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Can you please help me figuring out what the covariance structure of a model with random intercept and random slope is?

Here is my model $Y_{ij} = \beta_0 + \beta_1 t_{ij} + b_{0,i} + b_{1,i}t_{ij} + e_{ij}$

where $b_{0,i}$ and $b_{1,i}$ are random intercept and random slope and we know that $e_{ij}$ are mutually independent and also:

$b_i$ ~ $N(0,D)$ where $b_i = (b_{0,i},b_{1,i})$ and $e_{ij}$ ~ $N(0,\sigma^2)$

I appreciate your help.

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1 Answer 1

Using the properties of the covariance operator we have

$$$$

$$\begin{align*} \textrm{Cov}(Y_{ij}, Y_{ik}) & = \textrm{Cov}(\beta_0 + \beta_1 t_{ij} + b_{0i} + b_{1i} t_{ij} + e_{ij}, \; \beta_0 + \beta_1 t_{ik} + b_{0i} + b_{1i} t_{ik} + e_{ik}) \\ & = \textrm{Cov}(b_{0i} + b_{1i} t_{ij} + e_{ij}, \; b_{0i} + b_{1i} t_{ik} + e_{ik}) \\ & = \textrm{Cov}(b_{0i}, \; b_{0i}) + t_{ik} \, \textrm{Cov}(b_{0i}, \; b_{1i}) + \textrm{Cov}(b_{0i}, \; e_{ik}) \\ & \quad +\, t_{ij} \, \textrm{Cov}(b_{1i}, \; b_{0i}) + t_{ij} t_{ik} \, \textrm{Cov}(b_{1i}, \; b_{1i}) + t_{ij} \, \textrm{Cov}(b_{1i}, \; e_{ik}) \\ & \quad +\, \textrm{Cov}(e_{ij}, \; b_{0i}) + t_{ik} \, \textrm{Cov}(e_{ij}, \; b_{1i}) + \textrm{Cov}(e_{ij}, \; e_{ik}) \\ & = \textrm{Var}(b_{0i}) +\, (t_{ij} + t_{ik}) \textrm{Cov}(b_{0i}, \; b_{1i}) +\, t_{ij}t_{ik} \textrm{Var}(b_{1i}) +\, \textrm{Cov}(e_{ij}, \; e_{ik}) \end{align*}$$

$$$$

If $j \neq k$, then $$\begin{align*} \textrm{Cov}(Y_{ij}, Y_{ik}) & = \textrm{Var}(b_{0i}) + (t_{ij} + t_{ik}) \, \textrm{Cov}(b_{0i}, \; b_{1i}) + t_{ij} t_{ik} \textrm{Var}(b_{1i}) \end{align*}$$

If $j = k$, then $$\begin{align*} \textrm{Cov}(Y_{ij}, Y_{ik}) & = \textrm{Var}(Y_{ij}) \\ & = \textrm{Var}(b_{0i}) + 2 \, t_{ij} \, \textrm{Cov}(b_{0i}, \; b_{1i}) + t_{ij}^2 \textrm{Var}(b_{1i}) + \sigma^2 \end{align*}$$

$$$$

$\textrm{Var}(b_{0i})$, $\textrm{Var}(b_{1i})$, and $\textrm{Cov}(b_{0i}, \; b_{1i})$ are found in your matrix $D$.

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