TITLE: Ranking $ d, i, d^{(m)}, i^{(m)}, \delta$
QUESTION [1 upvotes]: 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 !!!

REPLY [0 votes]: Let us make the following assumptions regarding notation:

$d$ and $i$ are annual effective discount and interest rates, respectively
$d^{(m)}$ and $i^{(m)}$ are annual nominal discount and interest rates compounded $m$-ly, respectively
$\delta$ 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.
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