Is there a simple, intuitive derivation (using Taylor series) of the following approximation to Vega-weighted Implied Volatility?

Question

The approximation is:

$$\sigma \approx \frac{\sum V_j\sigma_j}{\sum V_j}$$

Background information from the first answer to this post:

"Say that you have a portfolio of options with prices $P_j$. Each one of them has a different pricing function $f_j$ (as function of vol) and a different implied vol $\sigma_j$. For each option $f_j(\sigma_j)=P_j$.

Now you put them together in a single product. If the implied vol of the product is $\sigma$ then $\sum f_j(\sigma)=\sum P_j$. Now, approximately each pricing function will satisfy $f_j(\sigma)\approx P_j+V_j (\sigma-\sigma_j)$ as a linear expansion around its price, with $V_j$ the Vega."

Answer 1

Intuitively, if an option has 0 Vega (k=0), it has no influence on the single volatility that will correctly price the portfolio. If you have one position with a \$100 vega per vol point, and another with only $50 vega per vol point, then using a volatility that is 1 point above the implied volatility of the first position, but 2 points below the implied volatility of the second results in offsetting mis-pricing for the individual options and an approximately correct priced portfolio.

More generally, portfolio pricing error is the sum of errors for each security (vega times the difference between vol used and true vol), or:

$\sum[V_i*(\sigma-\sigma_i)] = \sum[V_i\sigma]-\sum[V_i\sigma_i] = \sigma\sum V_i-\sum[V_i\sigma_i] $

Setting our error approximation equal to zero:

$0=\sigma\sum V_i -\sum[V_i\sigma_i] $

$\sigma= \frac{\sum[V_i\sigma_i]}{\sum V_i} $