment generating function (mgf) of X is the function MX : ℜ → ℜ given by MX (t ) = E [e X t ], defined for all t for which this expected value is finite. When X is a discrete random variable, then MX (t ) is obtained by MX (t ) =
j j ∫ x f X ( x ) dx = E [ X ] .
Thus, the j th moment of X can be found by computing the j th derivative of MX (t ) and then evaluating it at t = 0 . In particular, E[ X ] = M ′ X (0) E[ X 2 ] = MX ′′ (0)
dx e e dx
= e te = e te Thus, MX (t ) = e te
t 2 /2
×

∞ 2 1 e −( x −2 x ( + ∫ 2 π −∞
2 t ) + 2 )/(2 2 ) − t − 2 t2 /2
2 t 2 /2
× ∫ f Y ( x ) dx = e te
−∞
2 t 2 /2
k in Range X
∑ x j P( X = k)
E[ X j ] =
Range X
j ∫ x f X ( x ) dx
for X discrete
for X continuous.
But the mgf of X also can be used to obtain these moments. When X is discrete, then MX (t ) = ∑ e kt P ( X = k ) , and then the j th derivative of MX (t ) with respect to t is
k =1 ∞
we have MX (t ) = =
k in RangeX
∞ p ∞ ∑ e kt P ( X = k ) = ∑ e kt q k −1 p = ∑ ( qe t )k k =1
q k =1
p qe t pe t × = , for qe t < 1; i.e., for t < ln(1 / q). t t q 1 − qe 1 − qe

e xt
f X ( t ) = ∫ e xt
e bt − e at , for t ≠ 0. t( b − a ) The Exponential MGF
Let X ~ exp( ) with > 0 . We shall assume that integral below will converge. We then have MX (t ) = = =
Dr. Neal, WKU
The Poisson MGF Let X ~ Poi( ) . Because ∑ MX (t ) = xk = e x for all x , we have k =0 k!
k in RangeX ∞

e kt
P( X = k ) = ∑

e kt
k −
k=0
e k!
1 1 = , for t < 1/ t −1 1 − t
.
The Normal MGF Let X ~ N ( , f Y (x ) = = = ) . Then f X ( x ) = 1 2π 1 2π 1 2π e −( x − ( +
2 2 1 e − ( x − ) /(2 2π 2)
. Now let Y ~ N (
2
+ 2 t,
) with
t )) 2 /(2 2 )
=
2 1 e −( x − 2 x ( + 2π 2t + 4t 2) / ( 2 2 )
t ) + ( + 2 t) 2 )/(2 2 )
2 e −( x − 2 x ( + 2 e −( x − 2 x ( +
2 t ) + 2 +2 2
. .
n
n
n n = ∑ ( pe t )k qn − k = ( pe t + q )n , for all t . k k =0
A special case is the Bernoulli random variable X ~ b (1, p) for which MX (t ) = pe t + q . The Geometric MGF Let X ~ geo( p ) with p > 0 . Then 0 ≤ q < 1 and because ∑ x k = x / (1 − x ) for − 1 < x < 1 ,
2 Var ( X ) = M ′ ′ (0) − ( M ′ X X (0))
Example 1. Let X ~ b ( n, p) with MX (t ) = ( pe t + q )n , for all t . Then the first and second derivatives of MX (t ) are MX ′ (t ) = n ( pe t + q) n−1 × pe t and MX ′′ (t ) = n ( n − 1)( pe t + q )n − 2 × ( pe t )2 + n( pe t + q )n −1 pe t .