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I am generating some simulated data from a multivariate gaussian distribution with a covariance matrix sigma. To add some noise, I added an identity matrix to the covariance matrix which depicts gaussian white noise.

I then drew some samples from it, calculated the pairwise semivariance as

$\gamma(h)=\frac{1}{2N} \sum{|x_i-x_j|^2}$ where N is the total number of observations that are h distance apart.

Then I fitted a model variogram to it. I found out the nugget to be equal to 1 which makes sense because I added gaussian white noise of variance 1.

However, I then realized that it is the semivariance not variance so I should have got 1/2 i.e 0.5 as nugget isn't it(approx). I am plotting not the variogram but the semi variogram so I am a bit confused.

Suggestions? I am a bit confused. People are using the terms semi variogram and variogram with interchangeability. However, semi variogram is half of variogram. And people are like fitting variogram models(exponential, spherical etc) to the empirical semivariogram.

So I am confused. The graph of empirical semivariogram and variogram is definitely different. Variogram is double of that of semivariogram. So if I fit the model variorum, the sill param should be different as well.

kjetil b halvorsen
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user34790
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    Apparently it is very common to use "variogram" when you mean "semi-variogram" because no one uses full variograms. – J Kelly Mar 27 '14 at 15:06
  • Related: https://stats.stackexchange.com/questions/551359/what-is-the-difference-between-a-non-zero-nugget-and-a-noise-term-in-kriging-gpr – naught101 Nov 08 '21 at 01:03

1 Answers1

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The terminology around the (semi-)variogram can be confusing. However, what you are computing is entirely what is to be expected.

To see this, let $X_1, X_2$ be two independent copies of a random variable $X$. The expected squared difference between $X_1$ and $X_2$ is now given by \begin{align*} E\left[(X_1 - X_2)^2\right] &= E\left[X_1^2 -2X_1X_2 + X_2^2\right] \\ &= E\left[X_1^2\right] - 2E\left[X_1X_2\right] + E\left[X_2^2\right] \\ &= E\left[X_1^2\right] - 2E\left[X_1\right] E\left[X_2\right] + E\left[X_2^2\right] \\ &= 2E\left[X^2\right] - 2E\left[X\right]^2 \\ &= 2 \text{Var}(X) \end{align*} hence we have that $$\frac{1}{2}E\left[(X_1 - X_2)^2\right] = \text{Var}(X).$$

You estimate this quantity with $\gamma(h=0)$, so it makes perfect sense that you obtain the white noise variance $1$ rather than half of it. I hope this helps.

SamuelPawel
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