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I have to find the variance of $$ y= {\bf w^*n}$$ where ${\bf n}$ is a vector $(m\times 1)$ i.i.d Gaussian random variables with Covariance (Cov) matrix $$\text{Cov}({\bf n})=\sigma^2 {\bf I}$$ ${\bf I}$ being the identity matrix and where ${\bf w}$ is $(m\times 1)$ $${\bf w}=\sum_{i=1}^na(\phi_i)$$

Is it correct to say that $$\text{Var}({\bf w^* n})=\sigma^2 {\bf w^* I w}$$

Thanks

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The formula you have given is correct, though your notation for $w$ seems a bit over complicated: is your notation $$w = \sum_{i=1}^n a(\phi_i)$$ justified by some context? Why not write $w = (w_1,\ldots, w_m)$?

To derive the variance formula we use the following claim. If $X_1,\ldots, X_m$ are real valued random variables, and $w_1,\ldots, w_m \in \mathbb R$ are constants, then

$$\text{Var}\Big( \sum_{j=1}^m w_j X_j \Big) = \sum_{j,\,k=1}^m w_j w_k\, \text{Cov}(X_j,X_k)$$ Assuming this formula to be true then in the case you have given since the $X_j$ are all independent, we have $\text{ Cov}(X_j,X_k) = 0$ if $j\neq k$, and so the formula you have given becomes \begin{align} \text{Var}\Big( \sum_{j=1}^m w_j X_j \Big) & = \sum_{j=1}^m w_j^2 Var(X_j) \\ & = \sum_{j=1}^m \sigma^2 w_j^2 \\ & = \sigma^2 w^* w. \end{align}

To justify the variance formula we have used is an exercise in linearity of expectations: \begin{align} \text{Var}\Big( \sum_j w_j X_j \Big) & = \textbf E \Big[ \Big( \sum_j w_j X_j \Big)^2 \Big] - \textbf E \Big[ \sum_j w_j X_j \Big]^2 \\ & = \textbf E \Big[ \sum_{j,k} w_j w_k X_j X_k \Big] - \Big( \sum_j \textbf E[ w_j X_j] \Big)^2 \\ & = \textbf E \Big[ \sum_{j,k} w_j w_k X_j X_k \Big] - \sum_{j,k} \textbf E[ w_j X_j]\textbf E[w_k X_k] \\ & = \sum_{j,k} w_j w_k \Big( \textbf E[ X_j X_k] - \textbf E[X_j]\textbf E[X_k] \Big)\\ & = \sum_{j,k} w_j w_k\, \text{Cov}(X_j,X_k). \end{align}

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