The Kelly Criterion: Multiple Investment Opportunities

This is the second post in a four-part series exploring the Kelly criterion:

After the last post introduced the Kelly criterion and its application in deciding how much to invest in a single gamble, we'll investigate whether Kelly can help us choosing between multiple investment opportunities.

We'll start with a mathematical model:

V1=V0((1+r0)(1ili)+ili(1+ri))V_1 = V_0 \left( (1+r_0)(1 - \sum_i l_i) + \sum_i l_i(1 + r_i) \right)

Simple, right? :-)

  • V0,V1V_0, V_1 are the available money before and after the first round, as before.
  • r0r_0 is the risk-free return rate. This allows us to model opportunity costs (could put the money into a bank account instead of investing, r0=0.004r_0 = 0.004), or inflation (non-invested money loses value over time r0=0.02r_0 = -0.02).
  • lil_i is the fraction of the available money invested in stock ii.
  • rir_i is a random variable describing returns for stock ii.

The first term (1+r0)(1ili)(1+r_0) (1 - \sum_i l_i) represents all the money not invested in any stock, being invested at the risk-free return rate r0r_0. The second term ili(1+ri)\sum_i l_i(1 + r_i) represents the outcomes of the investments in individual stocks ii.

We will re-formulate a bit to simplify the expression:

V1V0=(1+r0)ili(1+r0)+ili(1+ri)=(1+r0)+ili(rir0)\begin{aligned} \frac{V_1}{V_0} & = (1+r_0) - \sum_i l_i (1+r_0) + \sum_i l_i(1 + r_i) \\ & = (1+r_0) + \sum_i l_i(r_i - r_0) \end{aligned}

We can further simplify by merging the lil_i and rir_i into the vectors l\vec l and r\vec r, with 1\vec 1 being a vector where all elements are 11:

V1V0=(1+r0)+l(rr01)\frac{V_1}{V_0} = (1+r_0) + \vec l \cdot (\vec r - r_0 \vec 1)

Let's move from a single game round to nn rounds:

VnV0=((1+r0)+l(rr01))n\frac{V_n}{V_0} = \left( (1+r_0) + \vec l \cdot (\vec r - r_0 \vec 1) \right)^n

Now we are again in a position where we can follow Kelly's prescription: invoking the law of large numbers and finding the ll which maximizes the gain Vn/V0V_n/V_0.

For large nn, we can approximate:

VnV0j(1+r0+l(rjr01))pjn\frac{V_n}{V_0} \approx \prod_j \left( 1+r_0 + \vec l \cdot (\vec r_j - r_0 \vec 1) \right)^{p_j n}

In this step we switched from the random variable r\vec r to its outcomes rj\vec r_j. Each possible outcome rj\vec r_j occurs with probability pjp_j. We justify this switch with the law of large numbers: the outcome rj\vec r_j will be observed proportionally to its probability, pjnp_j n times (for large nn). Consequently, we will have pjnp_j n factors involving rj\vec r_j in the overall product, with jj iterating over all potential outcomes of r\vec r.

Note that since rj\vec r_j is vector-valued, pjp_j is a joint probability distribution.

At this point we are nearly finished. We are looking for the vector lopt\vec l_{opt} that maximizes Vn/V0{V_n}/{V_0}. In analogy to last post, we arrive at:

lopt=argmaxlVnV0=argmaxllogVnV0=argmaxljpjlog(1+r0+l(rjr01))\begin{aligned} \vec l_{opt} & = \operatorname*{argmax}_{\vec l} \frac{V_n}{V_0} = \operatorname*{argmax}_{\vec l} \log \frac{V_n}{V_0} \\ & = \operatorname*{argmax}_{\vec l} \sum_j p_j \log \left( 1+r_0 + \vec l \cdot (\vec r_j - r_0 \vec 1) \right) \end{aligned}

This equation is not analytically solvable, but may be approximated as a quadratic programming problem as described in a paper by Vasily Nekrasov.

Conclusion

It should be obvious that the Kelly criterion is applicable in a wide range of scenarios, from gambling over investment decisions to whether to buy insurance. If you're interested in interactive plots that really helped me understand this material, you can find them in this Jupyter Notebook.

In the next post we'll discuss the origins of the logarithm in the Kelly formula.

The Kelly Criterion: Introduction

This is the first post in a four-part series exploring the Kelly criterion:

The Kelly criterion is a formula used to determine the optimal size of a series of bets in order to maximize wealth. It is often described as optimizing the logarithm of wealth, and will do better than any other strategy in the long run.

Let's look at a simple gamble: when you play, you win with probability pp and lose with probability q=1pq = 1-p. If you win, you get back your initial bet plus a fraction rr of that bet. If you lose, your bet is forfeited. The Kelly formula calculating the optimal fraction of your available wealth to bet is:

lopt=prqr=pr(1p)r=p1prl_{opt} = \frac{pr - q}{r} = \frac{pr - (1-p)}{r} = p - \frac{1-p}{r}

Here is a small demo to explore the formula:

Kelly calculation demo

p0.80
r0.40

Probability of winning: 80 %
Profit on win, as percentage of placed bet: 40 %
Optimal fraction of wealth bet in this game: 30 %

The main takeaways are:

  • In each game you invest a fraction ll of the total amount of money you have. If you play the game from the demo (p=0.8,r=0.4p=0.8, r=0.4) and you have $100 available, you should invest $30 (since l=0.3l=0.3). If you win, you will have $112 (100+30×0.4100 + 30 \times 0.4) and invest $33.60 (112×0.3112 \times 0.3) in the second round. If you lost the first round you'd have $70, and invest $21 in the second round. If you can not reinvest your earnings, Kelly does not apply.
  • This formula holds when a large number of games are played. If you play only few rounds of a game this may not apply.
  • When it applies, the Kelly strategy will generate the most wealth out of your initial starting capital.

Derivation of the Kelly criterion

We will take a short dive into mathematics to understand how Kelly came up with this formula. Let's compute the wealth V1V_1 after the first round of gaming, when starting from an initial wealth V0V_0:

V1=V0(1+WlrW+(1W)lrL)V_{1} = V_{0} (1 + Wlr_W + (1-W)lr_L)

Here, ll is the fraction of money bet, rWr_W is the profit on win, rLr_L is the fraction of the bet forfeited on loss (in the above scenario, rL=1r_L=-1). WW is a random variable that's 11 with probability pp and 00 with 1p1-p. So with probability pp we'll have W=1W=1, giving V1,win=V0(1+lrW)V_{1,win} = V_0 (1+lr_W), with probability 1p1-p we'll get V1,loss=V0(1+lrL)V_{1,loss} = V_0 (1+lr_L).

We can reformulate the expression in a way that will show an important generalization:

V1=V0(1+lrW)W(1+lrL)(1W)V_1 = V_0 (1+lr_W)^W(1+lr_L)^{(1-W)}

The two expressions look very different, but they are equivalent since WW can only take on the values 00 and 11. If W=1W=1 the second term will vanish since x0=1x^0 = 1, and we will get the same result for V1,winV_{1,win}. The same consideration holds for V1,lossV_{1,loss}.

If we repeat this game nn times, we arrive at

Vn=V0(1+lrW)Wn(1+lrL)(nWn)V_n = V_0 (1+lr_W)^{W_n}(1+lr_L)^{(n-W_n)}

with WnW_n being the number of wins in those nn games. Essentially, each won round adds a factor (1+lrW)(1+lr_W) to the expression, each lost round (1+lrL)(1+lr_L).

Now comes a crucial step: how many wins do we expect after playing many games? Asked mathematically, what value will WnW_n have for large nn? Because of the law of large numbers we expect W=pnW = pn. So for p=0.8p=0.8 and n=10000n=10000 we would expect to win 8000 out of the 10000 games.

Let's plug this into our formula:

Vn=V0(1+lrW)pn(1+lrL)(1p)nV_n = V_0 (1+lr_W)^{pn}(1+lr_L)^{(1-p)n}

The law of large numbers allowed us to eliminate the random variables WW, WnW_n and instead obtain an explicit formula in terms of p,n,rW,rLp, n, r_W, r_L. Since we are interested in maximizing our profit, we are looking for loptl_opt — the ll value that maximizes Vn/V0{V_n}/{V_0}.

lopt=argmaxlvnv0=argmaxllogvnv0=argmaxlplog(1+lrW)+(1p)log(1+lrL)\begin{aligned} l_{opt} & = \operatorname*{argmax}_l \frac{v_n}{v_0} \\ & = \operatorname*{argmax}_l \log \frac{v_n}{v_0} \\ & = \operatorname*{argmax}_l p \log(1 + l r_W) + (1-p) \log(1 + l r_L) \end{aligned}

We are free to introduce the logarithm and drop the nn factor because this does not change the value of loptl_{opt}.1 Differentiating the logarithmic expression and solving for ll gives:

lopt=prL1prWl_{opt} = -\frac{p}{r_L} - \frac{1-p}{r_W}

This is identical with the formula given in the introduction after substituting rL=1r_L = -1.

Extension to games with more than two possible outcomes

We can extend the Kelly argument to games that have a greater number of possible outcomes. Such games could be:

  • A game with 6 different rewards selected by roll of a 6-sided die, where rewards may be negative (money lost)
  • An investment in a stock, here we have a continuous range of outcomes.

If we have multiple outcomes rir_i, each happening with probability pip_i, we obtain:

lopt=argmaxlipilog(1+lri)l_{opt} = \operatorname*{argmax}_l \sum_i p_i \log(1 + l r_i)

While this equation is not easily solvable analytically, it is trivial to solve numerically. This allows us to extend the Kelly approach to a new kind of game with some interesting practical applications.

If you're interested in interactive plots that really helped me understand this material, you can find them in this Jupyter Notebook.

The Kelly formalism can be generalized further still, taking into account multiple parallel investment opportunities. We'll look at this in the next post.


1 The logarithm is a strictly monotonically increasing function, so α>βlogα>logβ\alpha \gt \beta \Leftrightarrow \log \alpha \gt \log \beta. Thus it doesn't change the location of any maxima: argmaxf(x)=argmaxlogf(x)\operatorname*{argmax} f(x) = \operatorname*{argmax} \log f(x). See Mark Schmidt's Argmax and Max Calculus paper for more information about properties of argmax\operatorname*{argmax}.

Exploring GDB's Python API with Jupyter

GDB — the most common console debugger on Linux systems — has a Python API for adding new debugger commands and pretty-printers for complex data structures, or to automate debugging tasks.

While Python scripts can be loaded from files, it is nice to interactively explore the API or the debugged program. IPython would be perfect for the job, but starting it directly inside GDB doesn't work well. Fortunately, it's easy to launch an IPython kernel and connect with an external Jupyter console.

Launching the kernel from the gdb prompt:

(gdb) python
>import IPython; IPython.embed_kernel()
>end

Gives the following message:

To connect another client to this kernel, use:
    --existing kernel-12688.json

We can start Jupyter on a separate terminal and connect to this kernel:

$ jupyter console --existing kernel-12688.json
In [1]: gdb.newest_frame().name()
Out[1]: 'main'

The GDB Python API is then available from the gdb module within this Python session. To get started, I'd suggest the API documentation or this series of tutorials.

Currently, only the console client can connect to existing kernels. Support in the Notebook or in Jupyter Lab is tackled in this Github issue. Even with the limited capabilities of the console client, it's a great way to explore the API and to tackle more complicated debugging problems that require automation to solve.

PyDays 2017

PyDays 2017 was Austria's first conference dedicated to the Python programming language. It took place on May 5 and May 6 and was graciously hosted by the Linuxwochen Wien at FH Technikum in Vienna. It was great on many levels: meeting new people excited about Python and where it's headed, over 20 talks, interesting hallway conversations, etc.

I helped out with the organization of the conference, mostly taking care of catering. We (the organization team) are very happy with how everything went. The atmosphere was welcoming and open, the quality of the talks and workshops was very good, attendance was great, and people seemed to have a really good time. From an organizational perspective, the talks mostly stayed on-time, the audio/video hardware worked fine, and the PyDays booth was well-received. Giving out snacks, coffee, soda, cake during breaks worked out nicely and was received very well. The feedback, both at and after the conference, was very positive.

In addition to the co-organization of the Conference, I also held a talk about Dask, a Python library for parallel computing. It provides data structures similar to NumPy Arrays or Python Dataframes that process operations in parallel and scale to out-of-memory datasets. Dask is exciting since it provides the analytical power and familiar mental model of Dataframes, but handles hundreds of gigabytes of data and allows seamless scaling from a single laptop to a computing cluster. The talk was well-received, and there was a lot of interest at the end and in the hallway afterwards. The slides are online, if you're interested as well.

Finally, I'd like to thank my co-organizers, Claus, Kay, Helmut and Sebastian, as well as the people who helped out during the conference, for all their great work to make the PyDays a success. I'd also like to thank our sponsors: the Python Software Foundation, the Python Software Verband e.V., T-Mobile, UBIMET, and Jetbrains.

Overall it was great to see so much community interest in Python. If you're passionate about Python as well, join us on the Python Austria Slack and our meetups!

GCC compatibility: inline namespaces and ABI tags

Keeping libraries binary-compatible with old versions is hard. Recently, GCC was in the unenviable situation of having to switch its std::string implementation.1 GCC used inline namespaces and ABI tags to minimize the extent of breakage and to ensure that old and new versions could only be combined in a safe way. Here, we'll have a look at those mitigation techniques: what they are and how they work.

First, some background: GCC used to have a copy-on-write implementation for std::string. However, C++11 does not allow this anymore because of new iterator and reference invalidation rules. So, GCC 5.1 introduced a new implementation of std::string. The new version was not binary-compatible with the old one: exchanging strings between old (pre-5.1) and new code would crash. We say the application binary interface (ABI) changed.

To understand the consequences of this, let's look at several scenarios:

  • A program uses only code compiled with GCC < 5.1: only old strings, works
  • A program uses only code compiled with GCC >= 5.1: only new strings, works
  • A program mixes code compiled with different GCC versions: both types of strings exist in the program, it could crash.

Let's dig into the last bullet point. When would it crash? Whenever "new" code accesses an "old" string or vice versa. Here are some examples where f() is compiled with an older GCC version and called from "new" code:

void f(int i);                 // (a) safe, no std::string involved
void f(const std::string& s);  // (b) will crash when f accesses s
std::string f();               // (c) will crash when the returned string is used

Given how common std::string is, such crashes would happen frequently if "old" and "new" code were combined. Unfortunately, it's surprisingly easy to end up with a program with some pre-5.1 parts and some newer ones. It's sufficient to link a "new" executable to an "old" library. Given the bad consequences, the GCC developers needed to solve this.

The solution is to change the symbol names of the GCC 5.1 std::string and all functions using it. Because the linker uses symbol names to resolve function calls into libraries, this would cause link-time errors for cases (b) and (c) while (a) would still work. Exactly the intended behavior.

How could this be done? The mechanism that converts C++ names into symbol names is called name mangling. The generated symbol names contain information about namespaces, function names, argument types, etc. Putting the new std::string into a separate namespace would change the symbol names of all functions accepting std::string as argument. But that's crazy—std::string needs to be in namespace std, right?

The solution for this is inline namespaces. All classes, functions, and templates declared in an inline namespace are automatically imported into the parent namespace. Their mangled name still references the original location, though.

Here's what GCC 5.1 does:

namespace std {
  inline namespace __cxx11 {
    template<typename _CharT, ...>  basic_string;
    typedef basic_string<char>      string;
  }
}

Looking at f(const std::string& s), what would the symbol names be when compiled with older and newer GCC versions?

Older GCCGCC 5.1
Symbol name_Z1fRKSs_Z1fRKNSt7__cxx1112basic_stringIcSt11char_traitsIcESaIcEEE
Decoded symbol name2f(std::basic_string<char, ...> const&)f(std::__cxx11::basic_string<char, ...> const&)

Indeed, the symbol names are different and the linker would give an error if GCC 5.1 code called f(const std::string &s) from a library compiled with an older GCC version. This solves the compatibility problem for all functions taking std::string as argument.

One problem remains: case (c) from above, std::string f(). The return value type of a function is not part of its mangled name.3 Thus, the symbol name wouldn't change for functions that return std::string, which could still lead to runtime crashes.

GCC 5.1 solves this with the use of ABI tags. From the documentation:

The abi_tag attribute can be applied to a function, variable, or class declaration. It modifies the mangled name of the entity to incorporate the tag name, in order to distinguish the function or class from an earlier version with a different ABI

[...]

When a type involving an ABI tag is used as the type of a variable or return type of a function where that tag is not already present in the signature of the function, the tag is automatically applied to the variable or function.

GCC applies the attribute __abi_tag__ ("cxx11")) to the std::__cxx11 namespace. This affects all classes therein, including the new version of std::string. The ABI tag propagates to all functions that return a string and changes their symbol name.

Let's look at the symbol names for std::string f() for different compiler versions:

Older GCCGCC 5.1
Symbol name_Z1fv_Z1fB5cxx11v
Decoded symbol name2f()f[abi:cxx11]()

Again, the symbol name is different, and the "cxx11" ABI tag is applied to std::string f() with GCC 5.1. This completes the second part of GCC's solution for the std::string ABI change.

These two methods to induce symbol name changes for anything using std::string make the migration path to GCC 5.1 much easier. If the program links correctly it should work, and be safe from difficult-to-debug runtime errors. There would be more to discuss about the GCC 5.1 changes, such as how libstdc++.so still exports the old std::string implementation for compatibility. But this post has gone on too long already, so let's leave it at that :-)


1 GCC 5.1 also introduced a new version of std::list. It is handled analogously to std::string.
2 Decoded using c++filt from the binutils package.
3 The return value type doesn't need to be part of the symbol name because it doesn't get used for overload resolution.

published October 29, 2016
tags c++