In this question it is shown that $\mathbb{E}_B[\|\bar{\mathbf{y}}_B-\bar{\mathbf{y}}\|_2^2] = \frac{N-|B|}{|B|}\frac{1}{N}e$ where $\mathbf{y}_1, \mathbf{y}_2, \dots, \mathbf{y}_N$ is a sequence of $N$ vectors in $\mathbb{R}^d$ and $\bar{\mathbf{y}}$ is the overall average, i.e., $\bar{\mathbf{y}}=\frac{1}{N}\sum_{i=1}^{N}\mathbf{y}_i$, $\bar{\mathbf{y}}_B$ is any sample average of size $|B|$ such that $\bar{\mathbf{y}}_B=\frac{1}{|B|}\sum_{i \in B}\mathbf{y}_i$ where $B$ is a random set such that $B \subseteq \{1, \dots, N\}$ and $|B|\leq N$, and $e=\frac{1}{N-1}\sum_{i=1}^N\|\mathbf{y}_i-\bar{\mathbf{y}}\|_2^2$. Note that elements in $B$ are drawn without replacement where $\mathbb{P}(i \in B)=\frac{|B|}{N}$ for all $i \in \{1, \dots, N\}$.
Question: Let $I$ be an index set of elements such that $|I|=s<d$ and $I$ is a function of $B$, i.e., $I(B)$. Can we write the folloiwng?: $$ \mathbb{E}_B[\|(\bar{\mathbf{y}}_{B})_{I(B)}-\bar{\mathbf{y}}_{I(B)}\|_2^2] = \frac{N-|B|}{|B|}\frac{1}{N}\frac{1}{N-1} \mathbb{E}_B\Big[ \sum_{i=1}^N\|(\mathbf{y}_{i})_{I(B)}-\bar{\mathbf{y}}_{{I(B)}}\|_2^2 \Big] $$ where $(\bar{\mathbf{y}}_{B})_{I(B)}$ is the coordinates of $\bar{\mathbf{y}}_{B}$ restricted to the random index set $I(B)$.
My try: I think it is possible to write the above but I am not sure how to show it. I believe it can be shown using the law of total expectation but I do not know how. The following steps has a shaky step but I think can be modified:
$$ \begin{aligned} \mathbb{E}_B[\|(\bar{\mathbf{y}}_{B})_{I(B)}-\bar{\mathbf{y}}_{I(B)}\|_2^2] &= \mathbb{E}_B\Big[\sum_{k\in I(B)}\Big((\bar{\mathbf{y}}_{B})_{k}-\bar{\mathbf{y}}_{k}\Big)^2\Big] \\ &= \mathbb{E}_B\Big[\sum_{k=1}^d\Big(\big((\bar{\mathbf{y}}_{B})_{k}-\bar{\mathbf{y}}_{k}\big)^2 \mathbb{P}(k \in I(B))\Big) \Big]\\ &= \sum_{k=1}^d \Big( \mathbb{E}_B\Big[\big((\bar{\mathbf{y}}_{B})_{k}-\bar{\mathbf{y}}_{k}\big)^2 \mathbb{P}(k \in I(B))\Big] \Big)\\ &= \sum_{k=1}^d \Big( \mathbb{E}_B\Big[\big((\bar{\mathbf{y}}_{B})_{k}-\bar{\mathbf{y}}_{k}\big)^2 \Big] \mathbb{P}(k \in I(B)) \Big)\\ &= \frac{N-|B|}{|B|}\frac{1}{N}\frac{1}{N-1} \sum_{k=1}^d \Big( \sum_{i=1}^N \Big( \big((\mathbf{y}_{i})_{k}-\bar{\mathbf{y}}_{k}\big)^2 \Big) \mathbb{P}(k \in I(B)) \Big)\\ &= \frac{N-|B|}{|B|}\frac{1}{N}\frac{1}{N-1} \mathbb{E}_B\Big[ \sum_{i=1}^N\|(\mathbf{y}_{i})_{I(B)}-\bar{\mathbf{y}}_{{I(B)}}\|_2^2 \Big] \end{aligned} $$