On financial advisors

I have become interested in finance relatively recently (say less than two years). At first, it was obviously because of the potential big pay, big bonuses, cash-cow financial industry. I don't think it's a shameful thing to say. So many people around me are clearly driven by the potential monetary returns of their respective field of study. However, the more I think about it, the more the whole financial sector seems like a huge corrupt monopoly with almost unsurmountable barriers of entry. This sorta makes things even more interesting to me.

The first doubt I had was back in summer 2008. I worked for a financial advisor in hopes to learn about stock picking, portfolio building, analyzing companies, and the such. I believed back then that one could, if smart enough, consistently outperform the market just by doing research, or whatever people did. To my surprise, almost none of the financial advisors at the firm had an exclusive finance education. My boss was a former engineer, and his colleague, a former criminologist. Sometimes, I would see the criminologist play poker on PokerStar. I wasn't long before I connected the dots.

Being a financial advisor has nothing to do with building a good stock portfolio. Most of their revenue is generated from attracting new clients to the company (that is where I came in, cold calling random people), and a bit of fees from their existing bank of clients. They have absolutely no incentive to do any research to pick good stocks (if they could actually do that). Briefly put, a financial advisor does not necessarily have more knowledge in how to outperform the market. They are mostly salesmen, with a four-month education of finance based on the Canadian Securities Course.

Now I have to say, to the financial advisor's defense, that they are not completely useless. In fact, most people have no clue about how things work in the financial industry (foreshadowing a future topic). Also, people's intuition on probabilities and exponentials is usually very weak. They do not know much about balancing risk and returns. I have seen people accepting to have a near zero rate of interest on their savings during my time at the firm, due the the sole fact that their investment was guaranteed. I have also seen people take huge bets on options, without seeing the incredibly leveraged potential losses.

Hence, a good financial advisor's job is to truly find his client's goals and risk tolerances, and find an appropriate balance of investment vehicles with his knowledge of risk management. They should also educate the client; but that's a bit much to ask. Obviously this is not your everyday financial advisor. They get paid in commissions when they sell products from the firm, and collect fees on transactions. When one say they want to put all their money in this, this and that stock; for the advisor, it's kind of hard to say no. Clients are still clients. And clients will go elsewhere if they are not satisfied. It is very difficult for a good advisor to show that a portfolio appreciation does not automatically validates the actions one took to get there. In an uneducated client's mind, or even anybody's mind, it is unnatural to frown on excessive risk-taking when given spectacular returns. These are the many problems an advisor must face.

[DM]

Discrete Fourier Transform, intro to

I have been reluctant on writing a post about the discrete Fourier transform (DFT) for a while because I don't feel comfortable with all the details. In fact, a few weeks ago was the first time I saw the DFT in an academic setting. But even then, the introduction given in CS371 was a high-level overview of the theory. So here it goes. Reader discretion is advised.

It can be shown that a "nice" function $f(x)$ (the true meaning of "nice" is beyond the scope of my current knowledge) defined on the interval $x \in [a,b]$ can be approximated by a Fourier series:

$$g(x) = \frac{a_0}{2} + \sum_{k=1}^{\infty}\bigg[a_k\cos(k\frac{2\pi x}{b-a})+b_k\sin(k\frac{2\pi x}{b-a})\bigg]$$

with

$$a_k = \frac{2}{b-a}\int_a^bf(x)\cos(k\frac{2\pi x}{b-a})dx$$

$$b_k = \frac{2}{b-a}\int_a^bf(x)\sin(k\frac{2\pi x}{b-a})dx$$

Just like the Taylor series, this result is quite miraculous if not more. The way I see it is that the set of all continuous functions $u: D \subseteq \mathbb{R} \mapsto \mathbb{R}$ form an infinite dimensional vector space. Then, we define the distance between two functions $u(x)$ and $v(x)$ to be $\sqrt{\int_a^b (u(x)-v(x))^2dx}$. Now that we have a notion of distance, we can take look at the set $B = \{1, \sin(2\pi x), \cos(2\pi x), \sin(4\pi x), \cos{4\pi x}, \cdots\}$ and grind it through Gram-Schmidt to get an orthonormal set of vectors. It's easy to see that the elements were linearly independent, and we could have started with an arbitrary finite number of them. Now, we just proceed to project our initial function $f(x)$ onto the subspace generated by $B$. The result is $g(x)$ the Fourier series of $f(x)$.

We can now interpret the coefficients of $g(x)$ as "how much of certain frequency of oscillation does the initial function contains". Now to finally define the Fourier transform, a FT is an operator $\mathcal{F}$ which takes in a "nice" function $f(x)$ and spits out two functions $a_n$ and $b_n$ defined on the non-negative integers representing the coefficients of the sines and cos.

In practice, one main problem. We never have continuous data in the real world. To deal with that, we simply discretize the integral into a $N$ part Riemann sum, where $N$ is the number of "equally distanced" data points we have, depending on the context. This is where the "discrete" part of the DFT comes in. Usually, when we perform a DFT, we don't care much about the higher frequency terms because they are usually small.

Now one will never see the Fourier transform under this form because it's so ugly to write out. We bring in Euler's Identity to simplify a few things with complex numbers. We can write:

$$g(x) = \sum_{k=-\infty}^{\infty}c_ke^{2\pi ikx}$$

$$c_k = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-2\pi ikx}dx$$

It is easy, but tedious to work out how the coefficients $c_k$ are related to $a_k$, $b_k$. One thing is for sure, the coefficients with the same $k$ correspond to the same frequencies.

Finally, we discretize and cut off high frequencies. Assume we get a time-series $f_k$ for $k = 1\cdots N$, we apply the DFT:

$$F_k = \frac{1}{N}\sum_{n=N/2+1}^{N/2}f_ne^{\frac{-2\pi ikn}{N}}$$

We go a bit further to make things look nice and substitute $\omega = e^{\frac{2\pi i}{N}}$. So,

$$F_k = \frac{1}{N}\sum_{n=N/2+1}^{N/2}f_n\omega^{-kn}$$

Before I end the post, I'd like to give one application of the DFT. The reason we want to perform the DFT is to distinguish between various frequencies by looking at the corresponding coefficients.

One application of this is filtering out noise from a signal. Usually, noise is of much higher frequency than the signal we wish to process. A simple procedure would be to take the DFT of this signal, multiply the coefficients corresponding to "high" frequencies by zero (the meaning of high depends on the context), and take the inverse DFT (I haven't talked about the inverse DFT, but this post is getting long). This will yield a nice signal devoid of most of the unwanted noise.

Other applications of the DFT will be subject of future posts (and maybe a description of how to perform the DFT efficiently, an algorithm called the Fast Fourier Transform).