Functions of Random Variables

=

Any image processing operation changes the signal g at the individual pixels. In the simplest case, g at each pixel is transformed into g' by a function p: g' p(g). Such a function is known in image processing as a point operator. Because g is a RV, g' will also be a RV and we need to know its PDF in order to know the statistical properties of the image after processing it.

It is obvious that the PDF f_g' of g' has same form as the PDF f_g of g

if p is a linear function: g' = p₀ + p₁g:

p1

f_g'(g') = f g(g) =

..

f_g((g' − p₀)/p₁),                                  (3.7)

p1

..

 

=              =  −

where the inverse linear relation g p− 1(g'): g (g' p₀)/p₁ is used to express g as a function of g'.

, = ±

From Eq. (3.7) it is intuitive that in the general case of a nonlinear function p(g), the slope p₁ will be replaced by the derivative p'(g_p) of p(g_p). Further complications arise if the inverse function has more than one branch. A simple and important example is the function g' = g2 with the two inverse functions g_{1, 2} g'. In such a case, the PDF of g' needs to be added from all branches of the inverse function.

=

Theorem 9 (PDF of the function of a random variable) If f_g is the PDF of the random variable g and p a diff erentiable function g' p(g), then the PDF of the random variable g' is given by

P f_g(g_p)

.

p

f_g'(g') =..p'(g

,                                    (3.8)

.

).

 

where g_p

p=1

are the P real roots of g’ = p(g).

A monotonic function p has a unique inverse function p− 1(g'). There- fore Eq. (3.8) reduces to

f_g'(g') =

f_g(p− 1(g'))

. ' −        .

p (p 1(g'))

 

.                                 (3.9)

In image processing, the following problem is encountered with re- spect to probability distributions. We have a signal g with a certain PDF and want to transform g by a suitable transform into g' in such a way that g' has a specifi c probability distribution. This is the inverse prob- lem to what we have discussed so far and it has a surprisingly simple solution. The transform

'

g

g' = F_g− 1(F

(g))                                           (3.10)

82                                                                         3 Random Variables and Fields

 

=

converts the f_g(g)-distributed random variable g into the f_g'(g')-dis- tributed random variable g'. The solution is especially simple for a transformation to a uniform distribution because then F− 1 is a constant function and g' F_g(g)).

Now we consider the mean and variance of functions of random vari- ables. By defi nition according to Eq. (3.3), the mean of g' is

∞

Eg' = µ_g' =

∫ g'f_g'(g')dg'.                                  (3.11)

− ∞

We can, however, also express the mean directly in terms of the function

p(g) and the PDF f_g(g):

 

Eg' = E.p(g)Σ  =

∫ p(g)f_g(g)dg.                              (3.12)

∞

− ∞

=

Intuitively, you may assume that the mean of g' can be computed from the mean of g: Eg' p(Eg). This is, however, only possible if p is a linear function. If p(g) is approximated by a polynomial

p(g) = p(µ_g) + p'(µ_g)(g − µ_g) + p''(µ_g)(g − µ_g)2/2 +... (3.13) then

g

µ_g' ≈ p(µ_g) + p''(µ_g)σ 2/2.                                           (3.14)

≈

From this equation we see that µ_g' p(µ_g) is only a good approximation if the both curvature of p(g) and the variance of g are small.

The fi rst-order estimate of the variance of g' is given by

σ

'

2

g' ≈ .p (µ_g

 

2

)

g

σ 2.                                      (3.15)

.

This expression is only exact for linear functions p.

The following simple relations for means and variances follow di- rectly from the discussion above (a is a constant):

E(ag) = aEg,       var(ag) = a2 var g,        var g = E(g2) − (Eg)2.          (3.16)

 

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