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I am trying to get a better picture of what is the intuition behind the Doléans measure.

Is it analogue to the measure used for random variables but it is for random process?

The only explicit way where I see the Doléans measure to appear is in definition of norms. Further, in the Lecture Notes, I see again an indirect appearance of this measure, when we define function $I$, $I: \mathcal{P} \to \mathcal{M}^2_c$. We have that $||X|| = ||X \cdot M||$, where the first norm uses Doléans measure, second norm - simple one.

Can somebody explain also how to read, understand this: $\mu_M$ on $([0,\infty) \times \Omega, \mathcal{B}([0,\infty))\times \mathcal{F})$ is defined by $\mu_M(A) = \int_\Omega \int^\infty_0 1_A(t,w) d \langle M \rangle _t(w)P(dw)$. I do not understand what the two products of sets mean. Where $\mathcal{B}$ -borel sigma algebra,$\Omega$ - sample space, $\mathcal{F}$ - sigma algebra.

StochasticIntegrationStudent
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  • There's no way we'll be able to explain that last thing without you telling us what $\beta$ is. $\mathcal{F}$ I can throw a guess (it's probably the sigma-algebra for the underlying probability space) but $\beta$ I have no clue. – Ian Mar 05 '17 at 17:25
  • Pardon, corrected the $\beta$ – StochasticIntegrationStudent Mar 05 '17 at 17:31
  • Ah, the symbol used in your reference was probably $\mathcal{B}$, written as \mathcal{B} in LaTeX. – Ian Mar 05 '17 at 17:56
  • You are right! Could not find appropriate one! – StochasticIntegrationStudent Mar 05 '17 at 18:02
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    Alright, then that last thing isn't so complicated: you want to turn the set set $[0,\infty) \times \Omega$ into a measure space. You give $[0,\infty)$ the Borel $\sigma$-algebra (as usual in probability theory) and you give $\Omega$ the $\sigma$-algebra $\mathcal{F}$. Then as usual in measure theory you give their product the product $\sigma$-algebra (which notably is not the Cartesian product of the two $\sigma$-algebras as sets). – Ian Mar 05 '17 at 18:07
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    (Cont.) Then you put a measure on that measurable space. The measure they choose is the one given by that integration; to know why they choose that one I would need some more context as to what they're trying to use $\mu_M$ for. You might gain a bit of intuition by recalling that $\int_\Omega X(\omega) P(d\omega)$ is the same as $E[X]$, so that integral is $E \left [ \int_0^\infty 1_A d \langle M \rangle_t dt \right ]$ – Ian Mar 05 '17 at 18:08
  • I see. So may I rephrase it in less rigorous way? So the first pair is the one that is transformed to create a product of sets used as measure space for a stochastic process, right? The next time author uses it is in definition of norm, $||X||T = ( \int X^2 1{[0,T] \times \Omega} d\mu_M)^{1/2}$ . I think there is typo with last formula, there should be no $dt$. – StochasticIntegrationStudent Mar 05 '17 at 18:36
  • Yes the $dt$ is a mistake, you're right. Also I assume there is a square root in the definition of the norm... – Ian Mar 05 '17 at 18:37
  • So similarly to probability theory, where we have triple of $(\Omega, \mathcal{F}, \mathbb{P})$ to descibe probability space where random variables live, now we have random processes, so for them we have another triple: $([0,\infty) \times \Omega, \mathcal{B}([0,\infty)\times \mathcal{F},\mu_M)$. Correct? Is there anything else to know about this measure? And could you please answer why the equality of different norms is meaningful? I think that as we have processes on both sides of $||X|| = ||X \cdot M||$, we should treat them with the same norm, should not we? Sorry if I am askying dull q. – StochasticIntegrationStudent Mar 05 '17 at 18:48
  • Yes, the resulting object is just a measure space. I don't know what they're doing with $\mu_M$ exactly so I can't comment on "other things to know about it". As for the norms, the point is just to define a norm for $X$ through $M$. Similarly in linear algebra we often define a norm on vectors through positive definite matrices: $| x |_A = (x^T A x)^{1/2}$. It's the same general idea, just a different way to measure the "size" of $X$. – Ian Mar 05 '17 at 20:55
  • Thank you! I appreciate the answer you did. Shall you copy it or put a link to your comments, I will accept it as an answer then. – StochasticIntegrationStudent Mar 05 '17 at 21:03

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I don't know what exactly they're doing with the norms, but the point is to give a norm for the process $X$ in relation to the process $M$. You might consider the analogous situation in linear algebra: if $A$ is a positive definite matrix then we often define $\| x \|_A = (x^T A x)^{1/2}$. It's just a different way to measure the size of a vector in relation to a matrix.

The last thing is just a way to make $[0,\infty) \times \Omega$ into a measure space. You have a "natural" choice of $\sigma$-algebra on each factor, namely $\mathcal{B}([0,\infty))$ and $\mathcal{F}$ respectively. So you give $[0,\infty) \times \Omega$ the product of these two $\sigma$-algebras (which, by the way, is not the Cartesian product of the two $\sigma$-algebras as sets). With that $\sigma$-algebra in hand you have a measurable space. You now give this measurable space the measure $\mu_M$ to make it into a measure space. I don't know what they're doing with $\mu_M$, but I can rewrite it as $\mu_M(A) = E \left [ \int_0^\infty 1_A d \langle M \rangle_t \right ]$, which might give you some intuition (or at least reduce the notational burden).

Ian
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