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Assume that the second-order Taylor expansion about point $x_0$ is given by $$g(x) = g(x_0) + g'(x_0)(x-x_0) + \frac{g''(x_0)(x-x_0)^2}2 + \text{remainder}$$ Let $\bf X$ be a $k\times1$ vector with finite second moments and infinitely differentiable. Define $Y = g({\bf X})$. Find expectation of $Y$ using second-order approximation given above about expectation of $\bf X$, i.e., use $x_0 = E({\bf X})$.

Therefore $E(Y) = E(g({\bf X}))$. I know that if $\bf X$ is a vector, $E({\bf X}) = \vec\mu$ which is also a $k\times1$ vector of expectations of all ${\bf X}$. Further, the second term in the approximation will be $0$ as $E({\bf X} - \vec\mu) = 0$. Therefore the expectation would be $$E(g({\bf X})) = E[g((\vec\mu))] + E\left[\frac{g''(\vec\mu)({\bf X} - \vec\mu)({\bf X}-\vec\mu)}2\right]$$

How does one simplify this further? The first term in above expectation does not have the same dimension as the second term. For example, if $g({\bf X})$ is a linear function of ${\bf X}$, the first term has dimensions $k\times1$ and the second term has dimensions $k\times k$.

Can someone help? Thank you :)

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  • $\begingroup$ Your post required intensive editing to its formatting. Please review this so that you may learn from it and ensure the edits I made did not corrupt the meaning you intended. $\endgroup$ – gen-z ready to perish Oct 9 '18 at 22:40

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