Let $\{\textbf{y}_t\}_{t=1}^T$ is a vector sequence. Summation of squared deviation wrt $\ell_2$-norm from vector mean of $\frac{1}{T}\sum \limits_{t=1}^{T}\textbf{y}_t$ is $$ \sum \limits_{t=1}^{T} \|\textbf{y}_t-\frac{1}{T}\sum \limits_{j=1}^{T}\textbf{y}_j \|_2^2 $$
What is the equivalent sum in terms of $\sum \limits_{t=1}^{T} \sum \limits_{j=1}^{T} \|\textbf{y}_t-\textbf{y}_j \|_2^2$. Is the following true? $$ \sum \limits_{t=1}^{T} \|\textbf{y}_t-\frac{1}{T}\sum \limits_{j=1}^{T}\textbf{y}_j \|_2^2=\frac{1}{2T}\sum \limits_{t=1}^{T} \sum \limits_{j=1}^{T} \|\textbf{y}_t-\textbf{y}_j \|_2^2 $$ The above formula is from the following paper (page 3): [http://proceedings.mlr.press/v23/chiang12/chiang12.pdf]
In the following picture from the paper, $\{\textbf{y}_t\}_{t=1}^T$ can be thought of as $\{\nabla f_t(x_t)\}_{t=1}^T$.
