0
$\begingroup$

I have the following problem: an equity portfolio allocated to 3 countries. Each country has an independent indicator (signal) which takes values from -4 to 4. The allocation for each country at neutral is 1/3 (33.33%) and based on the indicator can oscillate -20%/+20%.

the weights are: w1 +w2 +w3 =100% constraint w1,w2,w3 =33.33% (at neutral) w1,w2,w3 e [13.33%;53,33%] constraint the indicators : i1,2,3 e [-4;4]

the allocation can be linear, proportional. How can this be solved for 3 countries? because for 1 is easy: Like -4 represents -20% in a country. 0 represents neutral allocation , keeping 33.33% in that specific country. For 2 I made a compound indicator like the difference between countries indicators. But for 3+ countries I cannot find the solution.

$\endgroup$
  • $\begingroup$ Welcome to quant.SE @E.B.! Could you clarify your question a bit to make it easier to understand? You always force all three weights to sum up to 100%, right? So, in case all signals are bad (e.g. [-3, -3, -4]) you would still invest in all these assets? Or, if you get [-4, -4, -4], shouldn't this give you the neutral portfolio again? $\endgroup$ – muffin1974 Aug 30 '18 at 14:32
  • $\begingroup$ Yes, I must force them to be 100%. I am not allowed to keep in cash or to leverage the portfolio. So it must be 100%. Neutral means that the indicators are [0,0,0], so doesn't consider neither country to be nor positive nor negative. In case all the signals are negative I should still invest, and decide which one is less negative. And yes, [-4;-4;-4] should give me neutral again $\endgroup$ – E.B. Aug 31 '18 at 8:16
1
$\begingroup$

Here's a thought.

In 2-dim your score $(x, y) \in [-4,4]^2$ is best characterised as the minimal distance from the line $y=x$, along which your portfolio is balanced. I.e. wherever $y=x$ either at $(0,0), (-1,-1) (4,4)$ the weight is 50-50, since there is no minimal displacement vector. As a further example $(0,2)$ has minimal distance from the point $(1,1)$, whose displacement vector is $(-1,1)$. Note this is consistent with a difference mentality: $(0,2)$ having the same displacement as say $(1,3)$ or $(-2,0)$. Then you define some topology that maps a displacment vector to a new portfolio position.

In n-dim your score $\mathbf{x}=(x_1, ..., x_n) \in [-4,4]^n$ is reclassed as the minimum distance from the line $\mathbf{r} = t \mathbf{1} $ (i.e $x=y=z$ in 3-dim). And the vector defining the position from that central axis to your specific point score defines the displacement of the portfolio from neutrality. Note that in this case the minimum distance vector from the line to the point score is $\mathbf{x} - (\mathbf{x}\cdot \frac{\mathbf{1}}{||\mathbf{1}||})\frac{ \mathbf{1}}{||\mathbf{1}||}$. (https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line)

Yes, OK, you still need to define the topology, but this a mathematical construct from which to develop consistency in multiple dimensions. Presumably some scaling to satisfy your constraints.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.