Tax-deferred compounding

[BWV]

Trying to get some better mathematical insight on why tax-deferred compounding of an investment has an advantage over annual taxation
so R=return, T=tax rate and n=time (years)
If the full return is taxed every year, a portfolio grows at:

(1+R(1-T))ⁿ

if the portfolio is only taxed at the end of the multi-year period, the value is:

((1+R)ⁿ-1)(1-T) +1 or, alternatively (1+R)ⁿ-(1+R)ⁿT+T

For T >0 and R >0, and n > 1 the statement below is true, but not sure how I would prove it - Jenson's inequality perhaps?

(1+R(1-T))ⁿ < (1+R)ⁿ-(1+R)ⁿT+T

If you plot the growth of the two portfolios, for 12% return and 24% tax rate, the values diverge dramatically:

The y/y change on the 'Taxed every year' is simply (1+R(1-T)), however the 'Taxed Deferred' y/y change begins at (1+R(1-T)) but converges over time to (1+R) (at about 50 years in this example), - would like to understand this - taking derivatives with respect to n does not help much as you just get a log(1+r)(1+r)^n terms for both functions.

 

[pasmith]

If $0 < T < 1$ then $(1 - T)^n < (1-T)$ and hence $$\begin{split} (1 + R(1-T))^n &= \sum_{k=0}^n \binom{n}{k} R^k(1-T)^k \\ &\leq \sum_{k=0}^n \binom{n}{k} R^k(1-T) \\ &= (1+R)^n(1-T) \\ &< (1+ R)^n(1-T) + T.\end{split}$$ I asume for y/y you want $$\frac{(1+R)^{n+1}(1-T)+T}{(1+R)^n(1-T) + T} = (1 + R) \frac{(1-T) + T(1+R)^{-n-1}}{(1-T)+T(1+R)^{-n}}.$$ As $n \to \infty$ the right hand side tends to $(1 + R)\dfrac{1-T}{1-T} = 1 + R$.

 

[BWV]

I asume for y/y you want $$\frac{(1+R)^{n+1}(1-T)+T}{(1+R)^n(1-T) + T} = (1 + R) \frac{(1-T) + T(1+R)^{-n-1}}{(1-T)+T(1+R)^{-n}}.$$ As $n \to \infty$ the right hand side tends to $(1 + R)\dfrac{1-T}{1-T} = 1 + R$.

Thanks, that's perfect never fails that my bad algebra rather than my bad calculus is the impediment- although the T is just adjusting for the principal not being taxed, so the logic of that being the key issue is counterintuitive to me. But, it makes sense - if the whole amount, rather than just the gains were taxed (like in a 401K contribution) it would be

(1+R)ⁿ-(1+R)ⁿT , or more simply (1+R)ⁿ(1-T)
then if you compared it to a situation where $1 was taxed up front then on the gains every year after
(1-T) (1+R(1-T))ⁿ
then it is very simply (1+R)ⁿ vs (1+R(1-T))ⁿ

Dumb question, how did you get the
$$\frac {T(1+R)^{-n-1}}{T(1+R)^{-n}}.$$ term?

 

[BWV]

Another more basic math question - how do you get the binomial term here:

If $0 < T < 1$ then $(1 - T)^n < (1-T)$ and hence $$\begin{split} (1 + R(1-T))^n &= \sum_{k=0}^n \binom{n}{k} R^k(1-T)^k \\ &\leq \sum_{k=0}^n \binom{n}{k} R^k(1-T) \\ &= (1+R)^n(1-T) \\ &< (1+ R)^n(1-T) + T.\end{split}$$

 

[PeroK]

Trying to get some better mathematical insight on why tax-deferred compounding of an investment has an advantage over annual taxation
so R=return, T=tax rate and n=time (years)
If the full return is taxed every year, a portfolio grows at:

(1+R(1-T))ⁿ

if the portfolio is only taxed at the end of the multi-year period, the value is:

((1+R)ⁿ-1)(1-T) +1 or, alternatively (1+R)ⁿ-(1+R)ⁿT+T

The simple answer is that whoever takes the tax at the end of each year gets the subsequent growth on that tax. If you don't pay tax, you get subsequent growth on that extra capital (albeit the growth is eventually taxed). But, if you pay the tax, the capital is gone. After the first year we have:

a) Capital $1 + R - RT$

or

b) Capital $1 + R$

In a) you have less capital to grow and the IRS gets the growth on that RT they took at the end of year one. Whereas in b), the IRS eventually takes the tax on the extra RT, but only after you've benefited from the growth on it in subsequent years (even though you pay tax on that growth, it's generally better to pay tax on growth that not to have the capital to grow in the first place).

The numbers for $n = 2$ are easy to see:

a) $B_1 = (1 + R - RT)^2 = 1 + 2R + R^2 -2RT + R^2T^2 - 2R^2T$

b) $B_2 = 1 + \big [(1 + R)^2 - 1 \big ](1 - T) = 1 + (2R + R^2)(1-T) = 1 + 2R + R^2 -2RT - R^2T$

Hence $B_1 = B_2 + R^2T^2 - R^2T = B_2 - R^2T(1 - T) = B_2 - (RT)(R)(1-T)$

And we see that $B_2 > B_1$ by precisely the taxed growth on the extra capital at the end of the first year ($RT$). In other words, you get the growth on the tax you didn't pay at the end of year one and only have to pay the tax on that growth. In case a) you don't get that growth at all.

For $n > 2$ it's the same story of getting the growth (albeit taxed) on all the tax you didn't pay in the earlier years.