Probability in higher dimensionThere are n non negative numbers x i &lt; 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 &lt; c where 1 n &lt; c &lt; 1Assume 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 = cNow 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 → | | &lt; 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?).

Question

Probability in higher dimensionThere are n non negative numbers       x    i    &amp;lt;  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    &amp;lt;  c where      1    n    &amp;lt;  c  &amp;lt;  1Assume 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    =  cNow how to get the numerator integral, I&#039;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      →            |        |    &amp;lt;      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?).

glasinamaav · Accepted Answer

Step 1There 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 2Since 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              .

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.

Open question

Answer & Explanation

New Questions in High school geometry