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 $\ln (1+x)\le x$; 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 $R(0,2)=4\mathrm{\%}$ and $R(0,3)=5\mathrm{\%}$ and you want to calculate $F(0,2,3)$. By applying the formula above, you find that $F(0,2,3)=7.03\mathrm{\%}$.

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:

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

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

Using the properties of logarithms, you can write that:

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

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

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

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

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

For the general case, we can write:

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

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

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

For $t=0$, we get:

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

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

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

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

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

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

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!