I think the problem is with your expectations:) Note that when you added a random intercept for each individual, the standard error of the *intercepts* increased. Since each individual can have his/her own intercept, the group average is less certain. The same thing happened with the random slope: you are not estimating one common (within-group) slope anymore, but the average of varying slopes.

**EDIT: Why doesn't a better model give a more precise estimate?**

Let's think about it the other way around: why does the initial model underestimate the standard error? It assumes independence of observations that are not independent. The second model relaxes that assumption (in a way that affects the intercepts), and the third relaxes it further.

**EDIT 2: relationship with many patient-specific models**

Your observation is a known property (and if you had only two years, then the random effects model would be equivalent to a paired t-test). I don't think I can manage a real proof, but perhaps writing out the two models will make the relationship clearer. Let's ignore the grouping variable, as it would just complicate the notation. I will use greek letters for random effects, and latin letters for fixed effects.

The random effects model is ($i$ - subject, $j$ - replicate within subject):
$$Y_{ij} = a + \alpha_i + (b+\beta_i)x_{ij} + \epsilon_{ij},$$ where $(\alpha_i,\beta_i)'\sim N(0,\Sigma)$ and $\epsilon_{ij}\sim N(0,\sigma^2)$.

When you fit separate models for each subject, then
$$Y_{ij} = a_i + b_i x_{ij}+ \epsilon_{ij},$$ where $\epsilon_{ij}\sim N(0,\sigma_i^2)$.

*[Note: the following is really just handwaving:]*

You can see a lot of similarities between these two models with $a_i$ corresponding to $a+\alpha_i$ and $b_i$ to $b+\beta_i$. The average of $b_i$'s corresponds to $b$, because the random effects average to 0. The unconstrained correlation of the random intercept and slope leads to the fact that the models just can be fitted separately. I am not sure how the single $\sigma$ assumption meshes with the subject-specific $\sigma_i$, but I would assume that $\alpha_i$ picks up the difference.

## Best Answer

Fairly straightforward.

One way to express a random intercept model is in matrix notation like

$$ y = X\beta + Zu$$

Here, $X$ is a design matrix, $\beta$ are regression coefficeints, $Z$ is an indicator matrix for group membership, and $u$ are the random effects.

Using R, this made really easy for your problem