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I'm conducting a simulation of the evolution of a population and calculating values for heritability and confidence intervals for a type's frequency (the number of individuals of a type for each generation) as the program is running. The program runs for an x quantity of generations for each of a y quantity of times (maybe x=200 and y=5000), but the result is different each time because a random process is modeled.

Heritability=covariance of parent values and offspring values divided by variance of parent values

I believe that I am taking a sample because it is a random process, yet covariance and variance are defined in terms of expectations (in which the result is divided over all N pairs of values). So, is the denominator of the covariance N or N-1 and is the denominator of the variance N or N-1?

Thank you very much for your input.

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Asymptotically it makes no difference. Clearly $\frac{n}{n-1}$ converges to $1$ so provided your sample is sufficiently large, you'll get basically the same answer.

With that said, $n-1$ is normally used for the sample standard deviation in order to make the estimator unbiased. The covariance of two random variables can be expressed as the sum of two variances

$$Cov(X,Y) = Var \left(\frac{X+Y}{2} \right)-Var \left(\frac{X-Y}{2} \right)$$

Hence, in order to make the covariance estimator unbiased, you should also divide by $n-1$.

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  • $\begingroup$ Thank you very much. I appreciate your response a great deal. $\endgroup$
    – sterid
    Sep 30, 2018 at 2:33
  • $\begingroup$ @sterid Please accept this answer if it answers your question. $\endgroup$
    – Galen
    Dec 23, 2020 at 19:57
  • $\begingroup$ For added context, using $n-1$ instead of $n$ to reduce bias is called Bessel's Correction. As @Xiaomi alluded to, this correction is for reducing bias, but does not affect the consistency of the estimator. $\endgroup$
    – Galen
    Dec 23, 2020 at 20:00

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