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$f:A \times \Omega \rightarrow \mathbb{R} $ is a function defined as $f(a,\omega)=a_{1}X_{1}(\omega)+a_{2}X_{2}(\omega)+a_{3}X_{3}(\omega)$ such that $a_{1}+a_{2}+a_{3}=1$ with $a_{i} > 0.$

Here $A \subset \mathbb{R}^3$ and $\Omega$ is a measurable probability space.

Now since $a_{3}=1-(a_{1}+a_{2})$ we can write $$f(a,\omega)=a_{1}X_{1}(\omega)+a_{2}X_{2}(\omega)+(1-(a_{1}+a_{2}))X_{3}(\omega).$$

Now Can I say $f:A'\times \Omega \rightarrow \mathbb{R}$ where $A'\subset \mathbb{R}^2$ and $A'=(a_{1},a_{2})$ ??

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No, because the domain of function $f$ is not the set $A'\times\Omega$.

However, you are free to construct a function $g:A'\times\Omega$ by means of the prescription: $$\langle\langle a_1,a_2\rangle,\omega\rangle\mapsto f(a_1,a_2,1-a_1-a_2,\omega)$$

Then we have: $$g(\langle a_1,a_2\rangle,\omega)=a_{1}X_{1}(\omega)+a_{2}X_{2}(\omega)+(1-(a_{1}+a_{2}))X_{3}(\omega)$$

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  • $\begingroup$ Thanks for your help. Now I understand. $\endgroup$ Apr 18 '17 at 11:17

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