Overview :: Jane Street (2024)

OUR GROUPS

TRADING AND RESEARCH

At Jane Street, trading and research roles combine to form a collaborative quantitative team. Traders and researchers work together to develop the mathematical models and automated trading strategies that drive our business. We’ve been at the forefront of applying machine learning techniques to quantitative finance, tackling research problems ranging from sub-microsecond trading to uncovering long-term market inefficiencies across trillions of historic events.

TECHNOLOGY

Technology is central to everything we do. Most Jane Streeters writecode as part of their regular work, and we build almost all oursoftware in-house with a focus on productivity, real-time visibility,and reliability. Twenty years of using and applying the most advancedtools available has earned us a reputation for excellence in tech; ourtechnical infrastructure and software engineers drive Jane Street’sgrowth and generate our competitive edge.

INSTITUTIONAL SALES AND TRADING

Our sales traders and product specialists partner with a wide varietyof institutional clients to provide trading solutions and access toJane Street’s differentiated liquidity. By combining tradingexpertise, market knowledge, and a deep understanding of the liquidityneeds of asset managers, pensions, insurance companies, and others,these professionals help institutions achieve their trading goals.

INFRASTRUCTURE

Our infrastructure teams form the backbone of Jane Street, keeping theday-to-day operations of the firm running seamlessly. They handle ourlegal and compliance matters, see to our taxes and finances, manageour offices and human resources, and direct our creative projects.

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\(\def\V{\mathrm{Var}}\) Overview: the self-symmetry of the Sierpinski triangle leads to a linear equation that \(\Sigma\) must satisfy; specifically, \(\Sigma = \tfrac{1}{4}\Sigma + \tfrac{1}{24}I\), which gives \(\fbox{\(\Sigma = \tfrac{1}{18}I\)}\).

Without rigorously defining the uniform distribution over the Sierpinski triangle, we can at least recognize that there are certain properties it should have. The first relevant property is that it's invariant under rotation by 120°; it follows that \(\Sigma = kI\) for some \(k\). (Consider: The ellipse \(\mathbf{x}^*\Sigma^{-1} \mathbf{x} = 1\) will be invariant under this rotation, and the only ellipses invariant under this rotation are circles.) We have \(k = \Sigma_{11} = \V(X_1)\), so we have reduced the problem to finding \(\V(X_1)\).

The next relevant property of the distribution is that it's equal to a mixture of 3 equally weighted copies of itself, with each copy scaled by 1/2 and translated by \((1/4,\text{something}),(-1/4,\text{something})\), or \((0,\text{something})\). (The somethings are respectively \(-\sqrt{3}/12,-\sqrt{3}/12,\sqrt{3}/6\), but we don't need them.) In particular, if \(T\) is uniform over \(\{-1/4,0,1/4\}\) (and independent of \(X_1\)), then \(X_1\) has the same distribution as \(\tfrac{1}{2}X_1 + T\). We can directly compute \(\V(T) = 1/24\), and we then have \[ \begin{align} k = \V(X_1) &= \V(\tfrac{1}{2}X_1 + T)\\ &= \tfrac{1}{4}\V(X_1) + \V(T)\\ &= \tfrac{1}{4}k + 1/24, \end{align} \] and solving for \(k\) now yields \(k = 1/18\), so \(\Sigma = \tfrac{1}{18}I\).

A similar approach is to write \(X_1 = \sum_{n=0}^\infty \frac{1}{2^n}T_n\) where the \(T_n\) are independent copies of \(T\). Then \(\V(X_1) = \sum_{n=0}^\infty \V(\frac{1}{2^n}T_n) = \frac{1}{24} \sum_{n=0}^\infty \frac{1}{4^n} = 1/18\).

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Let \(C \subseteq \mathbb{R}^3\) be a closed bounded convex set.

Let \(B = \partial C\) be its boundary. Let \(D = B + B = \{\,p+q\mid p,q\in B\,\}\).

Prove that \(D\) is convex.

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Actually, let us start by considering the 2-dimensional case. Let \(E = \{\,(p+q)/2\mid p,q\in B\,\}\); this is just a scaled down \(D\), so it is enough to show that \(E\) is convex. It suffices to show \(E = C\). \(E\subseteq C\) follows by the convexity of \(C\). For \(C\subseteq E\), take any point \(p\in C\). If \(p\in B\), then \(p = (p+p)/2 \in E\). If \(p\notin B\), imagine drawing a vertical line through \(p\), and then continuously rotating that line counterclockwise (about \(p\)) by \(180^\circ\). It will intersect \(B\) in two points, one to either side of \(p\). By the Intermediate Value Theorem, there will be some angle at which these two points are equidistant from \(p\); then \(p\) is the average of those other two points, so \(p\in E\). We now have \(E = C\), so \(E\) is convex.

In 3 (or more dimensions) you can apply essentially the same argument. When you reach the \(p\notin B\) case, just pick any plane containing \(p\), and use the above argument within that plane.

(In the case \(C = \emptyset\), the above still applies, though some aspects are vacuous.)

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