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Prove that this set of expected values is convex

0

1

$begingroup$

I am facing the problem of proving that the following set $K$ is convex, it's part of a proof for the First Fundamental Theorem of Asset Pricing:
$$K := d cdot E_mathbbQ(boldsymbolX): mathbbQ sim mathbbP, ; d > 0 subset mathbbR^n$$
where $mathbbQ$ is a risk-neutral probability for the market $boldsymbolX$ which is a random vector of $n$ random variables on the probability space $(Omega, Sigma, mathbbP)$.

This is what I have accomplished until now:

Let $boldsymbolk_1, boldsymbolk_2 in K$ then $begincases boldsymbolk_1 = d_1 cdot E_mathbbQ_1(boldsymbolX):mathbbQ_1 sim mathbbP, ; d > 0, \ boldsymbolk_2 = d_2 cdot E_mathbbQ_2(boldsymbolX): mathbbQ_2 sim mathbbP, ; d > 0. endcases$

I have proved that, let $theta in [0,1], ; theta fracboldsymbolk_1d_1 + (1-theta) fracboldsymbolk_2d_2 = E_theta mathbbQ_1 + (1-theta) mathbbQ_2(boldsymbolX) in K.$

But I don't know how to prove that, let $lambda in [0,1], ; lambda boldsymbolk_1 + (1-lambda) boldsymbolk_2 in K.$

probability probability-theory convex-analysis finance convex-geometry

share|cite|improve this question

edited Apr 1 at 9:45

Jorge Garcia

asked Mar 31 at 8:20

Jorge GarciaJorge Garcia

215

$endgroup$

add a comment | 

0

1

$begingroup$

I am facing the problem of proving that the following set $K$ is convex, it's part of a proof for the First Fundamental Theorem of Asset Pricing:
$$K := d cdot E_mathbbQ(boldsymbolX): mathbbQ sim mathbbP, ; d > 0 subset mathbbR^n$$
where $mathbbQ$ is a risk-neutral probability for the market $boldsymbolX$ which is a random vector of $n$ random variables on the probability space $(Omega, Sigma, mathbbP)$.

This is what I have accomplished until now:

Let $boldsymbolk_1, boldsymbolk_2 in K$ then $begincases boldsymbolk_1 = d_1 cdot E_mathbbQ_1(boldsymbolX):mathbbQ_1 sim mathbbP, ; d > 0, \ boldsymbolk_2 = d_2 cdot E_mathbbQ_2(boldsymbolX): mathbbQ_2 sim mathbbP, ; d > 0. endcases$

I have proved that, let $theta in [0,1], ; theta fracboldsymbolk_1d_1 + (1-theta) fracboldsymbolk_2d_2 = E_theta mathbbQ_1 + (1-theta) mathbbQ_2(boldsymbolX) in K.$

But I don't know how to prove that, let $lambda in [0,1], ; lambda boldsymbolk_1 + (1-lambda) boldsymbolk_2 in K.$

probability probability-theory convex-analysis finance convex-geometry

share|cite|improve this question

edited Apr 1 at 9:45

Jorge Garcia

asked Mar 31 at 8:20

Jorge GarciaJorge Garcia

215

$endgroup$

add a comment | 

$begingroup$

I am facing the problem of proving that the following set $K$ is convex, it's part of a proof for the First Fundamental Theorem of Asset Pricing:
$$K := d cdot E_mathbbQ(boldsymbolX): mathbbQ sim mathbbP, ; d > 0 subset mathbbR^n$$
where $mathbbQ$ is a risk-neutral probability for the market $boldsymbolX$ which is a random vector of $n$ random variables on the probability space $(Omega, Sigma, mathbbP)$.

This is what I have accomplished until now:

Let $boldsymbolk_1, boldsymbolk_2 in K$ then $begincases boldsymbolk_1 = d_1 cdot E_mathbbQ_1(boldsymbolX):mathbbQ_1 sim mathbbP, ; d > 0, \ boldsymbolk_2 = d_2 cdot E_mathbbQ_2(boldsymbolX): mathbbQ_2 sim mathbbP, ; d > 0. endcases$

I have proved that, let $theta in [0,1], ; theta fracboldsymbolk_1d_1 + (1-theta) fracboldsymbolk_2d_2 = E_theta mathbbQ_1 + (1-theta) mathbbQ_2(boldsymbolX) in K.$

But I don't know how to prove that, let $lambda in [0,1], ; lambda boldsymbolk_1 + (1-lambda) boldsymbolk_2 in K.$

probability probability-theory convex-analysis finance convex-geometry

share|cite|improve this question

edited Apr 1 at 9:45

Jorge Garcia

asked Mar 31 at 8:20

Jorge GarciaJorge Garcia

215

$endgroup$

share|cite|improve this question

edited Apr 1 at 9:45

Jorge Garcia

asked Mar 31 at 8:20

Jorge GarciaJorge Garcia

215

share|cite|improve this question

edited Apr 1 at 9:45

Jorge Garcia

asked Mar 31 at 8:20

Jorge GarciaJorge Garcia

215

asked Mar 31 at 8:20

Jorge GarciaJorge Garcia

215

asked Mar 31 at 8:20

Jorge GarciaJorge Garcia

215