Let $B$ be a separable Banach space, and let $X$ be a Bochner integrable function taking values in $B$ such that $\int \Vert X\Vert^2\,\mathrm dP < \infty$. Consider any $x\in B$. Can we establish the following equality: $$E_P\Vert X - x\Vert^2 = E_P\Vert X-E_PX\Vert^2+\Vert x - E_PX\Vert^2,$$ where $E_PX := \int X\,\mathrm dP$ denotes the Bochner integral of $X$. This is known as variance-bias decomposition of the mean squared error.
What I tried so far:
"$\leq$'': Triangle inequalty gives $\Vert X - x\Vert^2 = \Vert X - E_PX + E_PX - x\Vert^2 \leq \Vert X-E_PX\Vert^2 + \Vert E_PX - x\Vert^2.$ Taking expectation shows lhs $\leq$ rhs.
"$\geq$": This is the difficult part. In the finite-dimensional case, one uses a property of the 2-norm which we don't have in general Banach spaces.
So I suppose that we don't have this decomposition. However, I would like to back this up by a reference (or counter example). Even better: does there exist a notion of variance-bias decomposition in Banach spaces?