# 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 $p(\theta)$

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