# Random variables and reference probability measure

I am self-studying probability theory, and I am having quite some problems with the very basic concepts of the theory that are seriously hampering any attempt to proceed further in my study.

Here there is one basic problem I am having, with my thoughts (written in italic) about it.

Assume the following:

• $(\Omega, \Sigma, \mu)$ is a probability space,
• $X$ is a continuous r.v.,
• $(\mathbb{R}, \mathcal{B}_\mathbb{R}, P)$ is the probability space induced by $X$ (where $\mathcal{B}_\mathbb{R}$ is the Borel $\sigma$-algebra),
• $f_X$ is the PDF of $X$,
• $F_X$ is the CDF of $X$.

When we talk about the expectation functional with respect to the r.v. $X$ we write $$\mathbf{E} (X) := \int_\Omega X d\mu. \hspace{4cm}(*)$$ Why is this the case? To me it looks we should rather write $$\mathbf{E} (X) := \int_\Omega X dP,$$ because the r.v. works on the induced probability space that has $P$ has the reference probability measure.

Any feedback is most welcome.

• From the set-up, $P$ maps elements of the Borel algebra to $[0,1]$, while $\mu$ maps the elements of $\Sigma$ to $[0,1]$. Integrating $P$ over $\Omega$ is then either meaningless or abuse of notation. However, if you set the second integral on $\mathbb{R}$, the measurability conditions on $X: \Omega \to \mathbb{R}$ ensure that the two expressions are equal, in a first year substitution in integration way. – stochasticboy321 Nov 15 '15 at 16:39
• You would think as in a more general case of $E[X]$, I mean $E[\phi(X)]$ for a borel-measurable function $f$ on $\mathbb R$, as I stated in my answer. – Fardad Pouran Nov 15 '15 at 18:02

Following the notations you gave, $E(X) = \int_\Omega X d\mu$ is correct, since $X$ is a function defined on the measure space $(\Omega, \Sigma, \mu)$, so more explicitly, we can write $$E(X) = \int_\Omega X(\omega) \mu(d \omega)$$ to highlight the integration is taken on the space $\Omega$ with respect to the measure $\mu$ defined on $\Sigma$. $E(X) = \int_\Omega X dP$ does not make sense in that $P$ is not a measure associated to the space $\Omega$.

There does exist an expression for $E(X)$ using the measure $P$ defined on the induced probability space $(\mathbb{R}, \mathscr{B}_{\mathbb{R}}, P)$, known as change of variable formula, as follows: $$E(X) = \int_\mathbb{R} x dP = \int_{\mathbb{R}} x P(dx),$$ which makes sense in that the function $x \mapsto x$ is a function defined on $\mathbb{R}$.

Remark: However, the notations you introduced are a little non-standard. Conventionally, we use $P$ as the probability measure defined on the original space $(\Omega, \Sigma)$, while use $\mu$ as the probability measure on the induced space $(\mathbb{R}, \mathscr{B}_{\mathbb{R}})$, which is exactly opposite to your notations.

• Thanks a lot for the very clear answer. I did correct what I saw as a typo in the first formula. – Kolmin Nov 15 '15 at 16:47
• $\mu(d\omega)$ is not a typo, To me, it even makes more sense than $d\mu(\omega)$. Of course, both notations are standard. – Zhanxiong Nov 15 '15 at 16:49
• Oh, ok. Then, I am really really sorry. It is just that I never saw that one anywhere, and I naively thought it was a typo. Sorry again, and thanks for the answer, in particular for the enlightening (for me) point that $\mathbf{E}(X)$ as defined in my second formula "does not make sense in that $P$ is not a measure associated to the space $\Omega$". – Kolmin Nov 15 '15 at 16:53
• You are very welcome. A few more words regarding to the notation: to my knowledge, in measure theory, the notation $dP$ is more used when both $\Omega$ and $\omega$ are suppressed, like $E(X) = \int X dP$, whereas $P(d\omega)$ is more used when you want to explicitly include the argument $\omega$, like $E(X) = \int_\Omega X(\omega) P(d\omega)$. – Zhanxiong Nov 15 '15 at 17:00
• Considering your nice answer to this problem I had, may I point you another question I have closely related that is creating me a lot of problems? math.stackexchange.com/questions/1530493/… – Kolmin Nov 15 '15 at 18:14

The way you have worded it, $(\Omega, \Sigma, \mu)$ is the underlying probability space. Thus $X:\Omega \to \mathbb{R}$ is $\Sigma/\mathcal{B}_{\mathbb{R}}$ measurable, and one talks about the integral of $X$ with respect to $\mu$: $$\int_\Omega X(\omega)\,d\mu(\omega) = E[X].$$ Next there is the law of $X$, $\mu_X := \mu\circ X^{-1}$, which in your notation is just called $P$. Now $P$ is a measure on $(\mathbb{R}, \mathcal{B}_{\mathbb{R}})$ that assigns to a subset $A \in \mathcal{B}_{\mathbb{R}}$ the chance that $X$ falls in that set. One can also talk about integration with respect to $P$, but the integral is over $\mathbb{R}$, not over the original probability space. The two are related by the fact that if $X \geq 0$ or $X$ is integrable, then $$\int_{\mathbb{R}}x\,dP(x) = E[X] = \int_\Omega X(\omega)\, d\mu(\omega).$$

Some of the confusion stems from the fact that one usually calls the underlying space $(\Omega,\Sigma,P)$, not $(\Omega, \Sigma, \mu)$.

• Thanks a lot for the answer. Actually, my confusion was not really coming from the usual convention regarding the probability measure as you write in the last paragraph of your answer, but rather on not properly seeing the very first formula that you wrote down, which basically settles the issue. – Kolmin Nov 15 '15 at 16:50

Lebesgue Integral, is just defined via the measure of the function's domain.

Moreover, there's not an "actual" difference between the measure $\mu$ on $\Omega$ and $P$ on $\mathcal{B}_\mathbb R$, since $P$ is defined by : $$\forall E\in\mathcal{B}_\mathbb R\;P(E):=\mu\left(X^{-1}(E)\right)$$ Note. $X^{-1}(E)\in\Sigma$, since $X$ is measurable.

Indeed, $P$ is just an induced measure of $\Omega$ on $\mathbb R$.

Furthermore, we have this equality which I hope clarifies the context : $$\mathbb E\left[X\right]=\int_\Omega X(\omega)\,d\mu(\omega)=\int_\mathbb R x\,dP(x)$$

There's a theorem which states :

For each Borel-Measurable function $\phi:\mathbb R\rightarrow\mathbb R$,

$\phi\,o\,X$ is also a r.v. and $\displaystyle\mathbb E\left[\phi\,o\,X\right]=\int_\Omega \phi \circ \!X(\omega)\,d\mu(\omega)=\int_\mathbb R \phi(x)\,dP(x)$.

You can just put $\phi=id$ to reach the above equality.