# Using Taylor series expansion of random vectors to find expectation

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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