Equivalence between expectation definitions $\int X \,d\mathbb{P} = \int x f(x)\,dx$

Question

Suppose I have a probability space $(A, \Omega, \mathbb{P})$. Traditionally, the expectation is defined as $$\mathbb{E}[X] = \int xf(x)\,dx \tag{1}$$ where $X$ is a random variable and $f$ is the probability density function (PDF).

In a measure theoretic setting we have $$\mathbb{E}[X] = \int X \,d\mathbb{P}. \tag{2}$$

I am new to probability, though I am comfortable working with measure theory, so it could be that some of these questions are relatively elementary. Basically I am unclear on the equivalence on these two integrals. Namely (2) makes no reference to the PDF, whereas (1) makes no reference to the random variable $X$. My best guess is that the PDF is somehow encoded within the random variable $X$, and that explains everything. But I am not sure if this is true and I could not find many sources that explain this directly.

Answer 1

First you change variables to get \begin{align} E(X):=\int_{\Omega}X\,dP=\int_{\Bbb{R}}x\, dP_X(x), \end{align} where $P_X:=X_*P=P(X^{-1}(\cdot))$ is the push-forward measure/law/distribution of $X$ under $P$. This is the general setting.

Now specialize to the case where the pdf of $X$ exists. By definition this means the Radon-Nikodym derivative of the measure $P_X$ with respect to Lebesgue measure $\lambda$ exists; this is what we call the pdf, i.e \begin{align} f_X:=\frac{dP_X}{d\lambda}. \end{align} In this case we can write the expectation integral as \begin{align} E(X)&=\int_{\Bbb{R}}x\, dP_X(x)\\ &=\int_{\Bbb{R}}x\cdot\frac{dP_X}{d\lambda}(x) \, d\lambda(x)\\ &\equiv \int_{\Bbb{R}}x f(x) \, d\lambda(x), \end{align} where the second equality is a standard measure theory exercise.

Answer 2

Consider a measure space $(X, \mathscr{X}, \mu)$ and a function $f:X \to \mathbf{R}$ which is Borel measurable (relative to $\mathscr{X}$). Consider the following function $\nu$ defined on the Borel sets of $\mathbf{R}$ $$ \nu (B) = \mu(f^{-1}(B)), $$ where $f^{-1}(B)$ is the preimage of $B$ by $f.$ This $\nu$ can be shown to be a measure on the Borel sets of $\mathbf{R}$ which is therefore called the image measure of $\mu$ by $f$ and denoted $f(\mu).$

Suppose now that we have a probability space $(\Omega, \mathscr{F}, \mathbf{P})$ and a random variable $X$ (which by definition is a Borel measurable real-valued function). The image measure of $\mathbf{P}$ by $X$ is known as the distribution law of $X$ and, by measure theoretic arguments, this measure is univocally identified with a function $F:\mathbf{R} \to [0, 1]$ such that $F$ is non-decreasing, right-continuous, with left-limits and $F(-\infty) = 0,$ $F(\infty) = 1,$ we call graciously this $F$ as a distribution function. By definition then $$ \mathbf{E}(\mathbf{1}_B(X)) = \mathbf{P}(X \in B) = \int_B dF(x) = \int_\mathbf{R} \mathbf{1}_B(x) dF(x), $$ where $dF$ now denotes the distribution law of $X$ (defined on the Borel sets of $\mathbf{R}$). Using measure theoretic arguments (linearity and monotone classes of functions), we can show that $$ \mathbf{E}(u(X)) = \int_\mathbf{R} u(x) dF(x), $$ for any function $u:\mathbf{R} \to \mathbf{R}$ that is Borel measurable (in the sense that either both integrals exist and are equal, or neither of them exist).

Sometimes it is the case that $dF$ has a density relative to Lebesgue measure, this means that there exists a Borel function $f:\mathbf{R} \to \mathbf{R}_+$ such that $$ dF(B) = \int_B f(x) dx. $$ In this case, it can be shown that $$ \int_\mathbf{R} u(x) dF(x) = \int_\mathbf{R} u(x) f(x) dx. $$

Putting all of these together, you get what you were asking, the expected value of $X$ is $$ \mathbf{E}(X) = \int_\mathbf{R} x f(x) dx. $$