I went with Bernoulli, Binomial, and Poisson, which felt right.
Start by naming three distributions: Bernoulli, Binomial, and Poisson. For each, clearly state the core assumption (e.g., single trial, fixed number of independent trials, or rare events in a fixed interval) and then provide the expectation and variance formulas. Emphasize how these distributions relate to ad click modeling and when each is appropriate.
Pro tip: Mention that in practice, ad clicks often exhibit overdispersion (variance > mean), so the Poisson assumption may be too restrictive; consider negative binomial or quasi-Poisson as extensions. This shows awareness of real-world data complexities.
Select three distributions: Bernoulli, Binomial, and Poisson. These cover single click, clicks out of impressions, and clicks per time period.
For each, articulate the key assumption: Bernoulli: one trial with constant click probability; Binomial: fixed number of independent trials with constant probability; Poisson: events occur independently at a constant rate in a fixed interval.
Write the formulas: Bernoulli: E[X]=p, Var(X)=p(1-p); Binomial: E[X]=np, Var(X)=np(1-p); Poisson: E[X]=λ, Var(X)=λ.
Explain how each models ad clicks: Bernoulli for a single impression, Binomial for multiple impressions with same click probability, Poisson for clicks over time or space when events are rare.
Note that real ad click data may violate assumptions (e.g., overdispersion, varying probabilities), and mention alternatives like negative binomial or beta-binomial.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Straightforward law of total expectation application.
Recognize that the overall expected CTR is a weighted average of the group CTRs, where the weights are the group proportions. Since both groups have the same CTR (0.30), the overall expected CTR is simply 0.30. State the calculation clearly and note that the result is independent of the mix when group CTRs are equal.
Pro tip: Even when the answer seems trivial, briefly mention that if the CTRs differed, you would compute a weighted average, and that this logic extends to any segmentation. This shows you understand the general principle and can handle more complex scenarios.
Restate the proportions (90% high-intent, 10% low-intent) and the CTR for each group (0.30).
The overall expected CTR is the weighted average: (proportion_high * CTR_high) + (proportion_low * CTR_low).
Compute (0.9 * 0.30) + (0.1 * 0.30) = 0.27 + 0.03 = 0.30.
Explain that because both groups have the same CTR, the overall CTR equals that common value, regardless of the mix.
Mention that if the CTRs were different, the overall CTR would be a weighted average that depends on the proportions.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
The sketch part was fine, decreasing curve starting at lambda, everyone knows that.
Start by sketching the exponential PDF with its characteristic decreasing curve and long right tail, clearly labeling the rate parameter λ and the mean 1/λ. Then explain that as sample size grows, the distribution of the sample mean (not the raw data) approaches a normal distribution due to the Central Limit Theorem, and relate this to why A/B tests rely on normal approximations for large samples.
Pro tip: Emphasize that the exponential distribution itself does not change shape with sample size—it's the sampling distribution of the mean that becomes normal. This distinction shows you understand the difference between the population distribution and the sampling distribution, which is crucial for experiment analysis.
Draw a curve starting at λ on the y-axis when x=0, decaying exponentially toward zero as x increases, with a long right tail. Label the x-axis as the variable (e.g., time between events) and the y-axis as density, and mark the mean at 1/λ.
Clarify that the shape of the exponential distribution itself remains unchanged regardless of sample size. Instead, as sample size grows, the distribution of the sample mean becomes approximately normal with mean 1/λ and variance 1/(nλ²), by the Central Limit Theorem.
Discuss how this justifies using normal-based methods (e.g., t-tests, z-tests) for large samples in A/B tests, even when the underlying metric (like time between clicks) is exponential. Mention that the skewness diminishes in the sampling distribution as n increases.
Note that for small samples, the exponential distribution's skewness can bias results, so transformations or non-parametric methods might be needed. For large samples, the CLT ensures robustness, but be aware of heavy tails or extreme outliers.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.