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Where the P represents the own performance, and the  the average own performance in the population. For example, if we consider rabbits and the trait under selection is body weight at 3 months of age, we need to start recording body weight at that age. The average rabbit in our population weighs 2.0kg0 kg. If we have a rabbit of 2.3 kg, then its phenotypic superiority would be 2.3 – 2.0 = 0.3 kg. Let’s assume that the heritability for this trait in this population is 0.2. Then the EBV for body weight for this rabbit is 0.2*0.3 = 0.06 kg. Note that the unity of the EBV is equal to the unity of the phenotype, in this case kg body weight.

We continue the example with the rabbits. It appeared that even though the rabbits were supposed to have been weighed at 3 months of age, this was not always the case. Due to holidays and weekends, some animals were weighed a bit younger and some a bit older. You can imagine that this has influenced the heritability because there was variation due to age that will have increased the error variance, and thus decreased the heritability. And remember, the closer to 1 the heritability is, the better the phenotype represents the underlying genetic superiority. The management was changed and now all rabbits were weighed at exactly 3 months of age and the heritability increased from 0.3 to 0.4. Consequently, we are better able to predict the underlying genetic superiority of the animal and the EBV increased to 0.4*0.3 = 0.12 kg.Thus:

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Thus: the EBV using own performance can be estimated as: EBVmass selection = h2 * (P-Pmean)   

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