Fulfill a matrix with a function

[homerjaysimpson]

I have the following vector in R :

p = c(1/4,1/8,1/6,1/16,3/16,1/12,1/8);p

which has length 7.

I want to c.reate a matrix that will be N times N, where here N=7 but th. e the entries to be with metropolis Hastings algorithm.
As the picture suggests:
So r(x,y)=1/7

and p(x,y) =r(x,y) \min(\frac{\pi(y)}{\pi(x)},1)

For example the p(1,2) = 1/7 \cdot \min(\frac{1/8}{1/4},1) =1/14

p(1,3) = 1/7 \cdot \min(\frac{1/6}{1/4},1) =2/21 apart from the entry p(1,1) = 1-\sum_{i=2}^{j=7}p(i,j).
This sum will be in all the diagonal entries.

r = 1/7
r*min(p[2]/p[1],1)

Any help ?

[mhakanda]

For diagonal items, your indexes are i and j, but inside you have p(x,y). See your p(1,1). What would be p(2,2) and p(3,3) and so on. Can you please clarify?

[homerjaysimpson]

p(i,j).I corrected it.The diagonal entries must have 1-sum(the values of the same row) .For example if row 4 is:
1,2,3,x,5,6,7 then x=1-sum(1,2,3,5,6,7)

[homerjaysimpson]

.

[mhakanda]

Please check its correctness. I assume there is a better, compact way to do that. But, i just follow the algorithm.

p = c(1/4,1/8,1/6,1/16,3/16,1/12,1/8)
mh<-matrix(nrow = 7,ncol = 7)
r<-1/7
for (i in 1:7) {
  for (j in 1:7) {
    if(i!=j){
      mh[i,j]<-r*min(p[j]/p[i],1)
    }
  }
  mh[i,i]<-1-sum(mh[i,],na.rm = TRUE)
}
mh
#>           [,1]       [,2]      [,3]       [,4]      [,5]       [,6]       [,7]
#> [1,] 0.5714286 0.07142857 0.0952381 0.03571429 0.1071429 0.04761905 0.07142857
#> [2,] 0.1428571 0.26190476 0.1428571 0.07142857 0.1428571 0.09523810 0.14285714
#> [3,] 0.1428571 0.10714286 0.3750000 0.05357143 0.1428571 0.07142857 0.10714286
#> [4,] 0.1428571 0.14285714 0.1428571 0.14285714 0.1428571 0.14285714 0.14285714
#> [5,] 0.1428571 0.09523810 0.1269841 0.04761905 0.4285714 0.06349206 0.09523810
#> [6,] 0.1428571 0.14285714 0.1428571 0.10714286 0.1428571 0.17857143 0.14285714
#> [7,] 0.1428571 0.14285714 0.1428571 0.07142857 0.1428571 0.09523810 0.26190476

[homerjaysimpson]

The rowsum must be 1

[mhakanda]

Yes, they are 1s.

rowSums(mh)
#> [1] 1 1 1 1 1 1 1

[mhakanda]

We can also see in this way.

#>      [,1]  [,2]  [,3]  [,4]  [,5]  [,6]  [,7] 
#> [1,]   4/7  1/14  2/21  1/28  3/28  1/21  1/14
#> [2,]   1/7 11/42   1/7  1/14   1/7  2/21   1/7
#> [3,]   1/7  3/28   3/8  3/56   1/7  1/14  3/28
#> [4,]   1/7   1/7   1/7   1/7   1/7   1/7   1/7
#> [5,]   1/7  2/21  8/63  1/21   3/7  4/63  2/21
#> [6,]   1/7   1/7   1/7  3/28   1/7  5/28   1/7
#> [7,]   1/7   1/7   1/7  1/14   1/7  2/21 11/42

[homerjaysimpson]

With the MASS::fractions() ?

[mhakanda]

It just shows number in fraction. Both is same matrix with different presentation. In later fraction matrix, we can easily see row sum is 1.

[homerjaysimpson]

thanks a lot @mhakanda I appreciate your help mate

[Leighton]

A huge thanks for this discussion. As a newbie I was very interesting to read