rnorouzian - 3 years ago 151

R Question

I'm completely baffled by the following simple R code. In the **first part**

`x`

`v`

But then strangely in the

`x`

`v`

`m1 = 5`

m2 = 1.3*m1

A = m1 + m2

x = 5

a <- function(m3){

abs((m1 - (A + m3)/3)^2 + (1.3*m1 - (A + m3)/3)^2 + (m3 - (A + m3)/3)^2 - 3*x) }

m3 = optimize(a, interval = c(0, 100), tol = 1e-20)[[1]]

v = var(c(m1, m2, m3))*(2/3) # gives "5" same as "x"

`eta.sq = .25`

beta = qnorm(c(1e-12, .999999999999))

q = c(0, 25)

mu.sig = solve(cbind(1L, beta), q)

m1 = mu.sig[[1]]

H = (mu.sig[[2]])^2

m2 = 1.3 * m1

A = m1 + m2

x = (H * eta.sq) / (1 - eta.sq) # "x" is: 1.052529

a = function(m3){

abs((m1 - (A + m3)/3)^2 + (1.3*m1 - (A + m3)/3)^2 + (m3 - (A + m3)/3)^2 - 3*x) }

m3 = optimize(a, interval = c(0, 100), tol = 1e-20)[[1]]

v = var(c(m1, m2, m3))*(2/3) # "v" is: 2.343749

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Answer Source

The difference is that for your first part, the function `a`

has two roots, and the optimize function finds one of them (`m3=10.31207`

). At this value of `m3`

, the fact that `a(m3)==0`

implies that the normalized sum of squares (SS) of `m1`

, `m2`

, and `m3`

is equal to `3*x`

:

```
> a(m3)
[1] 3.348097e-07
> ss <- function(x) { sum((x-mean(x))^2) }
> ss(c(m1, m2, m3))
[1] 15
> 3*x
[1] 15
>
```

By the definition of the sample variance, the variable `v`

is equal to one-third the SS, so you get `v==x`

.

In contrast, in the second part, your function `a`

has no roots. It attains a minimum at `m3=14.375`

, but at this value of `m3`

, the value of `a(m3)==3.87366`

is not zero, so the normalized sum of squares is not equal to `3*x`

, and so there's no reason to expect that `v`

(one-third the SS) should equal `x`

.

```
> a(m3)
[1] 3.87366
> ss(c(m1, m2, m3))
[1] 7.031247 -- actual SS value...
> 3*x
[1] 3.157587 -- ...couldn't be optimized to equal 3*x
>
```

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