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For a discrete set of scenarios, minimising value at risk can be formulated as a mixed integer linear programming problem. If each scenario has equal probability then this can be written as

\begin{align} &\text{minimize} &\gamma\\ &\text{subject to} &(-r^{s}){'}X &\leq \gamma + M\cdot Y_{s} &&\text{$s = 1,\dots,S$} \tag1\\ &&\frac{1}{S}\sum_{s=1}^{S} Y_{s} &\leq \alpha \tag2\\ &&Y_{s} &\in \{0,1\} &&\text{$s = 1,\dots,S$} \\ &&\sum_{i=1}^{n}x_{i} &= 1 \end{align}

where $\alpha$ is the confidence level say $0.05$, $M$ is a big constant, $r$ is the return on assets, $x_{i}$ is the percentage in asset $i$, and $S$ is the number of scenarios.

If we assume that scenarios do not have same probabilities then constraint $(1)$ can be formulated as: $(-r^{s}\cdot P_{s}){'}X \leq \gamma + M\cdot Y_{s}$ where $P_{s}$ is the probability of scenario $s$. But I am struggling with redefining constraint $(2)$.

How can this constraint/problem be formulated if scenarios have different probabilities?

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  • $\begingroup$ You shouldn't modify constraint (1). For (2), your probability is already there. You just need to move $\frac{1}{S}$ inside the sum and replace it by $P_s$. $\endgroup$ – Gabriel Gouvine Jul 24 at 18:01
  • $\begingroup$ Your suggestion works. Thank you! I did try this earlier with Open Solver, but it was recognising this constraint as non-linear, though I am certain I set it up correctly (as I was able to solve the case for equal probability). I ended up coding it on another platform where it worked. $\endgroup$ – Sam Jul 24 at 22:42
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How about \begin{align}\min&\quad\gamma\\\text{s.t.}&\quad(-r^s)^\top X\leq \gamma + M Y_s \qquad s=1,\ldots,S\\&\quad\sum_{s=1}^SP_sY_s \leq \alpha\\&\quad \sum_{i=1}^nx_i=1\\&\quad Y_s\in\{0,1\}\end{align}

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  • $\begingroup$ Thanks. This answer is along the lines suggested by @Gabriel Gouvine In the equal probability case, the RHS of constraint (2) $\alpha\cdot S$ is rounded down to the nearest integer. Would a similar consideration apply in the case of different probabilities i.e. when multiplying probabilities with the binary variable? $\endgroup$ – Sam Jul 24 at 23:16
  • $\begingroup$ As far as I can see this is not needed, since the left-hand-side does not necessarily sum to an integer value. $\endgroup$ – k88074 Jul 26 at 19:51

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