Leon Valentin Sierau

PhD candidate, Causality, Probability Theory & Statistics

About me

I am a PhD candidate and Research Associate at the Institute for Data Science Foundations at Hamburg University of Technology, where I work with Nihat Ay. My research is in causality, probability theory and statistics. I am currently also working on causal world models for reinforcement learning.

Before my PhD, I studied Data Science at TUHH and, before that, Psychology at Kiel University. My interest in cognition and intelligence has stayed with me, although I now approach these questions more from the mathematical side.

Publications

  • A Stronger Calculus of Intervention for Max-Linear Bayesian Networks
    Leon Sierau, F. Nowell, Nihat Ay.
    Uncertainty in Artificial Intelligence (UAI), PMLR 337, 2026. [paper] [code]

    We study how the structure of max-linear models allows us to identify causal effects beyond standard d-separation-based methods.

  • Common Cause Condition for Universal Approximation
    Leon Sierau, Nihat Ay.
    Extended version accepted for publication in Kybernetika. [WUPES 2025 conference version]

    A necessary common-cause condition for universal approximation of Markov kernels by feed-forward networks of binary stochastic units.

  • After-effects of 10 Hz tACS over the prefrontal cortex on phonological word decisions
    V. Moliadze, Leon Sierau, et al.
    Brain Stimulation 12:1464–1474, 2019. [paper]

Education

  • 2024–present: PhD candidate, Causality, Probability Theory & Statistics, Hamburg University of Technology.
  • 2021–2024: B.Sc. Data Science, Hamburg University of Technology (top of cohort).
  • 2012–2017: Diplom, Psychology, Kiel University.

Teaching

I teach Causal Inference and Machine Learning at TUHH to students in Computer Science, Data Science, and Technomathematics.

Fascinating maths

I like problems that are easy to explain but difficult to solve. What fascinates me is how a good idea can suddenly make them look simple, and how often that idea comes from a different part of mathematics. Here are a few examples.

The 100 prisoners problem

One hundred prisoners and 100 boxes are each numbered 1–100. The prisoners’ numbers are placed randomly in the boxes, one per box. Each prisoner may open 50 boxes, alone. They can agree on a strategy beforehand, but cannot communicate afterwards or leave the boxes changed. Everyone goes free only if everyone finds their own number. Can they give themselves a chance of more than 30%?

A strategy and why it works

Start with the box bearing your number, then open the box named by the number inside, and keep following. The arrangement is a permutation, so it splits into cycles. Everyone succeeds exactly when no cycle is longer than 50.

For k > 50, there can be at most one k-cycle. Choose its k numbers, arrange them in a cycle, then arrange the rest. This counts 100! / k of the 100! possible permutations, so the probability of such a cycle is 1 / k. Two cycles longer than 50 cannot coexist, so the failure probabilities simply add:

P(success) = 1 − (1/51 + 1/52 + ⋯ + 1/100) ≈ 31.18%.

Each prisoner still has a one-half chance of finding their number. The trick is to make their successes depend on one another. Independent guesses would give the group only a 2−100 chance.

Further reading: Peter Winkler’s explanation (PDF).

Is there always a “politician”?

In a finite group of at least three people, every pair has exactly one mutual friend. Must there be someone who is friends with everyone? Assume friendship is mutual and nobody is their own friend.

Proof sketch (with a little linear algebra)

Yes: this is the friendship theorem. Draw a graph with people as vertices and friendships as edges. Suppose, for a contradiction, that nobody is friends with everyone.

First, everyone has the same number of friends. Take two non-friends u and v. For each friend x of u, take the unique mutual friend of x and v. These are distinct friends of v: if two choices gave the same person y, then u and y would have two mutual friends. Hence v has at least as many friends as u; reversing the argument gives equality.

Call this number k, and let w be the mutual friend of u and v. Everybody except w is a non-friend of u or v, so has k friends too. By our assumption, w also has a non-friend, so the same holds for w. Every edge is part of a triangle, so k ≥ 2.

Now look at the adjacency matrix A. Its entries record friendships; entries of A2 count mutual friends. Thus

A2 = (k − 1)I + J,

where I is the identity matrix and J is the all-ones matrix. The all-ones vector has eigenvalue k. On its orthogonal complement, J vanishes, so the other eigenvalues are +√(k − 1) or −√(k − 1). If their multiplicities are r and s, the zero diagonal of A gives

0 = tr(A) = k + (r − s)√(k − 1).

So √(k − 1) is rational, hence an integer h. The same equation says h divides k = h2 + 1, forcing h = 1 and k = 2. But a graph in which everyone has two friends and every pair has one mutual friend must be a single triangle. Everyone is friends with everyone after all: a contradiction.

I like the unexpected turn here: a question about friendships ends with eigenvalues and a divisibility argument.

Following the proof in Aigner and Ziegler’s Proofs from THE BOOK, chapter 44. The theorem is due to Erdős, Rényi and Sós.

Counting without constructing: the probabilistic method

Twenty players each play every other player once, without draws. Can the results allow more than four trillion different line-ups of all 20 players in which each player beat the next?

A surprisingly short counting argument

Imagine deciding each match by an independent fair coin toss. For any fixed line-up, the 19 required results occur with probability 2−19. There are 20! line-ups, so the expected number that work is

20! / 219 = 4,640,392,319,062.5.

At least one set of match results must reach or exceed that average. That proves the claim without constructing a tournament or enumerating its paths. Only consecutive players in a line-up need to satisfy the condition, not every earlier and later pair.

This is the probabilistic method: introduce randomness to prove a definite existence statement. If something has a positive chance of happening, it must be possible. Here we use the fact that some outcome must be at least as large as the average. We can add the expectations even though the different line-ups are not independent.

Szele’s argument; see the introduction to Noga Alon’s paper on Hamiltonian paths in tournaments (PDF).