Median trick

The median trick is a generic approach that increases the chances of a probabilistic algorithm to succeed.^[1] Apparently first used in 1986^[2] by Jerrum et al.^[3] for approximate counting algorithms, the technique was later applied to a broad selection of classification and regression problems.^[2]

The idea of median trick is very simple: run the randomized algorithm with numeric output multiple times, and use the median of the obtained results as a final answer. For example, if an algorithm takes a set of data as input, and has sublinear runtime, then the same algorithm can be run repeatedly (or in parallel) over randomly sampled subsets of input data, and, per Chernoff inequality, the median of the results will converge to solution rapidly.^[4] Similarly, for the algorithms that are sublinear in space (e.g., counting the distinct elements of a stream), different randomizations of the algorithm (say, with different hash functions) may be used for repeated runs over the same data.^[5]

Statement

Given a set of independent random variables ${\textstyle X_{1},\dots ,X_{n}}$, and an unknown deterministic number ${\textstyle Y}$.

Suppose that each random variable ${\textstyle X_{i}}$ falls within ${\textstyle [Y\pm \epsilon ]}$ with probability ${\textstyle \geq p}$ where ${\textstyle p>1/2}$ is a constant, then the median trick states that ${\textstyle Med(X_{i})\in [Y\pm \epsilon ]}$ with probability ${\textstyle \geq 1-e^{-2n(p-1/2)^{2}}}$.

In other words, in order to ensure that ${\textstyle Y\in [Med(X_{i})\pm \epsilon ]}$ with probability ${\textstyle \geq 1-\delta }$, it suffices to use ${\textstyle {\frac {\ln {\frac {1}{\delta }}}{2(p-1/2)^{2}}}}$ samples.

Proof

Let ${\textstyle Z_{i}}$ be the indicator variable for the event that ${\textstyle X_{i}\in [Y\pm \epsilon ]}$. Then, the event ${\textstyle Med(X_{i})\in [Y\pm \epsilon ]}$ fails to occur only if at least half of ${\textstyle Z_{i}=1}$, that is, ${\textstyle {\frac {1}{n}}\sum _{i}Z_{i}\leq 1/2}$.

By Hoeffding's inequality, this event occurs with probability ${\textstyle \leq e^{-2n(p-1/2)^{2}}}$.