I am considering the following non-linear model

$$h(z) - \lambda_0 - \lambda_1 z - \lambda_2x = v$$

where $v \sim \mathcal N(0,\sigma^2)$ unobserved error and where $\lambda_j$ are unknown parameters, $z$ and $x$ are observed scalars. And I define $\lambda := (\lambda_0,\lambda_1,\lambda_2)^\top$.

Where the likelihood is said to be

$$ l(z,x,\lambda,\sigma) = \frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(- \frac{1}{2\sigma^2} v^2 \right) | \frac{\partial v(\lambda)}{\partial z}|$$

instead of the simpler

$$ l(z,x,\lambda,\sigma) = \frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(- \frac{1}{2\sigma^2} v^2 \right)$$

the text I am reading then makes the claim that confuses me:

(1) The likelihood is arrived at using a textbook application of a change of variables.

Which is fair but makes me wonder:

What assumption is it that is usually made - think for example standard application of linear regression or non-linear regression - that makes it the case that the latter term $| \frac{\partial v(\lambda)}{\partial z}|$ can somehow simply be ignored? And how is that related to the correlation between $v$ and $z$?

P.S. The author claims that the term $| \frac{\partial v(\lambda)}{\partial z}|$ explicitly controls for correlation between $v$ and $z$.

EDIT - Simulation Example

Motivated by Carlo's comment's I have constructed the following example: First assume that $h(z) = a_2z^2$ and for simplicity let $a_2=1$ $$a_2 z^2 + a_1 z + a_0 = v $$ Then draw $v \sim \mathcal N(0,\sigma^2)$ and solve for $z$ given the parameters $a_1$ and $a_0$, where $a_0$ is chosen in a sensible manner (ok the second order polynomial does not given a single solution but I simply impose one always being on one side of the parabola):

$$z = \frac{-a_1 + \sqrt{4*a_2*(a_0-v)}}{2*a_2} := Q(v)$$

Observing $z$ I then compute $y := z^2$. I then rewrite the structural equation $$z^2 + a_1 z + a_0 = v $$ to a structural regression equation $$y = - a_1 z - a_0 + v $$

Let $\hat b_1$ be the OLS estimate of the coefficient $-a_1$. Clearly

$$plim(\hat b_1) = var(z,z)^{-1}cov(z,y) = -a_1 + var(z,z)^{-1}cov(z,v)$$

in this eaxample the bias turn out to be pretty large (see the code).

I guess the estimation idea is then simply to say

$$F_V(v) := Pr(V \leq v) = Pr(Z^2 +a_1Z + a_0 \leq v)$$ then using $z=Q(v)$ as defined above

$$F_Z(z) := Pr(Z\leq z) = Pr(V \leq z^2 +a_1z + a_0) = F_V(z^2 +a_1z + a_0)$$

such that

$$f_Z(z) = f_V(z^2 +a_1z + a_0) \lvert \frac{\partial (z^2 +a_1z + a_0)}{\partial z} \lvert = f_V(z^2 +a_1z + a_0) \lvert 2z+a_1 \lvert$$

  • $\begingroup$ what is $h(z)$? $\endgroup$ – carlo Nov 3 '19 at 15:21
  • 1
    $\begingroup$ Lets assume it is a known one to one function. $\endgroup$ – Jesper for President Nov 3 '19 at 16:27
  • $\begingroup$ so it is an observed scalat value too, you may call it y. in that case yours is a pretty standard linear model formulation, that derivative doesn't have to stay there. althought it seems that you are trying to model x as a function of z as well. strange model you have $\endgroup$ – carlo Nov 3 '19 at 21:57
  • $\begingroup$ The »error« v is not assumed exogenous cov(z,v)!=0, but yes otherwise linear regression could work. I think that is the whole point of the extra term, as claimed by the author. $\endgroup$ – Jesper for President Nov 4 '19 at 6:41

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