∫ Discounted differentiation and integration

Kernel:

Notably changes to differentiation and integration to discount exponential growth.

In economics, wealth is a relative quantity. Especially when you are considering the value of your saving at different times, the value must be compute relative to the the inflation rate.

In this blog, I'm focused on the modification to calculus to work with the continuous discount factor. Especially when you are considering the value of your saving at different times under a inflation rate at an infinitesimal interval.

Suppose that you want to compute the true gaining in of saving at an infinitesimal interval and under an inflation rate of $ \iota $ where

$$ \begin{align*} \iota \gt 0. \end{align*} $$

How would you do that?

∂ Differentiation

Let function $ f(x) $ represents your earning at time $ x $. You can compute the relative earning rate compared to the inflation rate $ \iota $ at time $ x $ like this:

$$ \begin{align*} \frac{d_{\iota} f(x)}{d_\iota x} &= \lim_{\Delta x \to 0} \frac{\iota^{-\Delta x} f(x + \Delta x) - f(x)}{\Delta x} \\ &= \iota^{x}\lim_{\Delta x \to 0} \frac{\iota^{- x - \Delta x} f(x + \Delta x) - \iota^{- x}f(x)}{\Delta x} \\ &= \iota^{x} \frac{d \iota^{- x} f(x)}{d x} \\ \end{align*} $$

$$ \begin{align*} \frac{d_{\iota} f(x)}{d_\iota x} \triangleq \iota^{x} \frac{d \iota^{- x} f(x)}{d x} \triangleq f'_{\iota}(x) \end{align*} $$

For example, if your saving at time $ x $ follows function $ k\iota^{x} $ where $ k $ is a constant, then your relative earning rate is

$$ \begin{align*} \frac{d_{\iota} k\iota^{x}}{d_\iota x} &= \iota^{x} \frac{d \iota^{- x} k\iota^{x}}{d x} \\ &= \iota^{x} \frac{d k}{d x} \\ &= 0 \\ \end{align*} $$

This means that the relative earning rate is zero, which is consistent with the fact that althought the saving is numerically increasing, its value is not going anywhere relative to the inflation rate.

If your saving at time $ x $ never changes and stay $ k $, then your relative earning rate is

$$ \begin{align*} \frac{d_{\iota} k}{d_\iota x} &= \iota^{x} \frac{d \iota^{- x} k}{d x} \\ &= -\log(\iota) \iota^{x}k \iota^{-x} \\ &= - k \log(\iota) \\ \end{align*} $$

This means if the inflation rate is greater than 1, then the relative earning rate is negative. You're losing wealth.

Or in the case that your investment always gives you twice the inflation rate, then your relative earning rate is

$$ \begin{align*} \frac{d_{\iota} (2\iota)^{x}}{d_\iota x} &= \iota^{x} \frac{d \iota^{- x} (2\iota)^{x}}{d x} \\ &= \iota^{x} \frac{d 2^{x}}{d x} \\ &= \log(2) (2\iota)^{x} \\ \end{align*} $$

This means that the relative earning rate is always positive and you're gaining wealth.

∂ Integration

To find the infamous inverse of the derivative, let's try to expand the form:

$$ \begin{align*} f'_{\iota}(x) &= \iota^{x} \frac{d \iota^{- x} f(x)}{d x} \\ &= \iota^{x} \left[\iota^{- x} f'(x) - f(x) \iota^{- x} \log(\iota) \right] \\ &= f'(x) - f(x)\log(\iota) \end{align*} $$

This suspeciously looks like the first order non-homogeneous linear differential equation. Multiply both sides by $ e^{-\log(\iota)x} $:

$$ \begin{align*} f'_{\iota}(x)e^{-\log(\iota)x} &= f'(x)e^{-\log(\iota)x} - f(x)\log(\iota)e^{-\log(\iota)x} \\ f'_{\iota}(x)e^{-\log(\iota)x} &= \frac{d f(x)e^{-\log(\iota)x}}{d x} \\ f(x) &= Ce^{\log(\iota)x} + e^{\log(\iota)x}\int f'_{\iota}(x)e^{-\log(\iota)x} d x \\ f(x) &= C\iota^x + \iota^x\int f'_{\iota}(x)\iota^{-x} d x \end{align*} $$

Therefore,

$$ \begin{align*} f'(x) &= f'_{\iota}(x) + f(x) \log(\iota) \\ f(x) &= C\iota^x + \iota^x\int f'_{\iota}(x)\iota^{-x} d x \triangleq \int f'_\iota(x) d_\iota x \end{align*} $$

When $ f'_{\iota}(x) $ is a constant, the term can be further simplified.

$$ \begin{align*} f(x) &= C{\iota}^x - \frac{f'_{\iota}(x)}{\log(\iota)} \end{align*} $$

In this case, we don't need to do indefinite integration to find the inverse of the derivative!

Let's now try to integrate the previous three examples:

First example, $ f'_{\iota}(x) = 0 $:

$$ \begin{align*} f(x) &= C{\iota}^x - \frac{f'_{\iota}(x)}{\log(\iota)} \\ &= C{\iota}^x - \frac{0}{\log(\iota)} \\ &= C{\iota}^x \end{align*} $$

Which is true, any value that increases at the same rate as the inflation rate has zero relative earning rate.

Second example, $ f'_{\iota}(x) = - k \log(\iota) $:

$$ \begin{align*} f(x) &= C{\iota}^x - \frac{-k\log(\iota)}{\log(\iota)} \\ &= C{\iota}^x + k \end{align*} $$

Since the first term has zero relative earning rate, the constant part therefore contributes to the loss when the inflation rate is greater than 1.

Third example, $ f'_{\iota}(x) = \log(2) (2\iota)^{x} $:

$$ \begin{align*} f(x) &= C{\iota}^x + \iota^x\int \log(2) (2\iota)^{x} \iota^{-x} d x \\ &= C{\iota}^x + (2\iota)^{x} \end{align*} $$

We get back all the original saving functions.

For the sake of completion, let's end this with the definition of the definite integration.

$$ \begin{align*} \int_{a}^{b} f'_{\iota}(x) d_{\iota} x \triangleq f(b) - f(a) = C (\iota^b - \iota^a) + \iota^x \int f'_{\iota}(x) \iota^{-x} d x \Biggr|_{a}^{b} \end{align*} $$

We can use this to compute the total numerical increase in the saving from time $ a $ to $ b $ given the relative earning rate $ f'_{\iota}(x) $ and a boundary condition for solving the constant $ C $.