Upper limit for tax optimisation

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For a portfolio taxed on realisation, the portfolio value after \(y\) years is given by:

\[D_{y}\left(k,\left\{ a_{i}\right\} \right)=k\cdot\prod_{i=1}^{y}\left(1+a_{i}\right)-s_{p,q,g,y}\left(k\cdot\prod_{i=1}^{y}\left(1+a_{i}\right)-k,\left\{ r_{i}\right\} \right)\]

\(k\) is the starting capital, \(\left\{ a_{i}\right\}\) is a set of yearly returns and \(y\) is the number of years. Here, the tax is given by:

\[s_{p,q,g,y}\left(x,\left\{ r_{i}\right\} \right)=p\cdot\min\left(x,g\cdot\prod_{i=1}^{y}\left(1+r_{i}\right)\right)+q\cdot\max\left(0,x-g\cdot\prod_{i=1}^{y}\left(1+r_{i}\right)\right)\]

Here, \(x\) is the capital to be taxed, \(p\) is the lower tax rate, \(q\) is the higher tax rate, \(g\) is the progression limit and \(\left\{ r_{i}\right\}\) is a set of percentage increases of the progression limit. Since the portfolio value (including tax) only depends on the starting value and the final value of the portfolio, the following holds:

\[\prod_{i=1}^{y}\left(1+x_{i}\right)=\left(1+\bar{x}\right)^{y}\]

It therefore follows that:

\[D_{y}\left(k,\bar{a}\right)=k\cdot\left(1+\bar{a}\right)^{y}-s_{p,q,g,y}\left(k\cdot\left(1+\bar{a}\right)^{y}-k,\bar{r}\right)\]

with:

\[s_{p,q,g,y}\left(x,\bar{r}\right)=p\cdot\min\left(x,g\cdot\left(1+\bar{r}\right)^{y}\right)+q\cdot\max\left(0,x-g\cdot\left(1+\bar{r}\right)^{y}\right)\]

Given a portfolio that yields a return, the return can be realised or remain unrealised; this gives two situations for the return from the portfolio. If the portfolio is realised, the part from the return will develop in the following way:

\[\begin{eqnarray} D_{\mathrm{realiseret}} &=& \left.D_{y}\left(m\cdot k,\bar{a}\right)\right|_{g=g_{\mathrm{effektiv}}} \\ &=& m\cdot k\cdot\left(1+\bar{a}\right)^{y}-s_{p,q,g_{\mathrm{effektiv}},y}\left(m\cdot k\cdot\left(1+\bar{a}\right)^{y}-m\cdot k,\bar{r}\right) \end{eqnarray}\]

Here, \(m\) is one minus the tax rate paid at realisation, and \(g_{\mathrm{effektiv}}\) is the progression limit minus the part of the progression limit that is used up by the underlying portfolio. The other situation is that the return remains unrealised; this portfolio will develop in the following way:

\[\begin{eqnarray} D_{\mathrm{urealiseret}} &=& D_{y}\left(k,\bar{a}\right)+s_{p,q,g,y}\left(k\cdot\left(1+\bar{a}\right)^{y}-k,\bar{r}\right)-s_{p,q,g_{\mathrm{effektiv}},y}\left(k\cdot\left(1+\bar{a}\right)^{y},\bar{r}\right) \\ &=& k\cdot\left(1+\bar{a}\right)^{y}-s_{p,q,g_{\mathrm{effektiv}},y}\left(k\cdot\left(1+\bar{a}\right)^{y},\bar{r}\right) \end{eqnarray}\]

Note that tax is paid on the entire portfolio, since the “starting capital” is unrealised. In the situation where the underlying portfolio uses the entire progression limit, the tax is given by:

\[\left.s_{p,q,0,y}\left(x,\bar{r}\right)\right|_{x>0}=q\cdot x\]

The number of years that must pass before realisation for this tax optimisation to no longer pay off can be found by setting:

\[\begin{eqnarray} \left.D_{\mathrm{urealiseret}}\right|_{g_{\mathrm{effektiv}}=0} &=& \left.D_{\mathrm{realiseret}}\right|_{g_{\mathrm{effektiv}}=0} \\ k\cdot\left(1+\bar{a}\right)^{y}-q\cdot k\cdot\left(1+\bar{a}\right)^{y} &=& m\cdot k\cdot\left(1+\bar{a}\right)^{y}-q\cdot\left(m\cdot k\cdot\left(1+\bar{a}\right)^{y}-m\cdot k\right) \\ \left(1+\bar{a}\right)^{y}\cdot\left(k-q\cdot k\right) &=& \left(1+\bar{a}\right)^{y}\cdot\left(m\cdot k-q\cdot m\cdot k\right)+q\cdot m\cdot k \\ \left(1+\bar{a}\right)^{y}\cdot\left(k-q\cdot k-m\cdot k+q\cdot m\cdot k\right) &=& q\cdot m\cdot k \\ \left(1+\bar{a}\right)^{y} &=& \frac{q\cdot m\cdot k}{k-q\cdot k-m\cdot k+q\cdot m\cdot k} \\ y\cdot\log\left(1+\bar{a}\right) &=& \log\left(\frac{q\cdot m\cdot k}{k-q\cdot k-m\cdot k+q\cdot m\cdot k}\right) \\ y &=& \frac{\log\left(\frac{q\cdot m\cdot k}{k-q\cdot k-m\cdot k+q\cdot m\cdot k}\right)}{\log\left(1+\bar{a}\right)} \\ y &=& \frac{\log\left(\frac{q\cdot m}{1-q-m+q\cdot m}\right)}{\log\left(1+\bar{a}\right)} \label{eq1}\tag{1} \end{eqnarray}\]

The above equation is an upper limit for when tax optimisation can pay off. Note that the same equation is obtained with \(g_{\mathrm{effektiv}}=\infty\), i.e. if the progression limit is never reached. For this progression limit, \(q\rightarrow p\) and \(m=1-p\), and thereby:

\[\begin{eqnarray} y &=& \frac{\log\left(\frac{p\cdot\left(1-p\right)}{1-p-\left(1-p\right)+p\cdot\left(1-p\right)}\right)}{\log\left(1+\bar{a}\right)} \\ &=& \frac{\log\left(\frac{p\cdot\left(1-p\right)}{1-p-1+p+p\cdot\left(1-p\right)}\right)}{\log\left(1+\bar{a}\right)} \\ &=& \frac{\log\left(\frac{p\cdot\left(1-p\right)}{p\cdot\left(1-p\right)}\right)}{\log\left(1+\bar{a}\right)} \\ &=& \frac{\log\left(1\right)}{\log\left(1+\bar{a}\right)} \\ &=& 0 \end{eqnarray}\]

The lower limit is therefore 0 years, and thus not very useful. The upper limit from equation (\ref{eq1}) can now be plotted.

Often it will not pay off to do tax optimisation if there are more than 10 years until the money is needed.

The code used to make the graphs can be found here: skatteoptimering_limit.py

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