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This question is question 60 of chapter 1 of Introduction to Probability by Blitzstein and Hwang.

Basically the question walks you through reasoning about the bootstrap sampling procedure. Given $(a_1, ..., a_n)$ non-repeated numbers, $n\geq2$, you create a sample by sampling $n$ observations with replacement.

The question is: let $\textbf{b}_1$ be the most likely bootstrap sample, an $p_1$ the probability of getting $\textbf{b}_1$ and let $\textbf{b}_2$ be the least likely bootstrap sample, an $p_2$ the probability of getting $\textbf{b}_2$. What is $\frac{p_1}{p_2}$?


In my studies I found that:

a) there are $n^n$ ordered samples, since for each element of a sample, there are $n$ choices.

b) there are ${2n - 1 \choose n}$ unordered samples due to the stars and bars argument.

c) some samples from the ordered scheme are more represented in the unordered case.

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I think I have the question figured out, I just don't have a way of checking1.

When you compare the ordered samples to the unordered samples, one can see that some will be more represented and other less. For example take, with $n=2$ and $(a_1, a_2) = (3, 1)$. The ordered samples are $(1,1)$, $(3,1)$, $(1,3)$ and $(3,3)$. In the unordered case, the sample $(3,1)$ is more likely then $(1, 1)$ by a factor of 2. This gives us a hint that the number of permutations of the sample is the key.

Generically, the least likely sample $\textbf{b}_2$ is one where all the numbers are the same, since there is only one way of forming that sample. There are $n$ ways of getting a one element bootstrap sample, one for each element of $(a_1, ..., a_n)$. So $p_2 = \frac{n}{n^n} = \frac{1}{n^{n-1}}$.

Now, the most likely sample $\textbf{b}_1$ is the one where there are the most number of permutations. That is, the one where all elements of $\textbf{b}_1$ are different and therefore is equal to $(a_1, ..., a_n)$, since one repetition would cause the number of permutations to decrease. There is only one unordered sample where all elements are the same, that is $(a_1, ..., a_n)$. And there are $n!$ number of ways to sample it from the ordered samples. So $p_1 = \frac{n!}{n^n} = \frac{(n-1)!}{n^{n-1}}$.

Conclusively, $\frac{p_1}{p_2} = (n-1)!$ making $\textbf{b}_1$ $(n-1)!$ times more likely then $\textbf{b}_2$ during the sampling process.


Update with the probability of an arbitrary sample, as a suggestion by @Xi'an. A bootstrap sample boils down to the following:

  • which elements were sampled ($k$ as the number of elements)
  • how many times each element was sampled ($k_i$ as the number of times element $i$ was sampled, such that $\sum_{i=1}^{n}k_i = n$)
  • how many permutations can be made given the previous.

So, the number of ways to choose $k$ numbers from the $n$ $(a_1,..., a_n)$ is simply ${n \choose k}$. And so, given the vector of counts $$\textbf{v} = (k_1,...k_n) $$ such that $\sum_{i=1}^{n}k_i = n$, we can form the multinomial coefficient representing the permutations with repeated elements ${n \choose \textbf{v}}$ or equivalently ${n \choose k_1,...k_n} = \frac{n!}{k_1! \cdot ... \cdot k_n!}$ .

This leads to, by the multiplication rule, to the following probability:

$$ p_{k, \textbf{v}} = \frac{{n \choose k}\cdot{n \choose k_1,...,k_n}}{n^n} $$

For the cases of the sample $\textbf{b}_1$, $\textbf{b}_2$ the probabilities $p_1$ and $p_2$ are represented by $p_{n, (1,1,...,1)}$ and $p_{1, (n, 0,...0)}$, respectively. They simplify to:

$$ p_1 = p_{n, (1,1,...,1)} = \frac{{n \choose n}\cdot{n \choose 1,...,1}}{n^n} = \frac{1\cdot n!}{n^n} = \frac{(n-1)!}{n^{n-1}} $$

$$ p_2 = p_{1, (n, 0,...,0)} = \frac{{n \choose 1}\cdot{n \choose n,0,...,0}}{n^n} = \frac{n \cdot 1}{n^n} = \frac{1}{n^{n-1}} $$

1 I also just love the bootstrap, and take any chance of understanding it more technically.

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    $\begingroup$ Ah, yes! I was thinking of that. It would relate to the multinomial coefficient, since there would be repeated elements. I'll just think of a nice notation and update the answer! $\endgroup$ – Guilherme Marthe Mar 25 at 16:28
  • $\begingroup$ @Xi'an I think that my addition is what you suggested, right? $\endgroup$ – Guilherme Marthe Mar 25 at 20:33
  • $\begingroup$ Thanks for the addition. I am uncertain though that it is the correct answer. Check for instance Feller (1970, vol. 1, Section II.5, p.39, eqn (5.3)). $\endgroup$ – Xi'an Mar 26 at 7:34

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