Expectation via a Density: Pushforward Measures and Radon–Nikodym Derivatives
Summary
The document explains why the expectation of a measurable function of a random variable can be written as an integral against its density. First, the distribution of the random variable is represented by the pushforward of the probability measure onto the real line. If this distribution has a density with respect to Lebesgue measure, the Radon–Nikodym derivative converts integration against the distribution into integration against that density.
The key point is that the probability measure on the original sample space and Lebesgue measure are not directly measures on the same space. The pushforward measure places the random variable’s distribution on the real line, where it can be compared with Lebesgue measure and its density defined. The answer presents the identity through simple-function approximations, but does not discuss detailed integrability conditions; the expectation and integrals must be well-defined for the identity to apply.
Key ideas
- The distribution of a random variable is a pushforward measure on the real line.
- A probability density is the Radon–Nikodym derivative of that distribution with respect to Lebesgue measure.
- Integrating a function against the distribution is equivalent to integrating it against the density when the integrals are defined.
- The original probability measure and Lebesgue measure live on different spaces, so the distribution must first be transferred.
Tags
Full text
# Proving $\mathbb{E}(g(X)) = \int_{\mathbb{R}} g(x) f(x) dx$
# Proving $\mathbb{E}(g(X)) = \int_{\mathbb{R}} g(x) f(x) dx$
Let $X$ be a random variable on a probability space $(\Omega, \mathcal{F}, P)$ and let $g$ be a Borel-measurable function on $\mathbb{R}$. In Shreve II (p 28) he proves, using the standard machine, that $$ \mathbb{E}(g(X)) = \int_{\mathbb{R}} g(x)\, d(P \circ X^{-1})(x), $$ where $P \circ X^{-1}$ is the pushforward measure on $\mathbb{R}$. He then again uses the standard machine to prove that, for a continuous random variable $X$, that $$ \mathbb{E}(g(X)) = \int_{\mathbb{R}} g(x) f(x) d\lambda(x), $$ where $f$ is the probability density function of $X$ and $\lambda$ is the Lebesgue measure on $\mathbb{R}$.
My question is, is the standard machine really necessary for this second part? By definition of a continuous random variable, $f = \frac{d(P \circ X^{-1})}{d \lambda}$, and so $$ \int_{\mathbb{R}} g(x)\, d(P \circ X^{-1})(x) = \int_{\mathbb{R}} g(x)f(x) d\lambda(x) $$ since $f$ is the Radon-Nikodym derivative of $P \circ X^{-1}$ w.r.t. $\lambda$.
Perhaps I am overlooking some integrability conditions?
## Answer by Gordon (score 0, accepted)
https://quant.stackexchange.com/a/18640
As the Radon-Nykodim derivative is defined for measures on the same measurable space, and while probability $P$ is defined on the probability space $\Omega$, the standard machine is necessary to have the two measures $\lambda$ and $P \circ X^{-1}$ both defined on $\mathbb{R}$. That is, \begin{align*} \mathbb E(g(X)) &=\lim_{n\rightarrow \infty}\sum_{m=-\infty}^{\infty}\sum_{i=1}^{2^n}\Big(\frac{i-1}{2^n}+m\Big)P\Big(\frac{i-1}{2^n}+m \leq g(X) < \frac{i}{2^n}+m\Big)\\ &=\lim_{n\rightarrow \infty}\sum_{m=-\infty}^{\infty}\sum_{i=1}^{2^n}\Big(\frac{i-1}{2^n}+m\Big)P \circ X^{-1}\bigg(g^{-1}\Big[\frac{i-1}{2^n}+m, \ \frac{i}{2^n}+m\Big)\bigg)\\ &=\int_{\mathbb{R}} g(x)\, d(P \circ X^{-1})(x)\\ &= \int_{\mathbb{R}} g(x) f(x) d\lambda(x), \end{align*} where $f = \frac{d(P \circ X^{-1})}{d \lambda}$ is the Radon Nykodim derivative.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.