Expected Time to Cross Uniform Threshold

This quant interview question is about understanding how a random stopping time interacts with a random threshold in a simple continuous setting. It uses only basic distributions, but the dependence between the process and the threshold makes the expected value nontrivial. On MyQuantPartner, it falls into core quant prep for expectation and stopping-time intuition, with a clean, one-line final answer that hides subtle behavior.

It trains conditional expectation, the law of total expectation, and comfort integrating over a random parameter. It also builds intuition for when expectations blow up even though each conditional scenario looks reasonable, which is a recurring theme in stochastic processes and Monte Carlo analysis.

This matters in quant interviews because it tests whether you can move seamlessly between conditioning, averaging over a distribution, and assessing finiteness of expectations, all under time pressure, without getting lost in algebra.

What it tests

When a random process depends on a parameter that is itself random, the overall expectation is found by averaging the conditional expectation over the distribution of that parameter. This is the Law of Total Expectation, which states that $\mathbb{E}[Y] = \mathbb{E}[\mathbb{E}[Y \mid X]]$ for any random variables $X$ and $Y$. The key is that the inner expectation often depends on the realized value of the conditioning variable, and sometimes this dependence can cause the outer expectation to diverge, especially if the conditional expectation grows rapidly as the parameter approaches certain values. In this class of problems, the structure is governed by how the randomness of one variable (here, the threshold $U$) influences the distribution of another (the waiting time $N$), and whether the averaging process over all possible parameter values leads to a finite or infinite result. The divergence often arises when the parameter can get arbitrarily close to a value that makes the conditional expectation explode, as happens with $1/(1-u)$ as $u \to 1$.