Identify the parameters of the model $Y=\exp(\beta_0 + \beta_1 X + \beta_2 Z)+u_i$

Question

I have the model $Y_i=\exp(\beta_0 + \beta_1 X_i +\beta_2 Z_i) + u_i$ where we assume $\mathbf{E}[u_i|X_i,Z_i]=0$ and $Var(X_i)>0,Var(Z_i)>0$, and I need to show that $\beta_0,\beta_1,\beta_2$ are identified. I know that it's possible to take a log transformation somehow to make it linear but I can't seem to figure out how to get rid of the error term. I'm completely fumbling around here in the dark but I figured the best thing would be to try to write it as a product rather than as a sum, so I said let $v_i=1+\frac{u_{i}}{\exp\left(\beta_{0}+\beta_{1}X_{i}+\beta_{2}Z_{i}\right)}$ so that we can write the model as $$Y_{i}=v_{i}\exp\left(\beta_{0}+\beta_{1}X_{i}+\beta_{2}Z_{i}\right) $$Now I can write $$\ln Y_{i}=\beta_{0}+\beta_{1}X_{i}+\beta_{2}Z_{i} +\ln v_i$$

Now if I could show that $\mathbf{E}[\ln v_i]$ was 0 I would be done. Now, I can see, roughly, that $\mathbf{E}[v_i]$ is maybe 1, if I do some tricky stuff with the law of iterated conditional expectations (I don't know exactly how I would do it but I think I can figure that out) but even if it is, it's not true in general that $\mathbf{E}[\ln v_i]=\ln \mathbf{E}[v_i]$, so that wouldn't actually help me anyway.

Can anyone see how I should proceed? Has what I've done so far made sense?

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Edited to clarify

Apparently there is more than one way to say that a parameter is "identified". The definition that I am working with is this: a parameter is identified when, if we knew the distribution of observed variables, we could uniquely specify the parameter.

Answer 1

There appears to be some discrepancy here regarding what a proof of identification entails and what you are trying to prove. Let me rewrite your model as

$$ Y_i = \exp(\boldsymbol{X}_i'\boldsymbol{\beta}^0) + U_i $$ where $\boldsymbol{X}_i = [X_{1i},\ldots, X_{Ki}]'$, together with the assumption that $\mathbb{E}(U_i \mid \boldsymbol{X}_i)=0$. It appears you are trying to prove that $$ \mathbb{E}(Y_i \mid \boldsymbol{X}_i) = \exp(\boldsymbol{X}_i'\boldsymbol{\beta}^0) $$ in order to prove identification. This is tautological from your model, and does not amount to proving that the parameters are identified.

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Let us first work through some definitions of identification, isolating the exact form of identification that applies to nonlinear regression models, where only the conditional mean, and not the entire conditional distribution is specified.

Parametric identification of distributions

The most primitive notion of identification, is the notion of identification of distributions, usually just called identification. If the conditional density $Y_i \mid \boldsymbol{X}_i$ is written as $$ Y_i \mid \boldsymbol{X}_i \sim f(Y_i \mid \boldsymbol{X}_i; \boldsymbol{\theta}^0) $$ Then the conditional distribution is said to be identified if $$ \boldsymbol{\theta}^0 \neq \boldsymbol{\theta} \implies f(Y_i \mid \boldsymbol{X}_i; \boldsymbol{\theta}^0) \neq f(Y_i \mid \boldsymbol{X}_i; \boldsymbol{\theta}) $$

Parametric identification of the conditional mean function

In the case of regression models, we are typically interested in the identification of the parameters that enter the conditional mean. Let the conditional mean function be $$ \mathbb{E}(Y_i \mid \boldsymbol{X}_i) = m(\boldsymbol{X}_i;\boldsymbol{\beta}^0) $$ The parameters of the conditional mean function are said to be identified, or indeed, the conditional mean model is said to be identified if $$ \boldsymbol{\beta}^0 \neq \boldsymbol{\beta} \implies m(\boldsymbol{X}_i;\boldsymbol{\beta}^0) \neq m(\boldsymbol{X}_i;\boldsymbol{\beta}) $$ In your case, this is the notion of identification that you are interested in. Note that identification does not imply conditional mean identification.

Conditional mean identification in the exponential regression model

Now that the definitions are in place, we can solve the problem at hand, that is, specify primitive conditions for the conditional mean identification in the exponential regression model. Recall that the exponential regression model is a conditional mean model such that $$ m(\boldsymbol{X}_i;\boldsymbol{\beta}) = \exp(\boldsymbol{X}_i'\boldsymbol{\beta} ) $$ Make the assumption that $\mathbb{E}(\boldsymbol{X}_i\boldsymbol{X}_i')$ is nonsingular (this is essentially what your positive variance assumptions are). Then, we have that if $\boldsymbol{\beta}^0 \neq \boldsymbol{\beta}$, $$ \begin{alignat}{1} &\mathbb{E}((\boldsymbol{X}_i'(\boldsymbol{\beta}^0 - \boldsymbol{\beta}))^2) &= (\boldsymbol{\beta}^0 -\boldsymbol{\beta})'\mathbb{E}(\boldsymbol{X}_i\boldsymbol{X}_i')(\boldsymbol{\beta}^0 -\boldsymbol{\beta}) \\ &&>0 \\ \implies & \boldsymbol{X}_i'(\boldsymbol{\beta}^0 -\boldsymbol{\beta}) &\neq 0 \text{ on a set of positive measure} \\ \implies & \boldsymbol{X}_i'\boldsymbol{\beta}^0 &\neq \boldsymbol{X}_i'\boldsymbol{\beta} \text{ on a set of positive measure }\\ \implies & \exp(\boldsymbol{X}_i'\boldsymbol{\beta}^0) &\neq \exp(\boldsymbol{X}_i'\boldsymbol{\beta}) \text{ on a set of positive measure} \\ \end{alignat} $$

Thus, you have shown that under the nonsingularity of the $\mathbb{E}(\boldsymbol{X}_i\boldsymbol{X}_i')$ matrix, the conditional mean function (that is, its parameters) are identified.

Added explanation: The theory of integration (Lebesgue, in this case) states that if $\mathbb{E}(f)>0$ and $ f\geq 0$, then, $f>0$ on a set of positive measure. So, if I can prove that $\mathbb{E}((\boldsymbol{X}_i' (\boldsymbol{\beta}^0 -\boldsymbol{\beta}))^2)>0$, I can claim that $(\boldsymbol{X}_i' (\boldsymbol{\beta}^0 -\boldsymbol{\beta}))^2>0$ on a set of positive measure, in turn that $\boldsymbol{X}_i' (\boldsymbol{\beta}^0 -\boldsymbol{\beta})\neq0$ on a set of positive measure. Note that all this work has to be done to rule out the one case that could spoil the party, that is when the expectation and hence its argument is zero almost everywhere. That would mean that the conditional mean functions corresponding to the two parameter sets are the same almost everywhere.

Now, I know that for any nonsingular matrix $\mathbf{a}$ and nonzero vectors $\boldsymbol{x}$, $\boldsymbol{x}'\mathbf{a}\boldsymbol{x}>0$. So, I write the expectation of interest $\mathbb{E}((\boldsymbol{X}_i' (\boldsymbol{\beta}^0 -\boldsymbol{\beta}))^2)$ as $(\boldsymbol{\beta}^0 -\boldsymbol{\beta})'\mathbb{E}(\boldsymbol{X}_i\boldsymbol{X}_i')({\boldsymbol{\beta}}^0-\boldsymbol{\beta})$, from where, using the fact that $\boldsymbol{\beta}\neq\boldsymbol{\beta}^0$, and the nonsingularity of $\mathbb{E}(\boldsymbol{X}_i\boldsymbol{X}_i')$, I have the desired positivity of the expectation.

This is where you should end the answer to your homework question. From here, I discuss the links of conditional mean identification with the notion of asymptotic identification by estimators.

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(Asymptotic) parameter identification

Recall that (from Newey & McFadden (1994), pg. 2124), that the parameters of a model are (asymptotically) identified by an estimator (in this case, the NLS estimtator) if the limit of the objective function has a unique minimum (maximum) at the truth. We show that if the conditional mean of the distribution is identified in the sense above, then the parameters of the model are asymptotically identified by the NLS estimator.

Consider the nonlinear least squares objective function for the model at hand, where $\boldsymbol{Y}=[Y_1,\ldots, Y_n]$, and $\mathbf{X}=[\boldsymbol{X}_1, \ldots, \boldsymbol{X}_n]'$.

$$ \begin{align} q_n(\boldsymbol{Y},\mathbf{X}; \boldsymbol{\beta}) &= \sum_{i=1}^n\left(Y_i - \exp(\boldsymbol{X}_i'\boldsymbol{\beta})\right)^2 \\ &= \sum_{i=1}^n\left(\exp(\boldsymbol{X}_i'\boldsymbol{\beta}^0) + U_i - \exp(\boldsymbol{X}_i'\boldsymbol{\beta})\right)^2 \\ &=\sum_{i=1}^n \left(\exp(\boldsymbol{X}_i'\boldsymbol{\beta}^0) - \exp(\boldsymbol{X}_i'\boldsymbol{\beta})\right)^2+\sum_{i=1}^n U_i^2 \\ &\quad+ \sum_{i=1}^n U_i\left(\exp(\boldsymbol{X}_i'\boldsymbol{\beta}^0) - \exp(\boldsymbol{X}_i'\boldsymbol{\beta})\right) \end{align} $$

Given that each of the expectations is well defined, these converge to their expectations in probability, using appropriate laws of large numbers. A simple expedient that allows considerable simplifications is to write $\mathbb{E}_{U_i, \boldsymbol{X}_i}\equiv \mathbb{E}_{\boldsymbol{X}_i}\mathbb{E}_{U_i\mid \boldsymbol{X}_i}$, where the subscripts denote marginal densities with respect to which the expectations are taken, and the equivalence is by the law of iterated expectations. Then, $$ \begin{align} q_n(\boldsymbol{Y},\mathbf{X}; \boldsymbol{\beta}) &\to^p q_\infty(\boldsymbol{Y},\mathbf{X}; \boldsymbol{\beta})\\ &=\mathbb{E}_{\boldsymbol{X}_i}\left(\exp(\boldsymbol{X}_i'\boldsymbol{\beta}^0) - \exp(\boldsymbol{X}_i'\boldsymbol{\beta})\right)^2 + \mathbb{E}_{U_i\mid \boldsymbol{X}_i}(U_i^2) \\ &\quad+ \mathbb{E}_{\boldsymbol{X}_i}\left(\exp(\boldsymbol{X}_i'\boldsymbol{\beta}^0) - \exp(\boldsymbol{X}_i'\boldsymbol{\beta}\right)\underbrace{\mathbb{E}_{U_i\mid \boldsymbol{X}_i}(U_i) }_{=0} \end{align} $$

From here, it is easy to see that the last term is zero, and the second term is independent of the chosen parameter value. The first term is uniquely minimized to zero, when $\boldsymbol{\beta} = \boldsymbol{\beta}^0$. As is obvious, the last step assumes conditional mean identification.