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How can it be proved using the delta method that the Linear Taylor series expansion of a normal random vector containing independent but NOT identically distributed elements results in a random variable whose distribution is approximately normal?


marked as duplicate by whuber Jan 21 '15 at 19:14

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    $\begingroup$ Could you please be more specific about what you mean by "the Linear Taylor series expansion of a normal random vector"? Ordinarily one would expand a given function of a random variable in a Taylor series, but you haven't put any such function in evidence. $\endgroup$ – whuber Jan 21 '15 at 18:08
  • $\begingroup$ Indeed, I mean the Taylor series expansion of a function of random variables where we truncate the series to include only the linear terms, meaning the first two terms. $\endgroup$ – James_170 Jan 21 '15 at 18:11
  • $\begingroup$ So would it be fair to interpret your question as asking why a linear combination of a multivariate normal is normal? $\endgroup$ – whuber Jan 21 '15 at 18:12
  • $\begingroup$ Yes. Indeed, but in this case the function is not just Y = AX+B but in fact the first order Taylor series is given by Y=g(X) = g(mu_x)+[X-mu_x] dg(X)/dX (assuming that X is a univariate random variable). For this linear approximation is Y normal if X is a normally distributed random vector with independent but not identically distributed elements? $\endgroup$ – James_170 Jan 21 '15 at 18:18
  • $\begingroup$ By definition, the first order Taylor series is a linear function. You merely have $B=g(\mu_X) - \mu_x dg/dx$ and $A=dg/dx$, each of which is a number. $\endgroup$ – whuber Jan 21 '15 at 18:22