Mathematics Asked by AliBA on February 12, 2021
Consider a random permutation of numbers $1,…,31415$. Let $A$ be the number of its fixed points (i.e. numbers which are not moved by permutation). Let $B$ be the number of non-fixed points. What is the variance of $B-A$?
I have no idea.
Let $N = 31415$. So in this case $N = A + B$.
$implies Var(B-A) = Var(N-2A) = Var(-2A) = 4*Var(A)$
Therefore our primary task is to calculate $Var(A)$. We can use identity variables.
Let $X_i$ be identity random variable which becomes $1$ when ith element is at the ith position in the permutation.
$$A = X_1 + X_2 + .... + X_n$$
$X_i$ is $1$ with probability $1/N$ and $0$ with probability $(N-1)/N$
By linearity of expectation $E[A] = frac{1}{N} times N = 1$
$$E[A^2] = displaystyle sum_{n=1}^N E[X_i^2] + 2displaystyle sum_{i<j}E[X_iX_j] $$
Since $X_i$ is indicator variable $E[X_i] = E[X_i^2] = 1$
If $i<j, E[X_iX_j] = (n-2)!/n!$, because in $(n-2)!$ permutations $X_i = 1$ and $X_j = 1$.
$$E[A^2] = 1 + 2* n(n-1)/2 times (n-2)!/n! = 1 + 1 = 2$$
$$var(A) = E[A^2] - (E[A])^2 = 2 - 1 = 1.$$
$$var(B-A) = 4 times var(A) = 4$$
Answered by Sameer Pande on February 12, 2021
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