Ranking $ d, i, d^{(m)}, i^{(m)}, \delta$

Question

Any actuary or anyone studying mathematics of finance out there? Please help me out. How can I prove or show that $ d< d^{(m)}< \delta< i^{(m)}<i,$ for $m > 1$. Thanks a lot !!!

Answer 1

Let us make the following assumptions regarding notation:

1. $d$ and $i$ are annual effective discount and interest rates, respectively
2. $d^{(m)}$ and $i^{(m)}$ are annual nominal discount and interest rates compounded $m$-ly, respectively
3. $\delta$ is the force of interest

Let us further assume that $d, d^{(m)}, \delta, i^{(m)}, i$ are non-zero.

Then, by definition, $1+i = \left( {1+\frac{i^{(m)}}{m}} \right) ^m$ and $(1-d)^{-1} = \left( {1-\frac{d^{(m)}}{m}} \right) ^{-m}$.

Also, by definition, $1+i = (1-d)^{-1} = e^{\delta}$.

Now we know that $d^{(1)} = d$ and that $i^{(1)} = i$ and we also know that $d^{(\infty)}=i^{(\infty)}=\delta$.

We can show that $d<d^{(m)}<\delta<i^{(m)}<i$ for $m>1$ if we can show that $d^{(m)}$ is strictly increasing in $m$ and that $i^{(m)}$ is strictly decreasing in $m$ with $m>1$.

For clarity, recall that

$$ \left( {1+\frac{i^{(m)}}{m}} \right)^m = e^{\delta} \implies {1+\frac{i^{(m)}}{m}} = e^{\delta / m} \implies i^{(m)} = m\left(e^{\delta / m}-1\right)$$

So we have

$$\frac{d}{d m} i^{(m)} = e^{\delta / m}-1 + m\left(-\frac{\delta}{m^2} e^{\delta / m}\right) = e^{\delta / m}-1 -\frac{\delta}{m} e^{\delta / m} = e^{\delta / m}\left(1 -\frac{\delta}{m} - e^{-\delta / m}\right)$$

Recall the Taylor expansion

$$e^{-x} = \sum_{n=0}^{\infty} \frac{(-x)^n}{n!} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \dots $$

Hence $e^{-x} > 1-x$ for $x \neq 0$. For negative $x$ this should be clear; for positive $x$ notice how every negative term after $\frac{x^2}{2!}$ in the expansion is succeeded by an absolutely larger positive term.

So

$$ e^{-\delta / m} > 1 -\frac{\delta}{m} \implies 1 -\frac{\delta}{m} - e^{-\delta / m} < 0 \implies \frac{d}{d m} i^{(m)} < 0$$

So $i^{(m)}$ is strictly decreasing in $m$; in particular, $i^{(m)}$ is strictly decreasing for $m>1$.

Hence

$$\delta<i^{(m)}<i$$

Now

$$ \left( {1-\frac{d^{(m)}}{m}} \right)^{-m} = e^{\delta} \implies {1-\frac{d^{(m)}}{m}} = e^{-\delta / m} \implies d^{(m)} = m\left(-e^{-\delta / m}+1\right)$$

Differentiating with respect to $m$, we have

$$\frac{d}{d m} d^{(m)} = -e^{-\delta / m}+1 + m\left(-\frac{\delta}{m^2} e^{-\delta / m}\right) = -e^{-\delta / m}+1 -\frac{\delta}{m} e^{-\delta / m} = -e^{-\delta / m}\left(1 +\frac{\delta}{m} - e^{\delta / m}\right)$$

Now recall that

$$e^{x} = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots $$

So $e^{x} > 1+x$ for $x>0$. Hence

$$ e^{\delta / m} > 1 + \frac{\delta}{m} \implies 1 +\frac{\delta}{m} - e^{\delta / m} < 0 \implies \frac{d}{d m} d^{(m)} > 0$$

So we have shown that $d^{(m)}$ is strictly increasing in $m$. In particular, $d^{(m)}$ is strictly increasing for $m > 1$. Hence

$$\delta>d^{(m)}>d$$

Putting these inequalities together, we have $d<d^{(m)}<\delta<i^{(m)}<i$, as required.