This is a follow-up to Redirecting Taxable investments to Roth 401(k). The analysis is conducted in terms of capital appreciation and dividend yield rates. It also incorporates the additional basis created in a taxable account by dividend reinvestment.
Refer to the aforementioned post for the definitions of "Default strategy" and "Roth strategy" used below.
| Term | Description |
|---|---|
| $P_0$ | Amount withdrawn from the taxable brokerage in the Roth strategy |
| $B_0$ | Cost basis of $P_0$ |
| $B_N$ | Normalized cost basis; $B_N = B_0/P_0$ $(> 0)$ |
| $r_c$ | Annual capital appreciation rate |
| $r_d$ | Annual dividend yield |
| $T_c$ | Annual capital gains tax rate; $0 < T_c \le 1$ |
| $T_d$ | Annual dividend tax rate; $0 < T_d \le 1$ |
Assumptions
- $r_c$ and $r_d$ are annual returns measured relative to the beginning-of-year portfolio value
- Dividends reinvested are net dividend taxes
- For passive long-term investing it is reasonable to assume that both $r_c, r_d > 0$ and $0 < B_N \le 1$
Default strategy
"Net pay (after-tax) is used to fund living expenses"
The balance in the taxable account at the end of year $k$ $(\ge 1)$ is,
$$\begin{align*}P_k &= P_{k-1} + P_{k-1}r_c + P_{k-1}r_d(1-T_d)\\&=P_{k-1}[1+r_c + r_d(1-T_d)]\\&=P_0[1+r_c + r_d(1-T_d)]^k\\&=P_0 G_t^k\end{align*}$$
where $G_t = 1 + r_c + r_d (1 - T_d)$ is the yearly gain multiplier in the taxable account. At the end of the $m^{\text{th}}$ year,
$$P_\text{taxable, pre-tax} = P_m = P_0 G_t ^m$$
The additional basis created by dividend reinvestments in year $k$ is $\Delta B_k$ $=$ $P_{k-1}r_d(1-T_d)$ $=$ $P_0 G_t^{k-1} r_d (1 - T_d)$ and, therefore, after $m$ years, the basis is
$$\begin{align*}B_m &= B_0 + \sum_{k=1}^m \Delta B_k\\&= B_0 + P_0 r_d (1-T_d) \sum_{k=1}^mG_t^{k-1}\\&=B_0 + P_0 r_d (1-T_d) \frac{G_t^m - 1}{G_t - 1} \quad (\text{for } G_t \ne 1)\end{align*}$$
Final after-tax balance (net capital gains tax)
$$\begin{align*}P_\text{taxable, after-tax} &= P_m (1 - T_c) + B_m T_c\\&=P_0 G_t^m (1 - T_c)\\&\quad +\ [B_0 + P_0r_d(1-T_d)\frac{G_t^m - 1}{G_t - 1}] T_c\end{align*}$$
Roth strategy
"Cash equivalent to net pay is withdrawn from taxable brokerage to fund living expenses while the net pay - after setting aside some amount to pay capital gains tax - is used to contribute to the Roth via MBDR"
If $P_0$ is withdrawn at the start, then the initial principal available for Roth is $P_0 (1 - T_c) + B_0 T_c$. This principal grows without tax drag, i.e., with yearly gain multiplier $G_r$ $=$ $1 + r_c + r_d$, in a Roth to yield, at the end of the $m^{\text{th}}$ year,
$$P_\text{roth, pre-tax} = P_\text{roth, after-tax} = [P_0 (1 - T_c) + B_0 T_c ] G_r^m $$
Analysis of pre-tax balances
The advantage of the Roth over the taxable account for the pre-tax balance, normalized to $P_0$, is $(P_\text{roth, pre-tax} - P_\text{taxable, pre-tax})/P_0$, which, after some algebra becomes
The pre-tax balance is important for a legacy/inheritance goal as heirs get a step-up in basis in the taxable account. This can put the Roth strategy at a disadvantage if the Roth balance is lower than the pre-tax balance of the taxable account at inheritance (assuming the heirs liquidate the account immediately upon inheritance). If the heirs hold on to the account, it can give the Roth time to catch up. In general, once the Roth balance exceeds the pre-tax balance of the taxable account, the Roth strategy produces a larger inherited asset even after giving the taxable account the benefit of a full basis step-up.
The larger the $B_N$ the greater the $\Delta_\text{pre-tax}$ and the greater the benefit that the Roth strategy provides.
In the beginning, $m = 0$ and $\Delta_\text{pre-tax} = -(1-B_N)T_c < 0$ whenever $B_N < 1$ which is the case for most passive long-term investing. Thus, the Roth strategy starts with a disadvantage on the pre-tax balance. Given enough time, the sign of $\Delta_\text{pre-tax}$ can flip if $G_r > G_t$ which, in this model, requires a positive dividend yield and positive dividend tax rate. (Otherwise, the Roth has no advantage over taxable!) This happens in year $k_\text{flip}$ given by
The investing period must be at least as long as $k_\text{flip}$ for the Roth balance to exceed the pre-tax balance in the taxable account.
The larger the $B_N$, the smaller the $k_\text{flip}$, and the earlier the Roth strategy overcomes the initial deficit.
For a first-order approximation, using $\ln(1+x) \approx x$ when $x \ll 1$, and the fact that $G_t/G_r$ $=$ $1 - (r_dT_d/G_r)$,
$$k_\text{flip} \approx \lceil \frac{(1-B_N) T_c G_r}{r_d T_d} \rceil$$
(Caveat: This approximation is useful for estimating the scale of $k_\text{flip}$; the exact expression should be used to determine the actual 'crossover' year.) The factor $r_dT_d$ in the denominator determines the overall order of $k_\text{flip}$. If $r_d \rightarrow 0$ (low to zero dividend yield), then $k_\text{flip} \rightarrow \infty$. The Roth balance will never exceed the taxable pre-tax balance in this scenario. Under these circumstances, for a legacy/inheritance goal, the optimal strategy is the default strategy and not the Roth.
As an illustration, if $B_N = 0.7$, $r_c$ $=$ $7.177\%/\text{year}$, $r_d$ $=$ $2.75\%/\text{year}$, $T_c = 15\%$, and $T_d = 19\%$, then $k_\text{flip} = 10$ years. Thus, if the liquidation by heirs happens at least $10$ years after the first redirection of assets into the Roth, then the Roth strategy yields a better outcome in this case.
Discussion on $B_N$ for pre-tax
An alternate viewpoint is to inquire into what $B_N$ must be for the Roth strategy to be the optimal choice at the end of $m$ years. Imposing $\Delta_\text{pre-tax} \ge 0$ and assuming $T_c \ne 0$ gives,
As $m \rightarrow 0$, $G_t^m/G_r^m \rightarrow 1$, and $B_N|_{m=0} \ge 1$. The (normalized) cost basis must be high for the Roth strategy to be the optimal choice right from day $0.$
The right hand side of the expression monotonically decreases with $m$. Thus, a lower $B_N$ can be sustained as the investing period becomes longer.
As an illustration, if $r_c$ $=$ $7.177\%/\text{year},$ $r_d$ $=$ $2.75\%/\text{year},$ $T_c$ $=$ $15\%,$ $T_d$ $=$ $19\%,$ and $m$ $=$ $10,$ then $B_N$ $\ge$ $0.6898.$ Thus, the normalized cost basis must be at least $68.98\%$ for the Roth to yield a better outcome after $10$ years in this case.
Analysis of after-tax balances
The advantage of the Roth over the taxable account for the after-tax balance, normalized to $P_0$, is $(P_\text{roth, after-tax} - P_\text{taxable, after-tax})/P_0$, which, after some algebra becomes
The after-tax balance is important for a retirement spending goal.
- The first term $(1 - T_c)(G_r^m - G_t^m)$ captures the benefit of avoiding annual dividend taxation in the Roth and is nominally positive
- The second term $B_N T_c (G_r^m - 1)$ represents the tax-free growth of the transferred (normalized) basis and is nominally positive
- The higher the original (normalized) basis, the greater the benefit
- The third term $-r_d T_c (1 - T_d)$$(G_t^m - 1)$$/$$(G_t - 1)$ represents the correction for the capital gains tax benefit in the taxable account due to the additional basis resulting from reinvested dividends and is nominally negative
Discussion on $B_N$ for after-tax
Imposing $\Delta_\text{after-tax}$ $\ge 0$ and assuming $T_c \ne 0$, gives,
Based on the analysis in Appendix A, the right hand side of the above inequality decreases with $m$. Its limiting value, as $m \rightarrow 0^+,$ provides the worst-case lower bound, which is
$$B_N |_{m=0} \ge \frac{r_d (1 - T_d)}{G_t - 1} \frac{\ln(G_t)}{\ln(G_r)} - \left( \frac{1}{T_c} - 1 \right) \left( 1 - \frac{\ln(G_t)}{\ln(G_r)} \right)$$
This worst-case lower bound for $B_N$ must be met for the Roth strategy to prevail at all times. A lower $B_N$ can nevertheless be sustained if the investments are held in the Roth account for an appropriately longer period.
In particular, if $T_d = 0$, then $G_r = G_t$ and
$$B_N|_{m=0} \ge \frac{r_d}{r_c + r_d}$$
As an illustration, if $r_c$ $=$ $7.177\%/\text{year}$, $r_d$ $=$ $2.75\%/\text{year}$, $T_c = 15\%$, and $T_d = 19\%$, then $B_N|_{m=0} \ge -0.06033$. But since $B_N$ cannot be negative, it follows that even $B_N = 0$ guarantees that $\Delta_\text{after-tax} \ge 0$ from the start in this scenario. The same parameters, but with $T_d = 0,$ yield $B_N|_{m=0} = 0.2770$ which is the threshold that $B_N$ must exceed so that $\Delta_\text{after-tax} \ge 0$ from the start.
Additional insight into the lower bound of $B_N$ is gleaned with a first-order approximation. If both $r_c, r_d \ll 1$, then using $\ln(1+\epsilon)$ $\approx \epsilon$ when $\epsilon \ll 1,$ $\ln(G_r)$ $\approx$ $r_c + r_d$ $=$ $G_r -1$ etc.. Then $B_N|_{m=0}$ becomes, after some algebra,
$$B_N|_{m=0} \gtrapprox \frac{r_d}{r_c + r_d}\left(1 - \frac{T_d}{T_c} \right)$$
If $T_c < T_d$ then, as $B_N$ cannot be negative, it follows that $B_N \ge 0$ is sufficient to ensure $\Delta_\text{after-tax}$ $\ge 0.$ Conversely, if $T_c > T_d$, then the lower bound of $B_N$ must be satisfied for the Roth strategy to be the optimal choice.
Summary
The goal determines the relevant measure of interest: pre-tax wealth is the relevant measure for a legacy/inheritance goal because of the step-up in basis available to taxable assets upon inheritance, while after-tax wealth is the relevant measure when the assets are intended to fund the owner's retirement.
For a legacy/inheritance goal, the Roth strategy is not automatically superior. (If the taxable asset has a complete basis step-up upon inheritance and produces essentially no ongoing taxable distributions, the taxable account can be superior indefinitely.) In general, the redirected assets must be held in the Roth for at least $k_\text{flip}$ years for the Roth strategy to gain an advantage. The Roth starts with a lower nominal balance, but eventually overtakes the taxable account because the Roth compounds at the full return while taxable account compounds at the after-dividend-tax return. A higher (normalized) cost basis, a higher dividend yield, and a higher dividend tax rate all lower $k_\text{flip}$ and make the Roth strategy beneficial earlier. Further, the higher the (normalized) cost basis, the greater the advantage of the Roth strategy.
For a retirement spending goal, the Roth strategy is the optimal choice under the assumptions of this analysis, provided the (normalized) cost basis satisfies the condition for the intended investing period. The higher the (normalized) cost basis and the higher the dividend tax rate compared to the capital gains tax rate, the greater the advantage of the Roth strategy.
Appendix A - Dependency of the lower bound of $B_N$ on $m$ for after-tax
To understand how the lower bound of $B_N$ depends on the investing period $m$, it is sufficient to look at the behavior of each term. The lower bound of $B_N$ can be expressed as
$$B_N \ge \alpha \left( \frac{G_t^m - 1}{G_r^m - 1} \right) - \beta \left( \frac{G_r^m - G_t^m}{G_r^m - 1} \right)\quad \alpha, \beta > 0$$
where neither $\alpha$ nor $\beta$ depend on $m$.
For the first term, if $1 < G_t^m < G_r^m$ (a reasonable assumption for passive long-term investing), it follows that $(G_t^m - 1)/(G_r^m - 1) < 1$. Further, this term monotonically decreases with $m$ because the denominator increases faster than the numerator.
The second term, can be rewritten as
$$\frac{G_r^m - G_t^m}{G_r^m - 1} = 1 - \left( \frac{G_t^m - 1}{G_r^m - 1} \right)$$
As $(G_t^m - 1)/(G_r^m - 1)$ is already known to be monotonically decreasing with $m$, it follows that $(G_r^m - G_t^m)/(G_r^m - 1)$ monotonically increases with $m$.
Thus, as $m$ increases, the contribution from $\alpha$ reduces while the contribution from $\beta$ increases. Both serve to reduce the value of the right hand side. Hence, the lower bound of $B_N$ has a maximum value at $m=0$. The relative magnitudes of $\alpha$ and $\beta$ determine the sign of the lower bound of $B_N$.
Sidebar for $(G_t^m - 1)/(G_r^m - 1)$
Using $a^\epsilon$ $\approx$ $1 + \epsilon\ln(a)$, for $0 < \epsilon \ll 1$, it follows that $\lim_{m \rightarrow 0^+}$ $G_r^m$ $=$ $1 + m\ln(G_r)$ etc. Then,
$$\lim_{m \rightarrow 0^+} \frac{G_t^m - 1}{G_r^m - 1} = \frac{\ln(G_t)}{\ln(G_r)}$$
This is the upper bound for the term.
The lower bound for the term, as $m \rightarrow \infty$, is $0$ because $G_t / G_r < 1$.
Sidebar for $(G_r^m - G_t^m)/(G_r^m - 1)$
Using the same procedure as above,
$$\lim_{m \rightarrow 0^+}\frac{G_r^m - G_t^m}{G_r^m - 1} = 1 - \frac{\ln(G_t)}{\ln(G_r)}$$
This is the lower bound for the term.
The upper bound for the term is given by
$$\begin{align*}\lim_{m \rightarrow \infty} \frac{G_r^m - G_t^m}{G_r^m - 1} &= \lim_{m \rightarrow \infty} \left( 1 - \frac{G_t^m - 1}{G_r^m - 1} \right)\\&= 1 - 0 = 1\end{align*}$$
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