Expected number of record highs in ten days

Question

Each day a trader records a fresh figure, drawn independently from the same continuous distribution, so ties never happen. Day 1 counts as a record by default, and a later day sets a "record high" when its figure beats every earlier day.

Over the first 10 days, what is the expected number of record highs?

Accepted Answer

Focus on a single day k and ask: is it a record? Day k is a record exactly when its figure is the largest among the first k days. Since those k figures are exchangeable, each is equally likely to be the biggest, so day k is a record with probability \frac{1}{k}. The expected number of records is the sum of these individual chances (expectation adds up even though the events overlap): E[	ext{records}] = \sum{k=1}^{10} \frac{1}{k} = 1 + 	frac12 + 	frac13 + \cdots + 	frac{1}{10} = \frac{7381}{2520} \approx 2.929. Records get rarer over time because each new day faces stiffer competition, so …

Expected number of record highs in ten days

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