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Consider $Y_{ij} = \mu + u_i + e_{ij}$ where $\mu$ is a constant and $u_i ∼ N(0,\sigma_u^2 )$ is the subject $i$ random effect which is independent of the error term $e_{ij} ∼ N(0,σ_e^2)$. I can't figure out why $${\rm cov}(Y_{ij}, Y_{ik}) = \sigma_u^2$$

Can anyone please explain?

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Well,

$$ {\rm cov}(Y_{ij}, Y_{ik}) = {\rm cov}(u_{i} + e_{ij}, u_{i} + e_{ik}) $$

since the constant term, $\mu$, does not affect variances or covariances. By the bilinearity of covariance we have

$$ {\rm cov}(Y_{ij}, Y_{ik}) = {\rm cov}(u_{i}, u_{i}) + {\rm cov}(u_{i}, e_{ik}) + {\rm cov}(e_{ij}, u_{i}) + {\rm cov}(e_{ij}, e_{ik}) $$

By independence of $u_{i}$ and each $e_{ij}, e_{ik}$ and independence between $e_{ij}$ and $e_{ik}$ (which must be assumed for this model to make sense), the 2nd, 3rd and 4th terms are all zero above (since independence between two variables implies that their covariance is 0). So,

$$ {\rm cov}(Y_{ij}, Y_{ik}) = {\rm cov}(u_{i}, u_{i}) = {\rm var}(u_{i}) = \sigma^{2}_{u} $$

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  • $\begingroup$ To cover the case when $j=k$, you may add a Kronecker delta to Macros's answer: $\mathrm{Cov}[Y_{ij},Y_{ik}]=\sigma^2_u+\sigma^2_e\,\delta_{jk}$. $\endgroup$ – Zen May 21 '12 at 2:45
  • $\begingroup$ or you could just note that $${\rm cov}(Y_{ij},Y_{ij}) = {\rm var}(Y_{ij}) = \sigma^{2}_{u} + \sigma^{2}_{e}$$ $\endgroup$ – Macro May 21 '12 at 2:46
  • $\begingroup$ You are right. I'm just pointing a way to make the formula work for every $i,j,k$. $\endgroup$ – Zen May 21 '12 at 2:48

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