This was the centerpiece question and I spent way too long on S2 before realizing I needed to account for what the classifier actually does to the composition of shown impressions.
Start by clearly defining the notation and assumptions, then compute expected profit per 1,000 impressions for each strategy using the given probabilities and costs. For the threshold strategy, express profit as a function of the threshold, take the derivative, and solve for the optimal threshold where marginal revenue equals marginal cost. Finally, plug in the numbers to get the numeric value and compare strategies.
Pro tip: Emphasize that the optimal threshold is where the posterior conversion probability times revenue per conversion equals the cost per impression, i.e., p* = cost_per_impression / revenue_per_conversion. This shows you understand the underlying economics and can generalize beyond the specific numbers.
Clearly state the given probabilities (click and conversion for High-Intent and Low-Intent), revenue per conversion, cost per impression, and classifier precision and recall. Assume independence between click and conversion given segment, and that the classifier's predictions are based on observable features.
Calculate the overall click-through rate and conversion rate as weighted averages of the segment rates, using the segment proportions. Then compute expected profit per 1,000 impressions as 1000 * (conversion_rate * revenue_per_conversion - cost_per_impression).
Using the classifier's precision and recall, determine the proportion of users targeted and the conversion rate among them. Then compute expected profit per 1,000 impressions similarly, accounting for the fact that only predicted High-Intent users are shown the ad.
For the threshold strategy, express expected profit per impression as a function of the threshold t: profit(t) = P(p > t) * (E[p | p > t] * revenue_per_conversion - cost_per_impression). Show that the derivative leads to the condition that the optimal threshold is t* = cost_per_impression / revenue_per_conversion, assuming the posterior probability distribution is continuous and well-behaved.
Substitute the given values into the formulas to compute the numeric profit for each strategy and the optimal threshold. Compare the results to identify which strategy yields the highest expected profit per 1,000 impressions.
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I knew the law of total variance formula but applying it to a mixture Bernoulli is one of those things where you have to be careful about what's random.
First, define the mixture model: with probability p, an impression is High-Intent with conversion rate r_H, and with probability 1-p it is Low-Intent with rate r_L. Then apply the law of total variance to the number of conversions per 1,000 impressions, treating each impression as a Bernoulli trial with a random success probability drawn from the mixture. Finally, compare the resulting variance to the variance of a single Bernoulli with the population-average rate r_bar = p*r_H + (1-p)*r_L, and conclude that the mixture increases overdispersion (variance > Bernoulli variance) whenever r_H ≠ r_L.
Pro tip: Emphasize that the law of total variance decomposes the variance into 'within-component' and 'between-component' parts, and that the between-component term is exactly the extra overdispersion caused by the mixture—this shows you understand the mechanism, not just the formula.
Define the two intent segments with probabilities p and 1-p and conversion rates r_H and r_L. State that for a single impression, the conversion indicator Y has a Bernoulli distribution with random success probability R that equals r_H with probability p and r_L with probability 1-p.
Compute Var(Y) = E[Var(Y|R)] + Var(E[Y|R]) = E[R(1-R)] + Var(R). Show that E[R] = r_bar and Var(R) = p(1-p)(r_H - r_L)^2, and that E[R(1-R)] = r_bar(1-r_bar) - Var(R).
Assuming independence across impressions, multiply the single-impression variance by 1,000 to get the variance of the number of conversions per 1,000 impressions. Explicitly note the independence assumption and that the mixture is at the impression level.
The single Bernoulli with rate r_bar has variance 1,000 * r_bar(1-r_bar). The mixture variance is 1,000 * [r_bar(1-r_bar) + Var(R)] because the within-component term simplifies to r_bar(1-r_bar) - Var(R) and the between term adds Var(R). Thus the mixture variance exceeds the Bernoulli variance by 1,000 * p(1-p)(r_H - r_L)^2.
State that the mixture increases overdispersion whenever r_H ≠ r_L, because the between-segment variance term is strictly positive. If r_H = r_L, the mixture reduces to a single Bernoulli and there is no extra overdispersion.
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Honestly the most conversational part of the whole interview.
Start by restating the PM's proposal and framing the analysis around the two quantitative dimensions: reach impact and profit sensitivity to classifier errors. Then present two pros and two cons, each backed by specific numbers from the earlier analysis, and conclude with a balanced recommendation that acknowledges trade-offs.
Pro tip: Quantify the trade-off explicitly—e.g., 'Targeting high-intent users increases profit per impression by X% but reduces reach by Y%, and a 1% classifier error costs $Z in profit'—to show you can translate model metrics into business impact.
Briefly restate what targeting high-intent users means and identify the key metrics from the earlier analysis: reach, profit per impression, and classifier error rates.
State two pros with numbers: e.g., higher conversion rate leading to increased profit per impression, and better ROI due to reduced wasted spend on low-intent users.
State two cons with numbers: e.g., significant reduction in reach (e.g., 40% fewer impressions) and profit sensitivity to classifier errors (e.g., a 5% false negative rate could lose $X in potential profit).
Explain how the pros and cons interact, emphasizing the profit sensitivity to classifier errors and the diminishing returns of narrowing targeting.
Suggest a hybrid strategy, such as using a tiered targeting system or adjusting bids by intent score, to capture upside while mitigating reach loss and error sensitivity.
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Standard exponential distribution stuff, I was fine on the PDF/CDF and the CLT part.
Start by clearly stating the exponential distribution for inter-arrival times, then derive the PDF, CDF, mean, variance, and tail probability. Next, explain the Law of Large Numbers and Central Limit Theorem, and finally connect these to why average metrics stabilize in production dashboards.
Pro tip: Emphasize that the exponential distribution is memoryless and that the CLT explains the approximate normality of sample means, which is why dashboards can use confidence intervals and detect anomalies.
State that inter-arrival times follow an exponential distribution with rate λ, and write the PDF f(t)=λe^{-λt} and CDF F(t)=1-e^{-λt} for t≥0.
Derive the mean as 1/λ and variance as 1/λ^2, showing the integration or citing known results.
Compute P(T>10) = 1 - F(10) = e^{-10λ}, and plug in the given rate if provided.
State that by the Law of Large Numbers, the sample mean converges to the true mean 1/λ as sample size increases.
Describe how the Central Limit Theorem ensures the sample mean is approximately normal for large n, leading to stable average metrics and enabling statistical process control.
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This one tripped me up more than it should have.
Start by explaining that a mixture of two exponential distributions with different rates is not exponential; it is a hyperexponential distribution. Then describe its qualitative shape (heavier tail, higher probability of very short and very long inter-click times) and recommend a diagnostic plot such as a log-survival plot or a QQ plot against exponential to detect the mixture.
Pro tip: Mention that in practice you would also look at the hazard function: a decreasing hazard is a hallmark of unobserved heterogeneity (mixture), whereas a constant hazard indicates exponentiality. This shows you understand the underlying process, not just the math.
Clarify that if all users had the same click propensity, inter-click times would be exponential (constant hazard).
Describe that mixing two exponential distributions with different rates yields a hyperexponential distribution, which is not exponential.
Note that the mixture has a heavier tail and more mass near zero than a single exponential, and its hazard function decreases over time.
Suggest a log-survival plot (log P(T>t) vs t) which would be linear for exponential but convex for a mixture, or a QQ plot against exponential which would show curvature.
Briefly mention that detecting a mixture can inform user segmentation or model improvements, e.g., using a mixture model or adding covariates.
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