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I have the following $i$-th regressor function:

$\phi_i(x)$ in which $x$ is a vector with elements $x_1, \ldots, x_n$.

I cite from an article:

Let $e_i = \hat{x}_i - x_i$ and note that, since $\phi_i(\cdot)$ is continuously differentiable (an earlier made assumption) , we can write:

$\phi_i(x) = \phi_i(\hat{x}_1,\ldots,\hat{x}_{i-1},x_i,\hat{x}_{i+1},\ldots,\hat{x}_n) + \sum_{j=1}^{n} e_j\delta_{ij}(x,e), \qquad eq. (1)$

for some functions $\delta_{ij}(\cdot)$, with $\delta_{ii}(x,e) = 0$.

So a simple example, assume we have:

$\phi_2(x) = x_1x_2$, then we get:

$x_1x_2 = \hat{x}_1x_2 + e_1\delta_{21}(x,e)$ which can be written as:

$x_1x_2 = (x_1+e_1)x_2 + e_1\delta_{21}(x,e) $

So we simply find: $\delta_{21}(x,e)= -x_2$

Now I am wondering, why do we require $\phi_i(x)$ to be continuously differentiable in order for eq. (1) to be true in general?

Pietair
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1 Answers1

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Be careful: with $n=1$, we would have $\phi_1(x_1) = \phi_1(\hat{x_1}) + e_1 \times 0$. This cannot hold for all $x_1$ and $\hat{x}_1$, except if $\phi_1$ is constant.

It maybe that the article says $$ \tag{1} \phi_i(x_1,\dots,x_n) = \phi_i(\hat{x}_1,\dots,\hat{x}_{i-1},x_i,\hat{x}_{i+1},\dots,\hat{x}_n) + \sum_{j=1}^n e_j \, \delta_{ij}(x,e) $$ instead?

Equation $(1)$ now results from multiple applications of the Mean value theorem, which theorem requires differentiability of the function.

Siméon
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  • Thank you for your reply, you are indeed right. Have you might got a link to a website where they derive this equation? Cause so far I don't see the problem of applying eq. (1) to a regressor which is not continuously differentiable. – Pietair Aug 30 '14 at 12:13