I am trying to understand a proof of quite a long theorem that I report completely for the sake of completeness. This is From Jensen and Rahbek Asymptotic Inference for Nonstationary GARCH (2004). My question is actually basic and only regards the Taylor expansion done in the proof so one can skip to the end if he wishes.

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At the start of the proof through the mean value theorem $\lambda^*$ is proven to exist. Then a Taylor expansion is made.

$$\Big| \frac{\partial f(\lambda^*)}{\partial \lambda} \Big| = \Big| \sum_{i,j,l = 1}^k v_{1,i}v_{2,j} (\varphi_l - \varphi_{0,l}) \partial^3 \ell_T (\varphi_0 - \lambda^* (\varphi - \varphi_0))/ \partial \varphi_i \varphi_j \varphi_l \Big|$$

I think he cuts it off after the second term, to be clear I write the Taylor expansion that I think he does in the univariate form:

$$ f(\lambda^*)+\frac {f'(\lambda^*)}{1!} (\varphi-\lambda^*)+ \cdots. $$

My questions that I think are all connected are:

  1. Where does the first term go in the expansion? (where is $ f(\lambda^*)$ ?)
  2. Why introduce $\lambda^*$ at all? (notice that condition A.3 of Lemma 1 does not involve $\lambda^*$)?
  3. Why does equality hold in this Taylor expansion?
  • 1
    $\begingroup$ The series is not cut off in this exposition: it uses a mean-value form of the remainder in Taylor's Theorem. $\endgroup$ – whuber May 13 '15 at 21:27
  • $\begingroup$ @whuber Thanks! what would be the error function? $\endgroup$ – Monolite May 14 '15 at 9:38
  • $\begingroup$ @whuber Also is the Cauchy form of the remainder being used? I still can't seem to calculate it correctly. $\endgroup$ – Monolite May 14 '15 at 12:23

It is the Lagrange form of Taylor's theorem with remainder (https://www.math.hmc.edu/calculus/tutorials/taylors_thm/) I believe.

And it is an expansion of f(1) around f(0), ie $f(1)=f(0)+f'(\lambda^*).(1-0)$, for some $\lambda^*\in (0,1)$.

So hopefully it is clear that this explains your 3 questions.

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  • $\begingroup$ But where does $f(0) = v_2' [D^2 \ell_T( \varphi_0) ] v_2$ go? And if we are deriving with respect to $\lambda$ why is there the third derivative with respect to $\varphi$? Thanks by the way. $\endgroup$ – Monolite May 17 '15 at 10:21
  • $\begingroup$ I think he is just taking the derivative with respect to $\lambda$ but why is it written with respect to $\varphi$ ? Can we chat so you can explain it to me? $\endgroup$ – Monolite May 17 '15 at 18:21

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