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Let $R_k$ for $R_1, R_2, R_3... R_{10}$ be the cumulative inches of rainfall in year $k$ in Springfield. The amount of rainfall per year is independent of the rainfall during other years. On average, how many of these 10 years will have record rainfall? $\forall k \in [1, 10], R_k \sim Normal(\mu, \sigma)$ such that $\mu$ and $\sigma$ make it very unlikely for rainfall values to be negative.

Since each year is independent, would the answer be $10(\frac{1}{10}) = 1$?

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By record, do you mean an amount of rain that surpasses the preceding one? –  Jean-Sébastien Oct 15 '12 at 15:25
    
Sorry about that. I meant rainfall surpassing all previous rainfalls within the 10 years. –  John Hoffman Oct 15 '12 at 18:23

1 Answer 1

up vote 2 down vote accepted

We assume that record-keeping is confined to the $10$ years. In particular, Year $1$ automatically has record rainfall. We also assume that a tie between two years has probability $0$. This is a consequence of modelling rainfall as a random variable with continuous distribution.

Define the Bernoulli random variables $X_i$ by $X_1=1$, and for $i\gt 1$, $X_i=1$ if year $i$ has rainfall greater than years $1$ to $i-1$, and $0$ otherwise. The number of records $Y$ is the sum of the $X_i$.

We have $\Pr(X_i=1)=\dfrac{1}{i}$. This is because for years $1$ to $i$, the greatest rainfall is just as likely to be in one year as in another. So $E(X_i)=\dfrac{1}{i}$, and by the linearity of expectation $$E(Y)=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{10}.$$ Note that the $X_i$ are not independent, but that does not matter for expectation.

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