Math 123: Mathematical Aspects of Data Analysis – Fall 2026
The Basics of CQC Data
Data as Vectors
Data as Matrices
Remark (Rows versus columns). Two conventions:
- \(\bfX \in \Rbb^{n \times d}\): rows are observations, columns are features.
- \(\bfY \in \Rbb^{d \times n}\): columns are observations, rows are features.
These conventions are related by transposition: \(\bfY = \bfX^\top\).
Big-Data is Messed Up
Descent into Probability
Suppose we have a bunch of data, from the same ‘distribution’ \[ X_1,\ldots,X_n \] with a common mean \(\mu\). The average reading is \[ \overline{X} = \frac1n\sum_{i=1}^n X_i. \]
Our first probability question is the one that will guide both lectures:
Why should \(\overline{X}\) be close to \(\mu\), and how unlikely is it that it is not?
Probability enters data science whenever data are sampled, measurements are noisy, or an algorithm uses randomness. We will build only the vocabulary and tools needed to make statements about those situations.
Remark (Probability bootcamp I). By the end of this lecture, we want to be able to:
- identify random variables, distributions, events, and independent trials;
- compute simple expectations and use linearity of expectation;
- explain why averaging independent noise reduces variance;
- turn expectation or variance information into a tail bound using Markov’s or Chebyshev’s inequality.
Random Variables, Distributions, and Events
Expectation
Returning to repeated measurements with \(\Ebb[X_i]=\mu\), linearity gives \[ \Ebb[\overline{X}] = \frac1n\sum_{i=1}^n\Ebb[X_i] = \mu. \] Thus the average is centered at the right place.
Variance, Covariance, and Averaging
Covariance will reappear later as a matrix that describes the directions in which a point cloud varies. For now, its essential role is to show exactly where independence enters the averaging calculation.
Tail Probabilities
A tail probability measures the chance of an unusually large deviation, such as \[ \Pbb(|X-\Ebb[X]|\ge t). \] The event inside the probability is the bad event we want to control. Often we do not need its exact probability; an upper bound is enough.
Proof (idea). On the event \(\{X\ge t\}\), the variable \(X\) contributes at least \(t\). Therefore \[ \Ebb[X] \ge t\Pbb(X\ge t). \] Divide by \(t\).
Markov’s inequality is easy and informative, but often loose. Often, we can do a little better:
Proof (from Markov). Apply Markov’s inequality to the non-negative random variable \((X-\mu)^2\): \[ \begin{aligned} \Pbb(|X-\mu|\ge t) &= \Pbb((X-\mu)^2\ge t^2)\\ &\le \frac{\Ebb[(X-\mu)^2]}{t^2} = \frac{\sigma^2}{t^2}. \end{aligned} \]