Emanuel Winterfors

February 21, 2008

Mapping of a probability density

Filed under: Mathematics, Probability theory, Science — Emanuel Winterfors @ 11:59 am

Problem formulation

We have a probability density p_X (x), x \in X with respect to a measure dX on X.

If X is mapped onto another set Y by a function \Phi

\Phi :X \to Y

what will be the probability density p_Y (y), \quad y \in Y over Y, with respect to a measure dY on Y?

General solution

p_Y (y) = \int\limits_{x \in X} {p_X (x)\delta _Y \left( {y | \Phi (x)} \right)dX} \quad ,

where \delta _Y \left( {y | \Phi (x)} \right) is Dirac’s delta function at y=\Phi (x) so that

P(\Phi (x) \in B) \ = \ \int\limits_{y \in B} {\delta_Y \left( {y | \Phi (x)} \right) dY} \qquad,

for any measurable subset B of Y — in effect a conditional probability density with respect to the measure dY. If Y is a vector space, \delta_Y \left( {y | \Phi (x)} \right) can be written \delta_Y \left( {y-\Phi (x)} \right).

Proof

We need prove that A \subseteq X and B = \Phi (A) implies that P(A) = P(B) for any measurable subset A of X.

By definition, we have that  B \subseteq Y  P(A) = \int\limits_{x \in A} {p_X (x)dX}  and  P(B) = \int\limits_{y \in B} {p_Y (y)dY} .

From this, we can calculate

P(B)   = \int\limits_{y \in B} {p_Y (y)dY}
  = \int\limits_{y \in B} {\int\limits_{x \in X} {p_X(x)\delta _Y \left( {y | \Phi (x)} \right)dX}dY}
  = \int\limits_{x \in X} {p_X(x)\int\limits_{y \in B} {\delta _Y \left( {y | \Phi (x)} \right)dY} dX}
  = \int\limits_{x \in X} {p_X(x)dX}
  = P(A)

QED.

© 2008 Emanuel Winterfors

\LaTeX code can be used in comments: $latex p(\theta)$ gives p(\theta)
Advertisements

3 Comments »

  1. The change of order of integration in the proof poses some restrictions on the integrand p_X(x)\delta_Y(y|\Phi(x)), in order to be valid. I’m not sure what the minimal conditions are, though…

    Comment by winterfors — February 22, 2008 @ 11:36 am

  2. This is just a change of variables in Lebesgue integral?

    Comment by killua — March 27, 2008 @ 10:22 am

  3. Not necessarily – the probability density can be defined in relation to any measure on X, not necessarily the Lebesgue measure (even if it is certainly the most commonly used one).

    Having looked into the problem a bit, I think the Fubini’s theorem should provide a guarantee for change of order of integration to be valid, as long as the integral of the absolute value of the integrand | p_X(x)\delta_Y(y|\Phi(x)) | with respect to the Cartesian product measure dXdY exists. The existence and uniqueness of dXdY is in turn guaranteed by the Hahn-Kolmogorov theorem.

    Comment by Emanuel Winterfors — March 27, 2008 @ 2:00 pm


RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Create a free website or blog at WordPress.com.

%d bloggers like this: