Statistics (expectation)

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Let $I_i$ be a random variable that designates if the $i$th person got his name, namely $$ I_i \sim Ber(1/n), \quad i=1,...,n . $$ To be sure, you can check that $$ P(I_i=1)=\frac{n-1}{n}\frac{n-2}{n-1}\cdots\frac{n-i}{n-(i-1)}\frac{1}{n-i} =\frac{1}{n}. $$ The total number of people that got their names are $X = \sum_{i=1}^n I_i $, hence, $$ EX=nEI_i= \frac{n}{n}=1. $$

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Updated on August 01, 2022

Comments

  • HEM
    HEM over 1 year

    Each of $n\ge 2$ people puts his or her name on a slip of paper (no two have the same name). The slips of paper are shuffled in a hat, and then each person draws one (uniformly at random at each stage, without replacement). Find the average number of people who draw their own names.

    • lulu
      lulu over 6 years
      Any thoughts? Can you, say, solve it for modest $n$?
  • HEM
    HEM over 6 years
    yeah got it. thank you
  • BruceET
    BruceET over 6 years
    This is the famous 'item matching problem'. Also known by a variety of names, such as 'hat check problem.' Also, $Var(X) = 1,$ but that is not so easy to show because the indicator variables are not independent. For $n$ about 10 or more, the distribution of $X$ is approximately Pois(1); not exactly because $P(X = n-1) = 0$ and $P(X > n) = 0.$