0
$\begingroup$

I was doing a problem involving athletes competing at the high jump, with the jth jumper achieving height $X_j$, with $i = 1..n$ and $X_i$ i.i.d. from a continuous distribution. We say the $j$th jumper sets a record if $X_j > X_i$ for $i = 1..j-1$. The question is to calculate the expected number of "double records" among the first $n$ jumpers, with a "double record" occurring at time $j$ if both the $j$th and $j-1$th jumpers set records.

The solution is pretty straightforward - we calculate $\sum_1^n P(R_j)$ where $P(R_j)$ is the probability that a double record occurs at $j$. It assumes that the events $R_j$ are independent -- I don't think this is true, though. Isn't it the case that $P(R_j|R_{j-1}) = 1/j > P(R_j)$?

$\endgroup$
1
  • $\begingroup$ I'm not sure why $P(R_j|R_{j-1})$ should be $1/j$, but I do agree that they don't seem independent. $\endgroup$ Commented May 17, 2017 at 23:36

2 Answers 2

Reset to default
0
$\begingroup$

Thought about it a bit more and now it seems like a dumb question -- the expectation of the sum is just the sum of expectations regardless of whether the variables are independent.

$\endgroup$
0
$\begingroup$

The probability the $j$th jump is an individual new record is $\frac1j$ and it is independent of how many previous records there have been and which ones they were. But this is not the case with double records as the event $R_{j-1}$ requires the $j-1$th jump to have been a record and so increases the chance of the event $R_j$

You have $P(R_j \mid R_{j-1}) = \frac1j$ for $j \ge 1$, as you say

and $P(R_j)=\frac{1}{(j-1)j}=\frac1{j-1}-\frac1j$ for $j \gt 1$, and this is greater than $P(R_j \mid R_{j-1})$ when $j \gt2$

but, as you also say, this does not matter as you can use linearity of expectation

The expected number of double records is then $\sum\limits_{j=2}^n P(R_j) = 1 -\frac1n$

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.