There are n non negative numbers x_i<1. It is given that S_1=sum n_{i=1}^{n}x_i=1. What is the probability that S_2=sum_{i=1}^{n}x_{i}^{2}<c where 1/n<c<1.

Rosemary Burns

Rosemary Burns

Open question

2022-08-19

Probability in higher dimension
There are n non negative numbers x i < 1. It is given that S 1 = i = 1 n x i = 1. What is the probability that
S 2 = i = 1 n x i 2 < c where
1 n < c < 1
Assume that each x i is uniformly distributed.
Now I tried looking at this problem in terms of integrals:
P ( s q u a r e s | s u m ) = P ( s q u a r e s s u m ) P ( s u m ) = s o m e t h i n g 0 1 0 1 x n 0 1 x 3 x 4 . . . x n . . . 0 1 x 2 x 3 . . . x n d x 1 d x 2 . . . d x n 1 d x n
Now for the numerator, if we substitute the constraint in the inequality, we get
S 2 + ( 1 S 1 ) 2 = c
S 2 + ( 1 S 1 ) 2 = c
Now how to get the numerator integral, I'm not quite sure.
Also, is there any better, more elegant way of doing this problem? (I thought of using vectors, where the problem can be written as: x . 1 = 1 then P ( | | x | | < c ) = ?. But then we would have to model the distribution of | | x | | which I have no idea about)
Or maybe a more direct approach? Like the ratio of areas (or volumes?).

Answer & Explanation

glasinamaav

glasinamaav

Beginner2022-08-20Added 10 answers

Step 1
There is the following naive and non-rigorous approach based on low-dimensional geometric analogues.
The required probability is a relation V 1 V 2 of two ( n 1 ) dimensional volumes. Namely, V 2 is a volume of ( n 1 ) dimensional simplex S in R n with vertices e 1 , , e n , where for each i, the vector e i has i-th coordinate equal to 1 and other coordinates equal to zero; V 1 is a volume of the intersection of S with a ball B of radius c , centered at the origin ( 0 , , 0 ) of R n
Step 2
Since S is a regular simplex with side a = 2 , V 2 = n ( n 1 ) ! see, for instance. The set S B is an ( n 1 ) dimensional ball of radius c 1 n , centered at ( 1 / n , 1 , n ), so
V 1 = ( π ( c 1 n ) ) n 1 2 Γ ( n + 1 2 ) .
Thus, V 1 V 2 = ( π ( c 1 n ) ) n 1 2 ( n 1 ) ! Γ ( n + 1 2 ) n .

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