Quadratic Transaction Costs in Certainty-Equivalent Portfolio Rebalancing

# Maximum Certainty Equivalent Portfolio with Transaction Costs


# Maximum Certainty Equivalent Portfolio with Transaction Costs












Out of curiosity I tried to compute the portfolio weights of a maximum certainty equivalent allocation, however, by incorporating (quadratic) transaction costs. However, my result is not as intuitive as I thought =( I would be happy for each and every hint to solve this problem:

Let the parameters of the return distribution be $\Sigma$ and $\mu$. The current allocation vector is $\omega_c$. The risk aversion factor of the investor is defined as $\gamma$. When shifting his wealth to allocation $\alpha$, the investor pays a fee of the form $T = c(\alpha - \omega_c)'(\alpha - \omega_c)$ with some parameter $c$, therefore transaction costs increase quadratically by factor $c$. Therefore, at time point $t+1$ the investor expects the portfolio returns to be $$\mu_\text{PF} = \alpha'\mu - T(\alpha,\omega_c)$$ and the corresponding variance of the portfolio $$\sigma^2_\text{PF} = \alpha'\Sigma\alpha.$$ In one line, the allocation is chosen as the solution to the maximization problem $$\alpha^* = \arg \max _{\sum \alpha = 1} \alpha'\mu - T(\alpha,\omega_c) - \frac{\gamma}{2}\alpha'\Sigma\alpha.$$ Equivalently, we have: $$\alpha^* = \omega_c + \arg \max _{\sum \Delta = 0} (\omega_c+\Delta)'\mu - c\Delta'\Delta - \frac{\gamma}{2}(\omega_c+\Delta)'\Sigma(\omega_c+\Delta).$$ $$ \Delta^* = \arg \max _{\sum \Delta = 0} \underbrace{\omega_c '\mu - \frac{\gamma}{2}\omega_c'\Sigma\omega_c}_{CE(\omega_c)} +\Delta'\mu - c\Delta'\Delta - \frac{\gamma}{2}\Delta'\Sigma\Delta - \gamma \Delta'\Sigma\omega_c.$$ $$ \Delta^* = \arg \max _{\sum \Delta = 0} \Delta'\mu - \Delta'\underbrace{(c I + \frac{\gamma}{2}\Sigma)}_{:=A}\Delta - \gamma \Delta'\Sigma\omega_c.$$

The first-order conditions take the form: $$\mu - 2A\Delta - \gamma \Sigma\omega_c -\lambda\iota= 0$$ $$ \iota ' \Delta = 0$$ It follows that $$A^{-1} (\mu-\gamma \Sigma \omega_c - \lambda \iota) = 2\Delta$$ Evaluating $\iota'\Delta = 0$ with $\Delta$ as above results in $$\lambda = \frac{1}{\iota' A^{-1}\iota}\iota'A^{-1}[\mu - \gamma \Sigma \omega_c]$$ Plug-in gives $$\Delta = A^{-1} (I - \frac{1}{\iota'A^{-1}\iota} \iota' A^{-1}\iota) (\mu-\gamma \Sigma \omega_c ) = 0\iota.$$ In other words, no matter how sub-optimal the current allocation and irrespective of the sice of $c$, there will never be any rebalancing. I do not believe this result but I also do not see the mistake in my computations. Anyone an idea, where did I miss something/ did something wrong?

## Answer by krise (score 6, accepted)

https://quant.stackexchange.com/a/30550

Seems like a small mistake in the last equation. It should read

$\Delta^* = A^{-1} \left[\mu-\gamma \Sigma \omega_c - \frac{1}{\iota'A^{-1}\iota} \iota' A^{-1}(\mu-\gamma \Sigma \omega_c )\iota\right]$,

which is not equivalent to your result.

## Answer by Quantuple (score 4)

https://quant.stackexchange.com/a/30551

You would like to solve the following optimisation problem:

\begin{gather} \Delta^* = \arg \max_\Delta \Delta^T\mu - \Delta^T A \Delta - \gamma \Delta^T \Sigma\omega_c\\ \text{subject to:}\quad \Delta^T \mathbf{1} = 0 \end{gather}

Build the Lagrangian $$ \mathcal{L}(\Delta,\lambda) = \Delta^T\mu - \Delta^T A \Delta - \gamma \Delta^T \Sigma\omega_c - \lambda(\Delta^T \mathbf{1}) $$

First order KKT conditions then yield:

$$\partial_\Delta \mathcal{L}(\Delta^*,\lambda^*) = \mu - A\Delta^* - \gamma\Sigma\omega_c - \lambda^*\mathbf{1} = \mathbf{0} \tag{1}$$

along with $\partial_\lambda \mathcal{L}(\Delta^*,\lambda^*)=0$ (which is precisely the constraint equation)

$(1)$ is equivalent to writing $$ \Delta^* = A^{-1}(\mu-\lambda^*\mathbf{1} - \gamma\Sigma\omega_c) \tag{2}$$

Plugging this into the constraint equation then gives: \begin{align} &(\mu-\lambda^*\mathbf{1} - \gamma\Sigma\omega_c)^T(A^{-1})^T \mathbf{1} = 0 \\ &\mu^T (A^{-1})^T \mathbf{1} -\lambda^*\mathbf{1}^T (A^{-1})^T \mathbf{1} -\gamma\omega_c^T\Sigma^T (A^{-1})^T \mathbf{1} = 0 \end{align} Noting that $A := \frac{\gamma}{2}\Sigma + cI$ is symmetric, i.e. $(A^{-1})^T = A^{-1}$ further yields, as you mention: $$ \lambda^* = \frac{1}{\mathbf{1}^T A^{-1} \mathbf{1} } \mathbf{1}^T A^{-1} (\mu -\gamma \Sigma \omega_c ) \tag{3}$$ Plugging $(3)$ into $(2)$ finally gives: $$ \Delta^* = A^{-1}\left( \mu - \gamma\Sigma\omega_c - \frac{1}{\mathbf{1}^T A^{-1} \mathbf{1} } \mathbf{1}^T A^{-1} (\mu -\gamma \Sigma \omega_c ) \mathbf{1} \right) $$