# Roth vs Traditional, which one is better?

There are many websites describing the benefit of using Roth 401k versus traditional 401k, or Roth IRA versus traditional IRA. The common recommendation is that you should use Roth if you’d expect the income tax rate after retirement will be higher than what it is today. But why is that?

Let ${t}_{c}\in {\mathbb{R}}^{+}$ be the current tax rate and ${t}_{r}\in {\mathbb{R}}^{+}$ be the tax rate after retirement. Let $p\in {\mathbb{R}}^{+}$ denote the principle of an investment into either 401k or IRA before tax, and $n\in {\mathbb{Z}}^{+}$ be the number of compounding which happens between the time of the investment and the time of withdrawl after retirement. For simplicity, let us assume the constant rate of return at $r\in \mathbb{R}$. Furthermore, assume that $0\le {t}_{c}<1$ and $0\le {t}_{r}<1$ because if the tax rate were 100% or grater, there won’t be any money left to withdraw. Similarly, assume $-1 since 100% investment loss would imply there is nothing to withdraw.

In traditional 401k or IRA, the investment is made with pre-tax dollars $p$ so after $n$ investment periods, we have: $\left(1+r{\right)}^{n}p$ Upon widthdrawl, however, the income tax at the rate of ${t}_{r}$ is paid so the total amount available for withdrawal is: ${w}_{Traditional}=\left(1-{t}_{r}\right)\left(1+r{\right)}^{n}p$

In Roth 401k or IRA, the investment is made with after-tax dollars $\left(1-{t}_{c}\right)p$. Over the same $n$ invement periods at the constant return of $r$, we have: $\left(1+r{\right)}^{n}\left(1-{t}_{c}\right)p$ Upon widthdrawl, no income tax is paid on the investment return so the total amount available for withdrawal is: ${w}_{Roth}=\left(1+r{\right)}^{n}\left(1-{t}_{c}\right)p$

Now, ${w}_{Traditional}>{w}_{Roth}$ if and only if $\left(1-{t}_{r}\right)\left(1+r{\right)}^{n}p>\left(1+r{\right)}^{n}\left(1-{t}_{c}\right)p$ Removing the common factor $\left(1+r{\right)}^{n}p$ from both sides yield: $1-{t}_{r}>1-{t}_{c}$ Subtracting 1 from both sides and flipping the sign yields: ${t}_{r}<{t}_{c}$ Hence tradtional 401k or IRA would result in a grater amount to withdraw if and only if ${t}_{r}<{t}_{c}$.

Notice that this result is independent of $n$, $r$, and, $p$. In particular, making $r$ more generic by assuming returns of ${r}_{1},{r}_{2},...,{r}_{n}$ over $n$ investment periods, we see that the above inequality simply becomes and yields the same result:

$\left(1-{t}_{r}\right)\left(\prod _{i=1}^{n}\left(1+{r}_{i}\right)\right)p>\left(\prod _{i=1}^{n}\left(1+{r}_{i}\right)\right)\left(1-{t}_{c}\right)p$