# 7. Joint moment generating functions and properties.

```University of California, Los Angeles
Department of Statistics
Statistics 100B
Instructor: Nicolas Christou
Joint moment generating functions

Let X =





X1
X2
..
.




,







be a random vector and let t =
t1
t2
..
.



.


The joint moment generating
tn
Xn
Pn
t0 X
function of X is defined as MX (t) = Ee = Eexp( i=1 ti xi ).
Theorem
2
2
X (t)
X (t)
Let Mi (t) = ∂M∂tXi(t) , Mii (t) = ∂ M
, and Mij (t) = ∂ ∂tMi ∂t
.
∂ti 2
j
2
Then, EXi = Mi (0), EXi = Mii (0), and EXi Xj = Mij (0).
Corollary
2 ψ (t)
X
, and ψij (t) =
Let ψ(t) = logMX (t), ψi (t) = ∂ψ∂tXi(t) , ψii (t) = ∂ ∂t
2
i
Then EXi = ψi (0), var(Xi ) = ψii (0), and cov(Xi Xj ) = ψij (0).
Theorem
!
∂ 2 ψX (t)
.
∂ti ∂tj
Y
Let X =
. The marginal moment generating function of Y (Z) is the moment genZ
erating function of X ignoring the vector!Z (Y). This is expressed as MY (u) = MX (u, 0)
u
and MZ (v) = MX (0, v), where t =
.
v
Proof
Theorem
If Y and Z are independent then MX (t) =MY (u)MZ (v).
Proof
1
Example
1

X1

X=
 X2  have joint moment generating function
X3
MX (t1 , t2 , t3 ) = (1 − t1 + 2t2 )−4 (1 − t1 + 3t3 )−3 (1 − t1 )−2 .
Use the corollary on page 1 to find:
a. E(X1 ), E(X2 ), E(X3 ).
b. var(X1 ), var(X2 ), var(X3 ).
c. cov(X1 , X2 ), cov(X1 , X3 ), cov(X2 , X3 ).
d. ρX1 ,X3 .
Example 2
Let X and Y be independent normal random variables, each with mean µ and standard
deviation σ.
a. Consider the random quantities X + Y and X − Y . Find the moment generating
function of X + Y and the moment generating function of X − Y .
c. Find the joint moment generating function of (X + Y, X − Y ).
d. Are X + Y and X − Y independent? Explain your answer using moment generating
functions.
2
```