I'm reading Bernt Oksendal's "Stochastic Differential Equations" (edition 6) and I got quite confused on the conceptions. Please kindly help.

I don't understand what is an event in the definition of random variable

Definition: A random variable $X$ is an $\mathscr{F}$-measurable function $X:\Omega\ni \omega\mapsto X(\omega) \in \mathbb{R}^n$. (page 9)

Suppose there're zillions of pollens in a room ($\mathbb{R}^3$) doing brownian motion. We take a snapshot of the horizontal location of a particular Tulip pollen, and so have a random variable $X:\Omega\ni \omega\mapsto X(\omega) \in \mathbb{R}^2$ constructed for this special Tulip pollen.

Now, what is $\omega$, is it the Tulip pollen, or the physical location that the Tulip pollen happened to be at the split of the secone?

  • If $\omega$ is about the Tulip pollen, then the whole set $\Omega$ shall be the whole set of other pollen in the room.

    Consider the open circle $C(X(\omega), r)$, it's preimage $$X^{-1}(C) := \{pollen \in \{\text{All Pollens In The Room}\}; X(pollen) \in C(X(\omega, r)\}$$

  • Otherwise, if $\omega$ is about the physical position of the Tulip pollen, then the whole set $\Omega$ shall be the room.

    Consider the open circle $C(X(\omega), r)$, it's preimage $$X^{-1}(C) := \{position \in \{\text{The Room}\}; X(position) \in C(X(\omega, r)\}$$

I thought it shall be the pollen: the mapping from determined space in the room to $\mathbb{R}^2$ really does not sounds "random".

Am I correct here?

  • $\begingroup$ @did mind have a look? $\endgroup$
    – athos
    Sep 29 '13 at 12:56
  • $\begingroup$ @stefan-hansen mind have a look? $\endgroup$
    – athos
    Sep 29 '13 at 12:57

In your formulation, as long as one takes only one snapshot, both models are equivalent: if $\Omega=\{\text{pollens in the room}\}$ then $X:\Omega\to\mathbb R^3$ sends each pollen $\omega$ to its position $X(\omega)$ while, if $\Omega=\{\text{the room}\}\subset\mathbb R^3$ then $X:\Omega\to\mathbb R^3$ can be defined by $X(\omega)=\omega$ for every $\omega$ in $\Omega$.

As soon as one wishes to consider several snapshots, say at times $r$, $s$, $t$, etc., the second model becomes impractical. By contrast, in the first model, one can define random variables $X,Y,Z,\ldots:\Omega\to\mathbb R^3$ sending each pollen $\omega$ to its position $X(\omega)$ at time $r$, its position $Y(\omega)$ at time $s$, its position $Z(\omega)$ at time $t$, etc.

It is only natural to stumble on these questions when first becoming acquainted to the probabilistic formalism. As quickly as possible though, one should stop worrying about the exact identity of the probability triple $(\Omega,\mathcal F,P)$ one uses. All that matters is that there exists some such triple, adequate to model the experience or the experiences one has in mind. Everything is as if $(\Omega,\mathcal F,P)$ was hidden and one was only able to see, and interested in, the shadows it projects on the image space, that is, the distributions on $\mathbb R$ or $\mathbb R^3$ or whatever, which are the images of $P$ by the random variables $X$ or $(X,Y,Z)$ or whatever, defined on $\Omega$.

Said differently, probability theory allows to compute probabilities of events and expectations of random variables. If two random variables $X:\Omega\to\mathbb R$ and $Y:\Psi\to\mathbb R$ are defined on different probability spaces $(\Omega,\mathcal F,P)$ and $(\Psi,\mathcal G,Q)$ and if they have the same distribution $P_X=Q_Y$ on $(\mathbb R,\mathcal B(\mathbb R))$, then $E_P[u(X)]=E_Q[u(Y)]$ for every function $u$, and this is all that matters. Likewise for two processes $X=(X_t)_{t\in T}:\Omega\to\mathbb R^T$ and $Y=(Y_t)_{t\in T}:\Psi\to\mathbb R^T$ as long as the distributions $P_X$ and $Q_Y$ on $(\mathbb R^T,\mathcal B(\mathbb R^T))$ coincide.

  • $\begingroup$ thank you very much. Maybe I got a bit obsessive compulsive disorder -- being not a math guy, i'm willing to give up some math details, e.g. some tricky proof etc, but i don't sleep well just blindly using the formula to punch in number and get a result, i'd really like to understand the concept at least, so i feel more comfortable to move on. thanks again $\endgroup$
    – athos
    Sep 30 '13 at 16:42

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