Some applications of Dirac's delta function in Statistics for more than ...

52 Santanu Chakraborty
Then,
∞
∫
−∞
( n)
n
( n)
n ( n)
ψ ( 0)
= ( −1)
ψ ( x) δ ( x)
dx = ( −1)
〈 ψ , δ 〉 .
These two steps give us the relation (4).
When we move to the two-dimensional scenario, we shall have to use Taylor's series expansion
for a two-dimensional analytic function ψ about the point (0,0) which is given by
1 ∂ ∂ n
ψ ( x,
y)
= ∑ [ x + y ] ψ ( x,
y)
|
0≤
n<
∞ n!
∂x
∂y
1
= ∑
∂
x=
0,
y=
0
∞ ∞
n
n
k n−k
∑ Ck
ψ ( x,
y)
|
k n k
x=
y=
x y
−
0,
0
(5)
n=
0 n!
k = 0 ∂x
∂y
Now, here also, we follow the same technique and so we compute 〈 f , ψ 〉 which is defined as
∞ ∞
∫ ∫
〈 f , ψ 〉 = f ( x,
y)
ψ ( x,
y)
dxdy
−∞−∞
Now using Taylor’s series expansion of ψ ( x,
y)
about ( 0 , 0 ) y x from (5) and assuming that
interchange of integrals with summations permissible, we get
∞ ∞
1
∂
〈 f , ψ 〉 = ∫ ∫ f ( x,
y)[
∑ ≤n<
∞ ∑
ψ ( x,
y)
|
0 n!
=
=
−∞−∞
Now we define the
∞ ∞
n
1 ∂
k!
( n − k)!
∂x
∂y
n
n
C
0≤k
≤n
k k n−k
∂x
∂y
∑ ∑ ψ ( x,
y)
|
0≤n< ∞ 0≤k ≤n
k n−k
x=
0 , y=
0 ∫ ∫
∑ ∑
0≤n<
∞
th
(,)
0≤k
≤n
1
k!
( n − k)!
∂
k
∂x
∂y
x=
0,
y=
0
∞ ∞
−∞−∞
x
k
y
f ( x,
y)
x
n−k
n
k n−k
ψ ( x,
y)
|
−
x=
y=
〈 f x y 〉
n k
0,
0 , .
rs order non-central moment for the pair ( X , Y ) as
r s
r s
µ f ( x,
y)
x y dxdy = 〈 f , x y 〉 .
r,
s
= ∫ ∫
−∞−∞
Then from above,
n
1 ∂
〈 f , ψ 〉 = ∑ ≤ < ∞∑
ψ ( x,
y)
|
0 n 0 ≤k
≤n
k n−k
x=
0,
y=
0 µ k , n−k
k!
( n − k)!
∂x
∂y
1
n ( k ) ( n−k
)
= 〈 ∑ ∑
µ −
〉
≤ < ∞ ≤ ≤
− ( 1)
δ ( ) δ ( ) , ψ
0 n 0 k n
k , n k x y .
k!
( n − k)!
k
y
] dxdy
n−k
dxdy