Logarithmic approximation: application in CFA Level I and interview questions

Good afternoon everyone,

I’m almost done going through the CFA material for the Level I curriculum and I came across what they simply call an approximation in the Schweser resources but what actually is in fact a logarithmic approximation of numbers close to 1.

What is the logarithmic approximation for numbers close to 1?

Assuming you have a number which is relatively close to 1, say, 1.02, then you can write the following:

$$ x \simeq\ln (1+x)$$

Let’s plot two functions $f(x)=x$ and $g(x)=\ln(1+x)$ to see how they behave when $x$ is small:

As you cane see both function yield approximately the same value but $\ln(1+x) \leq x $; let’s visualize the error of the approximation (in absolute values):

The error is pretty small! As we can see on the graph it gets larger as $x$ gets away from 0, but this the important thing is to see that, as long as $x$ is close to 0, the approximation is pretty good.

Application to CFA Level I: Fixed Income

At some point of the curriculum they discuss about, they talk about the term structure of the interest rates. As this term structure is not flat; higher interest rates are (usually) required for bonds with higher maturity. This is why you have different yield for bonds with the same features but different maturity $T$. Let’s yields defined at time $t$ for maturity is defined as $R(t,T)$, they are the spot yields at time $t$.

There is also the possibility to enter a forward agreement, called forward yield, which allows you at time $t$ to have a certain rate between time $T$ and $S$, which we denote by $F(t,T,S)$.

Now, if you now the term structure of the interest rates at time $t$, for $T$ and $S$ (i.e. you know $R(t,T)$ and $R(t,S)$), you can then find the forward rate $F(t,T,S)$ as follows:

$$F(t,T,S)=\left( \frac{(1+R(t,S))^{S-t}}{(1+R(t,T))^{T-t}} \right)^{\frac{1}{S-T}}-1$$

Let’s take an example, you know that $R(0,2)=4\%$ and $R(0,3)=5\%$ and you want to calculate $F(0,2,3)$. By applying the formula above, you find that $F(0,2,3)=7.03\%$.

You could also have used the logarithmic approximation by noticing that $1+F(0,2,3)$ if close to 1, and since $ln(1+x) \simeq x$, well you can write:

$$\ln(1+F(0,2,3))=\ln \left( \left(1+ \frac{(1+R(0,3))^3}{(1+R(0,2))^{2}} \right)^{1}-1 \right)$$

$$\ln(1+F(0,2,3))=\ln \left( \frac{(1+R(0,3))^3}{(1+R(0,2))^{2}}  \right)$$

Using the properties of logarithms, you can write that:

$$\ln(1+F(0,2,3))=\ln\left( (1+R(0,3))^3\right) – \ln \left( (1+R(0,2))^{2}\right)$$

$$\ln(1+F(0,2,3))=2 \ln\left(1+R(0,3)\right) – 3 \ln \left( 1+R(0,2)\right)$$

Now you can apply the logarithmic approximation here : $\ln (1+R(t,T)) \simeq R(t,T)$ and it follows that:

$$\ln(1+F(0,2,3))=2 R(0,3)- 3 R(0,2)=0.07 \simeq F(0,2,3)$$

As you can see, both the real value, 7.03% and 7.00% are pretty close!

For the general case, we can write:

$$\ln \left(1+F(t,T,S)\right)=\ln \left(1+ \left( \frac{(1+R(t,S))^{S-t}}{(1+R(t,T))^{T-t}} \right)^{\frac{1}{S-T}}-1\right) \simeq F(t,T,S)$$

$$F(t,T,S)=\frac{1}{S-T} \left( (S-t)\ln \left(1+R(t,S)\right) – (T-t)\ln \left(1+R(t,T)\right) \right)$$

$$F(t,T,S) \simeq \frac{1}{S-T} \left( (S-t) R(t,S) – (T-t) R(t,T) \right) $$

For $t=0$, we get:

$$F(0,T,S) \simeq \frac{1}{S-T} \left( S \cdot R(0,S) – T \cdot R(0,T) \right) $$

Application to potential interview questions

In interviews, especially for a quantitative position, the interviewer might want to know if you’re aware of this logarithmic approximation. Therefore, they would ask you to compute without a calculator the following in a short period of time:

$$\frac{(1.02)^6 \cdot (1.01)^4}{(1.03)^3 \cdot (1.05)^2 }$$

The only way to perform this computation properly is to nice that result would be close to 1, and hence look like $1+x$. You can then approximate $x$ using the logarithmic approximation $\ln (1+x) \simeq x$:

$$x \simeq \ln \left( \frac{(1.02)^6 \cdot (1.01)^4}{(1.03)^3 \cdot (1.05)^2 } \right)$$

$$x \simeq \ln \left( (1.02)^6 \cdot (1.01)^4 \right) – \ln \left( (1.03)^3 \cdot (1.05)^2  \right)$$

$$x \simeq  6\ln (1.02) + 4 \ln (1.01)  – \left( 3 \ln  (1.03) + 2\ln (1.05)  \right)$$

$$x \simeq  6 \cdot 2\% + 4 \cdot 1\%  –  3 \cdot  3\% – 2 \cdot 5\%  = -3\%$$

Hence, you estimate that :

$$\frac{(1.02)^6 \cdot (1.01)^4}{(1.03)^3 \cdot (1.05)^2 } \simeq 1+x = 0.97$$

If you use a calculator, you find that the result is 0.9727388, which is pretty close to 0.97.

Pretty cool huh?

I’ll be back with more CFA footage soon!