Deviation with respect of the Mode, part 2 - Weighted linear combination
I a previous question I made, I wonder about How the Variance will behave if is defined in respect to the Mode "$\nu$" of a distribution (assuming here the distribution is unimodal and positively skewed, like the Log-Normal distribution): $$\text{Var}\nu[X] := E[(X-\nu)^2] = \text{Var}[X]+(E[X]-\nu)^2$$
In this question I want to know if the following its true: $$\text{Var}\nu\left[\sum_{i=1}^N a_i X_i\right] = \sum_{i=1}^N a_i^2\left[\text{Var}[X_i]+(E[X_i]-\nu_i)^2\right] + 2\sum_{1\leq i<j\leq N}a_ia_j\left[\text{Cov}[X_i,X_j]+\left(E[X_i]-\nu_i\right)\left(E[X_j]-\nu_j\right)\right]$$
Later I figure out the formula is wrong in general, so at the end I extend the question on how to find the proper expansion of $\text{Var}\nu\left[\sum_{i=1}^N a_i X_i\right]$.
Added later
As a test, I expanded the last formula using as if it were true that $\nu\left[\sum a_iX_i\right]\equiv\sum a_i\nu[X_i]$, and unfortunately for me, I was able to recover the same formula I am asking for at the beginning, so in general it should be wrong.
Could someone explain me how to properly expand the mode of the weighted sum properly $\nu\left[\sum a_iX_i\right]$? or better give me the proper expansion for $\text{Var}\nu\left[\sum_{i=1}^N a_i X_i\right]$?
PS: later I asked specifically how to spread or the mode or the median in this question.
Motivation_________
Following what it is said in the Wikipedia for Variance: Linear combinations and the Bienaymé's identity: $$\text{Var}\left[\sum_{i=1}^N a_i X_i\right] = \sum_{i=1}^N \sum_{j=1}^N a_i a_j\text{Cov}\left[X_i, X_j\right] = \sum_{i=1}^N a_i^2\text{Var}[X_i] + 2\sum_{1\leq i<j\leq N}a_ia_j\text{Cov}[X_i,X_j]$$
it looks like the operation happened only on paired terms (differently from, as example, the probability of not-necessarily mutually exclusive events formula), so I started by checking how the combination of two terms split the covariance: $$\begin{array}{r c l} \text{Cov}[a_1X_1,a_2X_2] & = & E[(a_1X_1-E[a_1X_1])(a_2X_2-E[a_2X_2])] \\ & = & E[(a_1X_1-a_1E[X_1])(a_2X_2-a_2E[X_2])] \\ & = & E[a_1a_2X_1X_2 -a_1a_2X_2E[X_1]-a_1a_2X_1E[X_2]+a_1a_2E[X_1]E[X_2]] \\ & = & a_1a_2\left(E[X_1X_2] -E[X_2]E[X_1]-\require{cancel}\cancel{E[X_1]E[X_2]}+\cancel{E[X_1]E[X_2]}\right) \\ & = & a_1a_2\left(E[X_1X_2] -E[X_1]E[X_2]\right) \\ & = & a_1a_2\text{Cov}[X_1,X_2] \end{array}$$ the formula for the Covariance which I am going to use later. Now, assuming it is true that the Mode follows the same affine transformations that the random variable: $\nu[aX]=a\ \nu[X]$ (not $100\%$ sure if is true), I could expand the following: $$\begin{array}{r c l} \text{CoVar}\nu[a_1X_1,a_2X_2] & := & E[(a_1X_1-\nu[a_1X_1])(a_2X_2-\nu[a_2X_2])] \\ & = & E[(a_1X_1-a_1\nu_1)(a_2X_2-a_2\nu_2)] \\ & = & E[a_1a_2X_1X_2 -a_1a_2X_2\nu_1-a_1a_2X_1\nu_2+a_1a_2\nu_1\nu_2] \\ & = & a_1a_2\left(E[X_1X_2] -\nu_1E[X_2]-\nu_2 E[X_1]+\nu_1 \nu_2 \pm E[X_1]E[X_2]\right) \\ & = & a_1a_2\left(E[X_1X_2] -E[X_1]E[X_2]+(E[X_1]-\nu_1)(E[X_2]-\nu_2)\right) \\ & = & a_1a_2\left(\text{Cov}[X_1,X_2]+(E[X_1]-\nu_1)(E[X_2]-\nu_2)\right) \\ \end{array}$$
Then finally I just pasted this formula for the pairs in the formula of the linear combinations of the variances, and I think it kind of make sense, but I am not sure if its right: maybe unintentionally I have assumed tacitly that $\nu\left[\sum a_iX_i\right]$ equals $\sum a_i\nu[X_i]$ which I believe is not true in general, but I am not sure about this neither. This because from the first formula I get: $$\text{Var}\nu\left[\sum_{i=1}^N a_i X_i\right] = \text{Var}\left[\sum_{i=1}^N a_i X_i\right]+\left(E\left[\sum_{i=1}^N a_i X_i\right]-\nu\left[\sum_{i=1}^N a_i X_i\right]\right)^2$$ and I don't know how to work with the term $\nu\left[\sum a_iX_i\right]$.
