A rumor that each person passes to 1.5 others

Question

On day 0, one person hears a rumor. Each person who newly hears the rumor passes it, the next day, to a random number of fresh people averaging 1.5 (everyone acts independently, and assume all the people reached are new). What is the expected number of people who first hear the rumor on day 4?

Accepted Answer

Each day's crop of new hearers is produced by the previous day's new hearers, so condition on the previous day and use the tower rule. Step 1, per-person spread. One new hearer generates, on average, m = 1.5 new hearers the next day. Step 2, the recursion. Let Xt be the number who first hear the rumor on day t. Given X{t-1}, each of those people independently generates 1.5 new hearers, so E[Xt \mid X{t-1}] = 1.5 \, X{t-1}, \qquad 	ext{hence} \qquad E[Xt] = 1.5 \, E[X{t-1}]. Step 3, iterate from X0 = 1. Each day multiplies the expected new count by 1.5: E[X4] = 1.5^4 = \frac{3^4}{2^4} = \frac…

A rumor that each person passes to 1.5 others

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