Quadratic utility is defined as: $$ U(W) = aW - bW^2 $$ where: - \(W\) = wealth, - \(a, b > 0\) are constants.

Why utility function is assumed this way?

If utility is quadratic: $$ U(W) = aW - bW^2 $$ then the expected utility is: $$ \mathbb{E}[U(W)] = a \mathbb{E}[W] - b \mathbb{E}[W^2] $$ Expand \(\mathbb{E}[W^2]\) using the formula: $$ \mathbb{E}[W^2] = (\mathbb{E}[W])^2 + \text{Var}(W) $$ This comes from the relationship: \(\text{Var}(X) = E(X)^2 - E(X^2)\)

thus: $$ \mathbb{E}[U(W)] = a \mathbb{E}[W] - b\left((\mathbb{E}[W])^2 + \text{Var}(W)\right) $$ Grouping terms: $$ \mathbb{E}[U(W)] = (a - 2b \mathbb{E}[W])\mathbb{E}[W] - b \, \text{Var}(W) $$ Interpretation: - You maximize expected utility by increasing expected return \(\mathbb{E}[W]\)
- and minimizing variance \(\text{Var}(W)\).

Thus, maximizing expected quadratic utility reduces to a trade-off between expected return and variance, which is exactly the objective of mean-variance optimization.

Why Mean Variance Optimization is a projection problem?

You are projecting the expected return vector \(\alpha\) onto the weight space, using a distance measured by the covariance matrix \(\Omega\).

The space (portfolio space or weight space) is \(\mathbb{R}^n\) for \(n\) assets. Instead of normal Euclidean distance, you use risk-based distance: $$ d(w) = \sqrt{w^T \Omega w} $$This risk-based distance stretches and twists the space — the geometry is no longer flat like Euclidean space.

"We are finding the weight vector \(𝑤\) that points in the best possible direction towards the return vector \(\alpha\), but measured using risk distance defined by \(\omega\).

Optimization Problem

$$ \max_w \quad \alpha^T w \quad \text{subject to} \quad w^T \Omega w \leq \sigma^2 $$ Lagrangian leads to: $$ \Omega w \propto \alpha \quad \Rightarrow \quad w \propto \Omega^{-1} \alpha $$ Thus, optimal weight is proportional to \(\Omega^{-1} \alpha\) — meaning alpha corrected by risk geometry.