← Two Sigma Interview Insights
The first part clicked pretty fast for me.
First, interpret the inequality as a minimax-type bound: for any fixed y, E_X[f(X,y)] ≤ E_{X,Y}[f(X,Y)] is not generally true without conditions. Instead, note that min_y E_X[f(X,y)] ≤ E_X[f(X, Y)] for any random Y, and then take expectation over Y to get min_y E_X[f(X,y)] ≤ E_{X,Y}[f(X,Y)]. For the second part, analyze when equality holds, especially under independence, and discuss conditions such as f being separable or achieving the minimum almost surely.
Pro tip: Emphasize the role of measurability and integrability to avoid technical pitfalls, and connect the inequality to the probabilistic method and minimax theorems, which are common in quantitative finance and algorithmic trading interviews.
State that X and Y are random variables on a common probability space, f is measurable, and expectations are well-defined (e.g., f is bounded or integrable).
For any fixed y, E_X[f(X,y)] ≤ E_X[f(X,Y)]? Not necessarily. Instead, use the fact that for any random variable Y, min_y E_X[f(X,y)] ≤ E_X[f(X,Y)] pointwise, then take expectation over Y to obtain the result.
Equality holds if and only if there exists a minimizer y* such that f(X, y*) = f(X, Y) almost surely, or more generally if the minimum is achieved almost surely by Y. Under independence, equality may hold if f is separable or if the minimum is constant.
Give a counterexample where the inequality is strict, e.g., f(x,y) = (x-y)^2 with X,Y independent standard normal, then min_y E[(X-y)^2] = Var(X) = 1, while E[(X-Y)^2] = 2.
Under independence, E_{X,Y}[f(X,Y)] = E_Y[ E_X[f(X,Y)] ]. Equality in the inequality requires that the minimum is achieved almost surely by Y, which is not generally true even under independence.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.