# CFA Level II: Quantitative Methods, Multiple Regression

Following my first post about the Level II and specifically correlation, I am now moving on to the main topic if this year’s curriculum: multiple regression. Before I get started, I want to mentioned that the program talks about regression in 2 steps: it starts by discussing the method with 1 independent variable and then with multiple variables. I will only talk about the multiple variable version, because it is generalized. If you understand the general framework, you can answer any question about the specific 1 variable case.

Multiple Regression is all about describing a dependent variable $Y$ with a linear combination of $k$ variables $X_k, k \in 1..K$. This is expressed mathematically as follows:

$$Y_i = b_0 + b_1 X_{1,i} + …. + b_K X_{K,i} + \epsilon_i$$

Basically, you are trying to estimate the variable $Y_i$ with the values of the different $X_{k,i}$. The regression process consists in estimating the parameters $b_0, b_1, … , b_k$ with an optimization method over a sample of size $n$ (there are $n$ known values for each of the independent variable $X_k$ and the dependent variable $Y$, represented by the $i$ index). When all $X_k=0$, $Y$ has a default value $b_0$ (called the intercept). The error term $\epsilon_i$ is there because the model will not be able to determine $Y_i$ exactly; there is hence a residual part of the value of $Y$ which is unexplained by the model which is normally distributed with mean 0 and constant variance $\forall i$.

So to sum up, the inputs are:

• $n$ known values of $Y$
• $n$ known values of each $X_i$

and the outputs are:

• $b_0, b_1, … , b_k$

So, say you have a set of new values for each $X_k$, you can estimate the value for Y, denoted $\hat{Y}$ by doing:

$$\hat{Y} = b_0 + b_1 X_1 + …. + b_K X_K$$

The most important part of this section is the enumeration of its underlying assumptions:

1. There is a linear relationship between the independent variable $X_k$ and the dependent variable $Y$.
2. The independent variable $X_k$ are not correlated with each other.
3. The expected value of $\epsilon_i$ is 0 for all $i$.
4. The variance of $\epsilon_i$ is constant for all $i$.
5. The variable $\epsilon$ is normally distributed,
6. The error terms $\epsilon_1, …, \epsilon_n$ are not correlated with each other.

If one of these assumptions is not verified for the sample being analyzed, then the model is misspecified and we will see in a subsequent post how to detect this problem and how to handle it. Note that point 2) only mentions colinearity between the independent variables; there is no problem if $Y$ is correlated to one of the $X_k$, it’s what we’re looking for.

Now, remember in my first post on the Quant Methods I said that one we would have to compute the statistical significance of estimated parameters. This is exactly what were are going to do now. The thing is, the output parameters of a regression (the coefficients $b_0, b_1, … b_K$) are only statistical estimates. As a matter of fact, there is uncertainty about this estimation. Therefore, the regression algorithm usually outputs the standard deviations $s_{b_k}$ for each parameter $b_k$. This allows us to create a statistical test to determine whether the estimate $\hat{b}_k$ is statistically different from some hypothesized value $b_k$ with a level of confidence of $1-\alpha$. The null hypothesis $H_0$, which we want to reject, is that $\hat{b}_k=b_k$. The test goes as follows:

$$t=\frac{\hat{b}_k – b_k}{s_{b_k}}$$

If the null hypothesis $H_0$ is verified, the variable $t$ follows a t distribution with $n-K-1$ degrees of freedom. So, you can simply look at the value in the t-distribution form for the desired level on confidence to find the critical value $c_\alpha$. If $t>c_\alpha$ or if $t<-c_\alpha$, then $H_0$ can be rejected and we can conclude that $\hat{b}_k$ is statistically different from $b_k$.

Usually, we are asked to determine whether some estimate $\hat{b}_k$ is statistically significant. As explained in the previous post, this means that we want to test the null hypothesis $H_0: \hat{b}_k = 0$. So, you can just run the same test than before with $b=0$:

$$t=\frac{\hat{b}_k}{s_{b_k}}$$

That’s it for today. The concepts presented here are essential to succeed the Quantitative Method part of the CFA Level II exam. They are nonetheless quite easy to grasp and the formulas are very simple. Next, we will look at the method to analyze how well a regression model does at explaining the dependent variable.

# CFA Level II: Economics, Exchange Rates Basics

Good evening everyone,

My weekly task is to go through the Economics part of the Level II curriculum. I was a bit afraid of it, because it was clearly my week point at the Level I and because I think that this topic covers a lot of material compared to its allocated number of questions.

In this level, the first challenge is to take into account the bid-ask spread for currency exchange rates. Just as we saw for security markets at Level I, exchange rate do not value a single “value”. That is, you cannot buy and sell a currency at the same price instantaneously. This is because you need to go through a dealer who has to make money for providing liquidity: this economical gain is provided by the bid-ask spread.

Let’s go back to the basics by looking at the exchange rate $\frac{CHF}{EUR}$. The currency in the denominator is the base currency; it is the asset being traded. The currency in the numerator is the price currency; it is the currency used to price the underlying asset which is in this case another currency. This is exactly like if you were trading a stock $S$. The price in CHF could be see as the $\frac{CHF}{S}$ “exchange rate”, i.e. the number of CHF being offered for one unit of $S$. Now as mentioned before, exchange are quoted with bid and ask prices:

$$\frac{CHF}{EUR} = 1.21 \quad – \quad 1.22$$

This means that $\frac{CHF}{EUR}_\text{bid}$ is 1.21 and $\frac{CHF}{EUR}_\text{ask}$ is 1.22. Again, you are trading the base currency: here Euros.

• The bid price is the highest price you can sell it for to the dealer.
• The ask price is the lowest price you can buy it for to the dealer.

If you want to make sure you got it right, just make sure  you can’t instantaneously buy the base currency at a given price (which you believe to be the ask) and sell it at a higher price (which you believe to be the bid). In this example, you can buy a EUR for 1,22 CHF and sell it instantaneously for 1.21 CHF making a loss of 1.22-1.21=-0.01 CHF. In fact, the loss can be seen as the price of liquidity which is the service provided by the dealer for which he has to be compensated. So the lower value is the bid, the higher value is the ask (also called the offer).

Recall from Level I that you could convert exchange rates by doing:

$$\frac{CHF}{EUR} = \frac{1}{\frac{EUR}{CHF}}$$

This is simple algebra and it works fine as long as you don’t have the bid-ask spread to take into account. The problem is that at Level II, you do. To invert the exchange rate with this higher level of complexity, you have to learn the following formula:

$$\frac{EUR}{CHF}_\text{bid} = \frac{1}{\frac{CHF}{EUR}_\text{ask}}$$

This might look complicated at first, but I got something in my bag to help you learning it. Look do the following steps:

• Define what you want on the left-hand side of the equation (currency in the numerator, currency in the denominator, bid or ask).

$$\frac{A}{B}_\text{side}$$

• On the right-hand side of the equation, write the inverse function:

$$\frac{A}{B}_\text{side}=\frac{1}{\cdot}$$

• On the right-hand side, replace the $\cdot$ by the inverted exchange rate:

$$\frac{A}{B}_\text{side}=\frac{1}{\frac{B}{A}_\cdot}$$

• Finally, replace the remaining dot on the right-hand side by the opposite side:

$$\frac{A}{B}_\text{side}=\frac{1}{\frac{B}{A}_\text{opp. side}}$$

Let’s take show how this work using our base example:

$$\frac{EUR}{CHF}_\text{bid}=\frac{1}{\frac{CHF}{EUR}_\text{ask}}$$

Simple. You can simply apply this method interchangeably to suit your needs. Actually, you might wonder what you need that to compute cross rates, which will be the subject of another post. Until that, grasp the concepts presented here and stay tuned on this blog!

# CFA Level II, Quantitative Methods: Correlation

Good evening everyone,

So I’m finally getting started to write about the CFA Level II material, and as I process in the classic order of the curriculum, I will start with the Quantitative Methods part. If you have a bit of experience in quantitative finance, I believe this part is quite straightforward. Actually, it talks about many different things and therefore I will make a different post for each topic to keep each of them as short as possible. This will hopefully make it as readable and accessible as possible for all types of readers.

As a brief introductory note, I would say that this section is very much like the rest of the CFA Level II curriculum, it builds on the concepts learnt in the Level I. Let me write that again: the concepts – not essentially all the formulas. For the Quant Methods part, you have to be comfortable with the Hypothesis Testing part I recapped in this post.

So let’s get started for the first post which will be about correlation. This part is actually quite easy because you’ve pretty much seen everything at the Level I. Let me restate the two simple definitions of the sample covariance and the sample correlation of two sample $X$ and $Y$:

$$\text{cov}_{X,Y} = \frac{\sum_{i=1}^n (X_i – \bar{X})(Y_i – \bar{X})}{n-1}$$

$$r_{X,Y} = \frac{\text{cov}_{X,Y}}{s_X s_Y}$$

where $s_X$ and $s_Y$ are the standard deviation of the respective samples.

The problem of the sample covariance is that it doesn’t really give you a good idea of the strength of the relationship between the two processes; it very much depends on each the samples’ variances. This is where the sample correlation is actually useful because it is bounded: $r_{X,Y} \in [-1,1]$. Taking the simple example, of using twice the same sample, we have $\text{cov}_{X,X} = s_X^2$ and $r_{X,X}=1$.

In general, I would say that the Quant Methods part of the Level II mainly focuses on understanding underlying models and their assumptions, not on learning and applying formulas – it was more the case in Level I. For correlation, there is one key thing to understand, it detects a linear relationship between two samples. This means that the correlation is only useful to detect a relationship of the kind $Y = aX+b+\epsilon$.

A good way to visualize whether there is a correlation between two samples is to look at a scatter plot. To create a few examples, I decided to use MATLAB as we can generate random processes and scatter plots very easily. So I will create 3 processes:

• A basic $X \sim \mathcal{N}(0,1)$ of size $n=1000$
• A process which is a linear combination of $X$.
• A process which is not a linear combination of $X$.
• And another process $Y \sim \mathcal{N}(0,1)$ of size $n=1000$ independent from $X$.

<br />
%Parameter definition<br />
n=1000;<br />
%Process definitions<br />
x=randn(n,1);<br />
linear_x=5+2*x;<br />
squared_x=x.^2;<br />
y=randn(n,1);<br />
%Plotting<br />
figure();<br />
scatter(x,y);<br />
figure();<br />
scatter(x,squared_x);<br />
figure();<br />
scatter(x,linear_x);<br />


The script presented above generates 3 different scatter plots which I will now present and comment. First, let’s take the example of the two independent process X and Y. In this instance, there is not correlation by definition. This give a scatter plot like this:

Now, let’s look at the scatter plot for the process which is a linear combination of X. Obviously, this is an example of two processes with positive correlation (actually, perfect correlation).

Graphically, we can say that there is correlation between two samples if there is a line on the scatter plot. If the line has a positive slope (it goes from down-left to up-right), the correlation will be positive. If it has a negative slope (it goes from top-left to bottom-right), the correlation will be negative. The magnitude of the correlation is determined by how much the points lie on a straight line. In the example above, all the points perfectly line on a straight line.  So a general framework to “estimate” correlation from a scatter plot would be the following:

• Do the points lie approximately on a straight line?
• If no, then there is no correlation, stop here.
• If yes, then there is correlation and continue:
• Is the slope of the line positive or negative?
• If positive, then the correlation $r \geq 0$
• If negative, then the correlation $r \leq 0$
• If there is no slope (vertical or horizontal straight line), then one of the sample is constant and there is no correlation $r=0$.
• How much are the points on a straight line?
• The more they look to be on a straight line, the more $|r|$ is close to 1.

In the example above, the points look like a straight line, the slope is positive, and the points are very much on a straight line, so $r \simeq 1$.

Finally, let’s look at the third process which is simply the squared values of X, so the relationship is not linear:

If we apply the decision framework presented above, we can clearly see that the points do not lie on a straight line, and hence we conclude that there is no correlation between the samples.

Now let’s look at the values given by MATLAB for the correlations:

Samples Correlation
Independent processes X and Y 0.00
X and Linear Combination of X 1.00
X and squared X 0.02

As we can see, MATLAB confirms what we determined looking at the scatter plots.

I now want to explain a very important concept in the Level II curriculum: the statistical significance of an estimation. Quite simply, given the fact that we estimated some value $\hat{b}$, we say that this estimation is statistically significant with some confidence level $(1-\alpha)$, if we manage to reject the hypothesis $H_0 : b=0$ with that confidence level. To do so, we will perform a statistical test which will depend on the value we are trying to estimate.

For the given post, we would like to determine whether the sample correlation $r$ we estimate is statistically significant. There is a simple test given by the CFA institute:

$$t = \frac{r \sqrt{n-2}}{\sqrt{1-r^2}}$$

This variable $t$ follows a Student-t law, with $n-2$ degrees of freedom. This means that, if you know alpha, you can simply look at the critical value $t_c$ in the distribution table and reject the hypothesis $H_0 : r=0$ if $t < -t_c$ or $t > t_c$. Although you are supposed to know how to compute this test yourself, you can also be given the p-value of the estimated statistic. Recall from Level I, the p-value is the minimal $\alpha$ for which $H_0$ can be rejected. Quite simply, if the p-value is smaller than the $\alpha$ you consider for the test, you can reject the null hypothesis $H_0$ and conclude that the estimated value is statistically significant.

Let’s look at the p-values MATLAB gives me for the correlation we discussed in the example:

Samples Correlation P-Value
Independent processes X and Y -0.01 0.77
X and Linear Combination of X 1.00 0.00
X and squared X 0.02 0.53

If we consider an $\alpha$ of 5%, we can see that we cannot reject the null hypothesis for the independent and squared process, so they are not statistically significant. On the other hand, the linear combination of X has a p-value close to 0, which means that it is statistically significant for virtually any $\alpha$.

One last word on the important points to keep in mind about correlation:

• Correlation detects only a linear relationship; not a nonlinear one.
• Correlation is very sensible to outliers (“weird” points, often erroneous, included in the datasets).
• Correlation can be spurious; you can detect a statistically significant correlation whereas there is no economic rationale behind it.

I hope you enjoyed this first post on the CFA Level II, and I’ll be back shortly with more!