# Interest Rate Parity & Linear Approximations

• Sep 18th 2007, 12:16 PM
NYKarl
Interest Rate Parity & Linear Approximations
Hello,
Can anyone explain how my finance professor arrives at the following linear approximation of the interest rate parity principle? My log skills are failing me.
Thanks for the help!
B

[1+ (Intfc/(360/k)] = (F/S)x[1+ (IntUSD/(360/k)]

Taking natural logs, linear approximation is:

Intfc - IntUSD = ((F-S)/S)*(360/k)

ps:

Intfc = interest rate of foreign currency
IntUSD = interest rate of US Dollar
F = Forward exchange rate
S = Spot exchange rate
K = # of days
• Sep 21st 2007, 02:10 PM
TKHunny
The idea takes quite a bit more than logarithms. This one needs Taylor Series.

$\displaystyle ln(1+x) = x - \frac{1}{2}x^{2} + \frac{1}{3}x^{3} - \frac{1}{4}x^{4} + ...$

The Linear approximation, then, $\displaystyle ln(1+x)\; \approx\;x$, simply discarding all the higher order terms. If x is close to zero, this is not an unreasonable approximation. I would check that stipulation before jumping in with both feet! (Note: Since these are all interest rates, or differences of interest rates, it seems a rational process.)

I'll simplify your notation a bit, using IFC and IUS for the interest rates.

Also, I realize there is a point to be made by writing the ratios that way, but $\displaystyle \frac{k*IFC}{360}$ is wonderfully simpler to handle.

That's all the background.

We have $\displaystyle \left(1+\frac{k*IFC}{360}\right)\;=\;\frac{F}{S}*\ left(1+\frac{k*IUS}{360}\right)$

Now, it's time for the logarithms...

$\displaystyle log\left(1+\frac{k*IFC}{360}\right)\;=\;log\left(\ frac{F}{S}*\left(1+\frac{k*IUS}{360}\right)\right)$

That was the easy part! This can be slightly rewritten, using log rules:

$\displaystyle log\left(1+\frac{k*IFC}{360}\right)\;=\;log\left(\ frac{F}{S}\right)\;+\;log\left(1+\frac{k*IUS}{360} \right)$

We are almost ready to use the linear approximation. The only problem is that pesky F/S term. It doesn't quite look like what we want. Let's just beat it up until it does.

$\displaystyle \frac{F}{S}\;=\;1 + \frac{F}{S} - 1\;=\;1 + \frac{F-S}{S}$

Is this starting to look familiar? We now have:

$\displaystyle log\left(1+\frac{k*IFC}{360}\right)\;=\;log\left(1 +\frac{F-S}{S}\right)\;+\;log\left(1+\frac{k*IUS}{360}\righ t)$

This lends itself directly to the linear approximation of all three terms, simply

$\displaystyle \frac{k*IFC}{360}\;=\;\frac{F-S}{S}\;+\;\frac{k*IUS}{360}$

This is now quite easily solved for IFC - IUS. I'll leave the algebra to the interested and motivated student.
• Sep 22nd 2007, 10:02 AM
NYKarl
Thanks and Follow-Up

I have two follow ups- first, can you help me see what point my professor is trying to make in clumsily putting the (360/k) term in the denominator? And second, how are you able to type such nice looking fractions and brackets?

Many thanks again- you've really been tremendously helpful.
B
• Sep 22nd 2007, 10:42 AM
TKHunny
The notation emphasizes $\displaystyle \frac{Annual\;Interest\;Rate}{Portion\;of\;a\;Year }$, That's all. It is good to emphasize this until the student "gets it". After that, it serves no purpose.

The fancy parentheses and various structures are courtesy of LaTeX. There are various code generators available for free download, but I normally just type it all myself. It can be a bit tedious, but is well worth the effort in clarity.

Note: You don't necessarily need the Taylor Series. That is the generalized principle, but other, simpler solutions also exist. In particular, one concept that usually appears WAY before Taylor Series often is called "Linearization". It is just using the first derivative to create a local and linear approximation. However, should the day come that you want a quadratic estimate, or a cubic estimate, nothing additional is required if you have the Taylor Series.
• Sep 29th 2007, 06:36 PM
NYKarl
Thanks
Thx TKHunny. You've been really helpful.take care
B