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There are $n$ vectors each containing exactly $q$ random variables as elements. Each vector is denoted I$_k$.

Each variable within the vector has its own (normal) probability distribution, and the kinds of variables you find in each of the $n$ vectors are identical. In other words, each of the $n$ vectors has an identical set of expected values for its various $q$ random variables as every other vector, which also enjoy the same variance-covariance matrix amongst their $q$ random variables as in all the other $n-1$ random vectors.

The key fact, however, is that all of the variables $x$ on each row $j$ across the $n$ random vectors I$_k$ (where $k=1,2,...,n$) have a covariance with one another. They tend to covary together.

Now, for each of the $n$ random vectors, there is a specific vector W$_k$ of $q$ weightings that it is multiplied by, such that each element in the vector I$_k$ is multiplied by a particular element in the vector W$_k$.

The resulting values after this multiplication step are summed up, across all the $n$ vectors, to yield the scalar value $R$.

I would really like to know how would you work out the variance $\operatorname{var}(R)$ for this kind of problem. I'm not sure how you would go about constructing the covariance matrix, or indeed where to start...! Could you be my saviour? Thanks in advance if so!

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  • $\begingroup$ Is each $\mathbf I_k$ multiplied by just one separate scalar weight? Your way of writing about this is vague. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jan 13 '16 at 23:12
  • $\begingroup$ Woops, apologies about that! Each $\mathbf{I}_k$ is multiplied by a vector $\mathbf{W}_k$ of the same length, such that each value in $\mathbf{I}_k$ is multiplied by a specific value in $\mathbf{W}_k$, rather than every value in the vector being multiplied by a single scalar weight. Sorry for my vagueness! $\endgroup$ – Sprog Jan 13 '16 at 23:15
  • $\begingroup$ So now it seems that you want a diagonal matrix $\mathbf{W}_k$ instead and the multiplication will be just $\mathbf{W}_k\mathbf{I}_k$. And is the resulting $R$ a scalar or vector? If it is a scalar, are you summing all the entries in $\sum_{k=1}^n \mathbf{W}_k\mathbf{I}_k$? $\endgroup$ – BGM Jan 14 '16 at 4:57
  • $\begingroup$ Exactly, thanks. $R$ is a scalar, which is simply the sum of all the entries in each product of the multiplication step. I realise I need to be clearer in how I write, sorry. $\endgroup$ – Sprog Jan 14 '16 at 10:56

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