ISI 2008 ME1 doubt

[varnika1880]

A square matrix of order n is said to be a bistochastic matrix if all of its entries are
non-negative and each of its rows and columns sum to 1. Let 1
1 . ×
× = n
n×n
P
y where
elements of y are some rearrangements of the elements of x. Then n x
(a) P is bistochastic with diagonal elements 1,
(b) P cannot be bistochastic,
(c) P is bistochastic with elements 0 and 1,
(d) P is a unit matrix.

[varnika1880]

Question 26 and 27

[Dr. Strange]

26.

r(x)= x* f(x)/(1-F(x))

Numerator = x*f(x)  -> we can check using calculus that for x< e^ mu this function is increasing in x

Denominator =1-F(x) -> always  decreasing in x
Since numerator is increasing in x and denominator is  decreasing in x so whole function is increasing in x

[Dr. Strange]

27. Let y=[ y1 y2 ... yn ], x=[ x1 x2 ... xn ], P=[ a11 a12 a13... a1n]
                                                                         a21     ......       a2n    
                                                                          an1....              ann
Now y1 = a11 x1+ a12 x2 +..... a1n*xn
 Since y is rearrangement of x
So,
y1= xi ( i canbe 1 - n)
xi=a11 x1+ a12 x2 +...a1k*xi....+ a1n*xn

This becomes an identity only when a1k=1 and all other coefficents =0

So P has exactly one non zero value in every row and column and that value is 1

So  P is bistochastic with elements 0 and 1
   

[varnika1880]

Thanks a lot