Variance Stabilization:
(a) Introduction:
Suppose
Yi   0   1 X i1   2 X i 2     p 1 X ip1   i   i   i ,
where  i   0   1 X i1   2 X i 2     p 1 X ip1 . Note that in standard linear
regression,
E (Yi )   i , Var (Yi )   2 .
That is  i is independent of  2 .
However, Yi might not be normally distributed for some type of data. Thus,  i and
 2 might be dependent.
Example:
Yi is distributed as Poisson random variable, Yi ~ Poisson (i ). Thus,
Var(Yi )   i2  E(Yi )  i   0  1 X i1   2 X i 2     p1 X ip1 .
Example:
Yi is distributed as binomial random variable, Yi ~ Binomial (1,i ). Thus,
Var (Yi )   i2  i (1  i ) .
In the above two examples,  i2 ’s are dependent on  i . Also, the variance  i2 ’s for
every observation are different.
(b) Derivation of the transformation:
Objective: find some transformations to stabilize the variance (make all variance
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 i2 equal). Then, the standard least square method can be employed.
Suppose Var (Yi )   i2  f 2 ( i ) and the transformation to stabilize the variance
is h(Yi ) . Thus, by Taylor’s first order approximation,
h(Yi )  h( i )  h ' ( i )(Yi   i ) .
Furthermore,

Varh(Yi )  Varh(Yi )  h(i )  Var h' (i )(Yi  i )






 h' (i ) Var (Yi )  h' (i ) f 2 (i )  h' (i ) f (i )
2
2

2
So, if
2
 1

1
h ' ( x) 
, Varh(Yi )  
f ( i )  1 .
f ( x)
 f ( i )

Result:
for the observation Yi with E (Yi )  i and Var (Yi )   i2  f 2 ( i ) ,
then the transformation h(Yi ) satisfying
h ' ( x) 
h( x)
1

,
x
f ( x)
is the transformation to stabilize the variance.
Example:
Yi ~ Poisson (i ).
 h ' ( x) 
 Var (Yi )   i  f 2 ( i )  f ( x)  x1 / 2 .
1
1
 1/ 2
f ( x) x
 h( x)   h ( x)dx  
'
1
x1 / 2
dx   x
Therefore,
2
1 / 2
dx  x
1/ 2
.
h(Yi )  Yi1 / 2
can be used to stabilize the variance.
Example:
Yi ~ Binomial (1,i ).
 Var (Yi )   i (1   i )  f 2 ( i )  f ( x)  x1 / 2 (1  x)1 / 2 .
1
1
 1/ 2
f ( x) x (1  x)1 / 2
.
1
1 1 / 2
 h( x)   h ' ( x)dx   1 / 2
d

sin
x
x (1  x)1 / 2
 h ' ( x) 
Therefore,
h(Yi )  sin 1 (Yi1 / 2 )
can be used to stabilize the variance.
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