# Why is expectation defined by $\int xf(x)dx$?

I recently found out that the expectation of a random variable $X$ in a probability space $(\Omega, \mathcal F, \mathbb P)$, $\mathbb E(X)$, is just the term used in probability theory for the measure of the function $X$; i.e.:

$$\mathbb E(X)=\mathbb P(X)$$

or, for a real-valued random variable $X:\Omega\to\mathbb R$:

$$\mathbb E(X)=\int_\Omega X(x)\mathbb P(dx)$$

The thing is, I am also familiar with the definition of the expectation of a random variable $X:\Omega\to\mathbb R$ with probability distribution function $f$ to be:

$$\mathbb E (X)=\int_\mathbb R xf(x)dx$$

Here, the integral is with respect to the Lebesgue measure on $\mathbb R$.

We can define the function $f$ in the following way: we have $F(x)=\int_\mathbb R f(x)dx$, where $F(x)=\mathbb P(X^{-1}((-\infty,x]))$.

I can't see any way to show that these two definitions are equal to one another. The only transformation between integrals with respect to different measures that I know is that $\mu(f^{-1}(g))=\mu(g\circ f)$, which gives that $\mathbb E(g)=\mu_X(g)$, where $\mu_X$ is the image measure $\mathbb P\circ X^{-1}$, but that doesn't seem to help.

Why can we write $\mathbb E(X)=\int_\mathbb R xf(x)dx$? Why are these two definitions consistent?

• $\int_\mathbb{R} X(x)\mathbb{P}(dx)$ does not make sense since $\mathbb{P}$ is a measure on $\Omega$ and not on $\mathbb{R}$. – Stefan Hansen Nov 17 '13 at 16:40
• @StefanHansen - thanks - yes, that should be an $\Omega$, not an $\mathbb R$. – John Gowers Nov 17 '13 at 16:49

Let $$(\Omega,\mathcal{F},P)$$ be a probability space and $$X:\Omega\to\mathbb{R}$$ a random variable, i.e. a $$(\mathcal{F},\mathcal{B}(\mathbb{R}))$$-measurable mapping. Then $$X$$ induces a probability measure on $$(\mathbb{R},\mathcal{B}(\mathbb{R}))$$ defined by $$P_X(B):=P(X^{-1}(B)),\quad B\in\mathcal{B}(\mathbb{R}),$$ which is well-defined since $$X$$ is measurable. This is called the distribution of $$X$$ or the pushforward measure of $$P$$ under $$X$$. The definition of the expectation of $$X$$ is the following Lebesgue integral on $$\Omega$$: $${\rm E}[X]:=\int_{\Omega} X\,\mathrm dP=\int_\Omega X(\omega)\,P(\mathrm d\omega),$$ given that this integral exists. This integral can always be transformed into a Lebesgue integral on $$(\mathbb{R},\mathcal{B}(\mathbb{R}))$$. The following holds:

For any integrable random variable $$X$$ one has $${\rm E}[X]=\int_{\mathbb{R}} x\,P_X(\mathrm dx).\tag{1}$$

In the special case where $$X$$ admits a density function, i.e. $$P_X(B)=P(X\in B)=\int_B f_X(x)\,\mathrm dx$$ for all $$B\in\mathcal{B}(\mathbb{R})$$ and for some measurable, non-negative function $$f_X$$, we can simplify $$(1)$$ even further: $${\rm E}[X]=\int_{\mathbb{R}}xf_X(x)\,\mathrm dx. \tag{2}$$

A standard technique for showing the results in $$(1)$$ and $$(2)$$ is to a) show that it holds for indicator functions, i.e. $$X=\mathbf{1}_A$$ for $$A\in\mathcal{F}$$, b) show that if it holds for $$X$$ and $$Y$$ then it also holds for $$\alpha X+Y$$, $$\alpha\in\mathbb{R}$$, and c) if it holds for a sequence $$(X_n)$$ then it also holds for $$\lim X_n$$.

It is not true in general, the distribution of $X$ must be absolutely continuous. I probably should elaborate a bit:

Theorem (Radon-Nikodym): Let $\mu$ be a $\sigma$-finite measure on a measurable space $(X,S)$, and $\nu$ a finite measure which absolutely continuous wrt $\mu$. Then there is a $h\in L^1(X,S,\mu)$ with $$\nu(E)=\int_E h d\mu$$ for all $E\in S$. (Any two such $h$ are equal a.e. $\mu$). $h$ is said to be the Radon-Nikodym derivative or density of $\nu$ wrt $\mu$ and denoted $d\nu/d\mu$.

This implies that if $f\in L^1(\nu)$ then $$\int f\frac{d\nu}{d\mu} d\mu=\int fd\nu,$$ which includes as a special case what you are looking for (take $\mu$ to be Lebesgue measure).

• Indeed, and then the magic Radon-Nikodym theorem jumps in. – Miha Habič Nov 17 '13 at 12:40
• Indeed, I have expanded my answer. – Alexander Grothendieck Nov 17 '13 at 12:50

If you start with probability space $\left(\Omega,\mathcal{F},\mathbb{P}\right)$ and a measurable random variable $X:\Omega\rightarrow\mathbb{R}$ where $\mathbb{R}$ is equipped with the $\sigma$-algebra of the Borel sets $\mathcal{B}$ then $E\left(X\right)$ is denoted by $\int Xd\mathbb{P}$ or $\int X\left(\omega\right)\mathbb{P}\left(d\omega\right)$. A probability measure $P$ is induced by $X$ on $\left(\mathbb{R},\mathcal{B}\right)$ by $P\left(A\right)=\mathbb{P}\left(X^{-1}\left(A\right)\right)$. Now start with probability space $\left(\mathbb{R},\mathcal{B},P\right)$ and define $Y:\mathbb{R}\rightarrow\mathbb{R}$ by $x\mapsto x$. Then $X$ and $Y$ have the same distribution so $E\left(X\right)=E\left(Y\right)$. Applying the mentioned technique we now find $E\left(X\right)=E\left(Y\right)=\int YdP=\int Y\left(x\right)P\left(dx\right)$. Since $Y\left(x\right)=x$ this is also written as $\int xP\left(dx\right)$ or $\int xdF\left(x\right)$ where $F\left(x\right)=P\left\{ X\leq x\right\}$. If $F\left(x\right)=\int_{-\infty}^{x}f\left(x\right)dx$ for some $f$ then also $E\left(X\right)=\int xf\left(x\right)dx$.