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In this paper, the authors make a simple model with:

(1) A global bank, who is risk-neutral but has a Value-at-Risk constraint:

$$\max_{x_t^B} E_t[x_t^B\prime R_{t+1}]$$ s.t. $$\alpha (Var(x_t^B\prime R_{t+1}))^{\frac{1}{2}} <= 1$$ where $R_{t+1}$ is a (n x 1) vector of returns, $x_t^B$ is a (n x 1) vector of weights $\alpha$ is a parameter, and $Var$ is the variance operator.

I tried setting up a lagrangean which should yield:

$$ \mathcal{L} = E_t[x_t^B\prime R_{t+1}] - \lambda_t (1 - \alpha (Var(x_t^B\prime R_{t+1}))^{\frac{1}{2}}) $$

The first order conditions w.r.t. $x_t^B$ are yielding me:

$$ E_t(R_{t+1}) - \lambda_t \alpha Var(x_t^B\prime R_{t+1})^{-\frac{1}{2}} Var(R_{t+1}) x_t^B = 0 $$

In comparison with the solution given on the paper it seems that I have an extra term $ Var(x_t^B\prime R_{t+1})^{-\frac{1}{2}}$.

Can anyone help me on this? Thanks.

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  • $\begingroup$ This is not a Value-at-risk constraint. Take square of your constraint and it becomes an old-fashioned mean-variance optimization. It is neither risk neutral if you ask me: whether the constraint is set by the bank or the regulator is practically the same. It seems like old fashioned Markowitz dressed up to be served as something else. What about the weights? Don't they have a constraint to be positive and sum up to one (or allowed negative and the absolute weights to sum up to one)? Do they have a 'value at risk constraint' but are allowed to run an infinite long-short book? $\endgroup$ – Kiwiakos Mar 25 '16 at 17:42
  • $\begingroup$ This is supposed to get the same result as a standard mean variance. They are indeed allowed to have a long-short book of infinite amounts, as long as the constraint is respected. Probably squaring the constraint is the way to go. Thanks $\endgroup$ – phdstudent Mar 25 '16 at 18:08
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Regarding your question, you appear to treat variance as a linear operator, which it is not. But hard to tell as your parantheses don't match.

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Completely agree with @Kiwiakos' remark, this is not a 'VaR' (Value-At-Risk) constraint but rather a 'Var' (Variance) constraint.

Here's how I would do it, I'll drop the superscript $B$ to keep notations uncluttered.

First, square the original constraint to obtain an inequality involving variance instead of standard deviation. Next build the Lagrangian

$$ \mathcal{L}(x_t^\prime) = E_t[ x_t^\prime R_{t+1} ] - \lambda \left( 1 - \alpha^2 \text{Var}(x_t^\prime R_{t+1}) \right) $$

Assuming no further constraints on the portfolio weights, KKT stationary condition writes:

$$ \frac{\partial \mathcal{L}}{\partial x_t^\prime}(x_t^\prime) = \frac{\partial}{\partial x_t^\prime} \left( E_t[ x_t^\prime R_{t+1} ] - \lambda \left( 1 - \alpha^2 \left( E_t[\left(x_t^\prime R_{t+1}\right)^2] - E_t[x_t^\prime R_{t+1}]^2 \right) \right) \right) = 0$$

By linearity of the expectation operator, the RHS equally writes $$ \frac{\partial \mathcal{L}}{\partial x_t^\prime}(x_t^\prime) = E_t[ R_{t+1} ] + \lambda \alpha^2 \left( E_t[2(x_t^\prime R_{t+1})R_{t+1}] - 2 E_t[x_t^\prime R_{t+1}] E_t[R_{t+1}] \right) $$

Finally, because the portfolio weights $x_t$ are known at time $t$ \begin{align} \frac{\partial \mathcal{L}}{\partial x_t^\prime}(x_t^\prime) &= E_t[ R_{t+1} ] + 2 x_t^\prime \lambda \alpha^2 \left( E_t[R_{t+1}^2] - E_t[R_{t+1}]^2 \right) \\ &= E_t[ R_{t+1} ] + \tilde{\lambda} x_t^\prime \text{Var}(R_{t+1}) \end{align}

Is that the result given in you referenced paper? I can't quite confirm since it appears to be behind a paywall.

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