Assume $\boldsymbol{Y}=\boldsymbol{X} \boldsymbol{\beta}+\boldsymbol{\varepsilon}$, where $\boldsymbol{\varepsilon} \sim \mathcal{N}_n\left(\mathbf{0}, \sigma^2 \boldsymbol{I}_n\right)$ and $\operatorname{rank}(\boldsymbol{X})=p$
To find confidence intervals for $\boldsymbol{a}_{\boldsymbol{i}}^{\prime} \boldsymbol{\beta}$ with confidence level at least $1-\alpha$, we can first stimate $k$ linear transformations $\boldsymbol{a}_i^{\prime} \boldsymbol{\beta}$ by $\boldsymbol{a}_i^{\prime} \hat{\boldsymbol{\beta}}$, for $i=1,2, \ldots, k$. And then for each $i$, $$ \boldsymbol{a}_i^{\prime} \hat{\boldsymbol{\beta}} \sim \mathcal{N}\left(\boldsymbol{a}_i^{\prime} \boldsymbol{\beta}, \sigma^2 \boldsymbol{a}_i^{\prime}\left(\boldsymbol{X}^{\prime} \boldsymbol{X}\right)^{-1} \boldsymbol{a}_i\right) $$
I want to understand the quantity $\boldsymbol{a}_i^{\prime}\left(\boldsymbol{X}^{\prime} \boldsymbol{X}\right)^{-1} \boldsymbol{a}_i$ intuitively.
I can think I was given $k$ many new observations and each has feature vector $a_i, i= 1,...,k$. Recall $X$ is the design matrix consist training (old) data points. Intuitively, the variance of the estimated mean response $\boldsymbol{a}_i^{\prime}\boldsymbol{\hat \beta}$ means the uncertainty.
Consider $\boldsymbol{a}_i$ and $X$ are real valued, is it correct to say - the more "different" my new data point $\boldsymbol{a}_i$ is from the old data point, the higher the uncertainty we have about the new estimated mean value? I am being vague about "different" because I don't know how to define that.
How about $\boldsymbol{a}_i$ and $X$ are random? Suppose the features of observations in my training dataset follows a certain distribution. And now $\boldsymbol{a}_i, i=1,...,k$ are iid but follow a different distribution. Then can I ask the same question as above?