Random variables and reference probability measure

Question

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.
Thank you for your time.

Answer 1

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 $\mathbb{R}$.

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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.

Answer 2

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 $(\Omega, \Sigma, \mu)$.

Answer 3

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 : $$\mathbb E\left[X\right]=\int_\Omega X(\omega)\,d\mu(\omega)=\int_\mathbb R x\,dP(x)$$

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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.