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$$ Y_i = \sum_j B_{ij} X_j$$

$$ covar(X_i, X_j) = V_X = \delta_{ij} var(X_i)^2 $$

$$ covar(B_{ij},B_{kl}) \neq 0 $$

$$ V_Y = ? $$

I know, that if $B$ was fixed, it is straight forward, but I would like to generalize it. I managed only by inventing some notation myself, which a physicist should not do I guess. I will check my method later with Monte Carlo, but maybe there is a formal way to write it down?

Maybe the answer should be generalized from variance-of-product-of-2-independent-random-vector?

Edit

We can also assume that $covar(X_i,B_{kl})=0$.

Edit2

Replaced Einstein Summation rule with a sum. I don't just skip the indexes as I want to to stress the dimensions and make the latter more readable.

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  • $\begingroup$ Physicists have created very powerful methods using their own notation, so you shouldn't hold back. I presume there's an implied summation over $j$ in the definition of $Y_i$, for otherwise there are no difficulties. But maybe it's worth pointing out that since $Y_i$ is a bilinear form in $B_{i{*}}$ and $X$, polarization will re-express it as a linear combination of squares. After that the calculations can get hairy... . It's also a little easier when the $X_k$ and $B_{ij}$ are uncorrelated. $\endgroup$
    – whuber ♦
    Commented Jun 10, 2014 at 19:42
  • $\begingroup$ Thanks! Added your comments to Q. But I am not sure if I get the comment about the polarization. 'polarization' is not mentioned in your link. $\endgroup$
    – Johu
    Commented Jun 10, 2014 at 19:54
  • $\begingroup$ Sorry--I actually found that thread with a search on "polarization"! It turns out I linked through a similar answer at stats.stackexchange.com/a/71143. (BTW, I would recommend not using Einstein summation here--it is not a statistical convention. Statistical theory so frequently needs to reference individual terms within the sums that the economy of notation is not worth the additional ambiguity and confusion it can cause, especially when modern mathematics affords even more compact notation for sums, such as $Y=\mathbb{B}X$.) $\endgroup$
    – whuber ♦
    Commented Jun 10, 2014 at 20:01

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