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This is from Pliskas book in mathematical finance. I do not know what was best to write the question so I included the pages from the book. He has not written what form of the separating hyperplane theorem he uses, and why this follows. If someone understands it, could you please explain it? I have outlined in red the sentence I do not understand.

They refer to 1.3 which is: $V_t^*=V_t^*/B_t$, $t=0,1$

and 1.16: There are no arbitrage oppurtunities if and only if there exists a risk neutral probability measre on Q.

He also uses a single period model.

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In our lecture, we were told to omit the proof because it was too difficult. Maybe it will help you though if you can read it here:

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After applying the theorem you can take an $Y$ orthogonal to the separation plane, pointing in the right direction, you will easily have the property $X.Y>0$ for all $X$.

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  • $\begingroup$ Hello, could you please xplain the details for me? The theorem says that there is a plain that separates $A^+$ And $W$. How do they then find the $Y$ in $W^{\perp}$? And why can you choose Y orthogonal to the separation plane? $\endgroup$
    – user119615
    Commented Aug 24, 2014 at 14:36
  • $\begingroup$ I see how you can choose a vector that is perpendicular to the separating plane. But how do you get that this vector is in $W^{\perp}$, and how do you get that the inner product with every vector in $A^+$ is positive? $\endgroup$
    – user119615
    Commented Aug 24, 2014 at 15:36
  • $\begingroup$ well this is the particular case where the separation plane between two set is one of those set (W) so the orthogonal of the separation plane coincide with the orthogonal of one of the sets ($W^T$). The product will be positive with the use of the theorem. The formulation of the theorem is more complex than "there exist a plane that separate the two sets." In fact in the proof we build a particular vector e such that e.x > 0 for every x in the first set and e.x<0 for every x in the second set, the separation plane is then defined as the boundary (ie such that e.x = 0). $\endgroup$
    – Lucas Morin
    Commented Aug 24, 2014 at 16:20
  • $\begingroup$ Thank you very much. I just have two more questions. 1. Do you have a link to the proof you mentioned? And 2 how do you see that in this case the plane coincides with W? Is W also a plane? $\endgroup$
    – user119615
    Commented Aug 24, 2014 at 20:26
  • $\begingroup$ I can't manage to find one (I would have linked to it if i could). W is a subspace of $R^K$ with the first coordinate set to 0 and the k-1 others free, so it is a subspace of dimention k-1. Imagine W as a plane touching in only one point some kind of a sphere. Theyre are not stricly separated here, meaning that we accept $x.e<=0$ or $x.e>=0$ for each sets. The touching point has to be on the separation (only way to satisfy bot equations). Then the orientation: if we choose another plane than W they would cross, the only way to separate the two object is to take W as the border. $\endgroup$
    – Lucas Morin
    Commented Aug 24, 2014 at 21:07

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