I have the following problem:
Athletes compete one at a time at the high jump. Let $X_j$ be how high the $j^{th}$ jumper jumped, with $X_1, X_2, \dots$ i.i.d. with a continuous distribution. We say that the $j^{th}$ jumper sets a record if $X_j$ is greater than all of $X_{j - 1}, \dots, X_1$.
- Find the mean number of records among the first $n$ jumpers. What happens to the mean as $n \to \infty$?
The solution provided is as follows:
By linearity, the expected number of records among the first n jumpers is $\sum_{i = 1}^j = 1/j$ which goes to $\infty$ as $n → \infty$ since the harmonic series diverges.
I understood everything except for where the $\sum_{i = 1}^j = 1/j$ came from. I would greatly appreciate it if people could please take the time to explain this.
Consider the sequence $(A_j)$ where $A_j$ denotes the event that the $j^\text{th}$ jumper sets a record, and then consider their indicator random variables $$1_{A_j} = \begin{cases}1 & \text{ if } A_j \text{ occurs} \\ 0 & \text{ otherwise}\end{cases}$$ Now, you should convince yourself that the number of records among the first $n$ jumpers is $$\sum_{j=1}^n 1_{A_j}$$
By linearity and the definition of the expectation of a bernoulli random variable, we have $$E\left[\sum_{j=1}^n 1_{A_j}\right] = \sum_{j=1}^n E\left[1_{A_j}\right] = \sum_{j=1}^n P(A_j)$$
Therefore, to finish the computation, it suffices to compute the probability that the $j^\text{th}$ jumper sets a record. Since the jumps are iid with a continuous distribution, this can be rephrased as "what is the likelihood that the best jumper out of jumpers $1,2,\ldots, j$ is in position $j$? Since the best jumper out of the first $j$ jumpers is equally likely to be given any position, this probability is $1/j.$
Therefore the mean number of records in the first $n$ jumpers is $$\sum_{j=1}^n \frac{1}{j}$$