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Looking for some help to understand the following problem and how to use the reduced form in econometrics

Consider a model for the health of an individual: $$health = b_0 + (b_1)age + (b_2)weight + (b_3)height + (b_4)male + (b_5)work + (b_6)exercise + u$$

assume that all variables in the equation with the exception of exercise are uncorrelated with u.

A)Write down the reduced form for exercise, and state the conditions under which the parameters of the equation are identified.

B)How can the identification assumption in part c be tested?


Is it correct to assume:

$$exercise = b_0 + (b_1)age + (b_2)weight + (b_3)height + (b_4)male + (b_5)work + u$$ as the reduced form?

and is the condition for identification of parameters simply

$E(exercise|u)=0$

and how can I test it? But moreover what is it good for?

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This is the very standard question on Instrumental Variables of Single-Equation Linear models. Given the primitives of your question, the only endogenous variable is exercise. In order to answer this particular question, you need an exogeous variable, z, that satisfies two conditions:

  1. cov(z,u)=0.
  2. There must be a relationship between the endogenous variable and this exogenous variable you are proposing but that it wasn't part of the true postulated model (the structural model). In other words, $$ exercise=\beta_0+\beta_1 age +\beta_2 weight + \beta_3 height + \beta_4 male + \beta_5 work + \phi z + \varepsilon_{exercise} $$ with $\phi\ne 0$, $\mathbb{E}\,( \varepsilon_{exercise})=0$ and orthogonal to all your explanatory variables (other than exercise) and to z.

Before moving on, a remark. By structural model I mean, following Wooldridge and Goldberger convention, the postulated model. That is, the model that states the causal relationship between health and your covariates. This is a key difference and a disagreement with previous answers.

Now, back at the problem at hand, condition 2 is what in the simultaneous-equations literature call the reduced form equation, which is nothing but a linear projection of the endogenous onto all exogenous variables, including z.

Now, plug the reduced form into your postulated model and you'll get

$$ health=\alpha_0 + \alpha_1 age + \alpha_2 weight + \alpha_3 height + \alpha_4 male + \alpha_5 work + \delta z + \nu $$ where $\alpha_i = b_i + b_6\beta_i,\: \forall i \in \{1,\dots,5\}$, $\delta=b_6\phi$ and $\nu = u+b_6\varepsilon_{exercise}$. By the definition of linear projection, $\nu$ is uncorrelated with all explanatory variables and thus OLS of this last equation will produce consistent estimates for $\alpha_i$ and $\delta$, not the underlying $b_i$ in the true model.

Identification requires a bit of manipulation in matrix form but essentially it reduces to the so-called rank condition. Define $\mathbf{b}=(b_0,\dots,b_6)'$ and $\mathbf{x}=(1, age, \dots, exercise)'$ so that your structural model is $health=\mathbf{x}'\mathbf{b}+u$. Now define $\mathbf{z}\equiv(1,age,\dots,work,z)'$. By condition 1 (cov(z,u)=0 so that E(z,u)=0), $$ \mathbb{E}(\mathbf{z}u)=0 $$ If you multiply bot sides of the structural model by $\mathbf{z}$ and take expectations you have $$ \mathbb{E}(\mathbf{z}\mathbf{x}')\mathbf{b}=\mathbb{E}(\mathbf{z}y) $$ Rank condition states that $\mathbb{E}(\mathbf{z}\mathbf{x}')$ is full column rank. In this particular example and given conditions on z this is equivalent to $rank(\mathbb{E}(\mathbf{z}\mathbf{x}')=6$. Therefore we have 6 equations in 6 unknowns. Hence there exists a unique solution for the system i.e. $\mathbf{b}$ is identified and equals to $[\mathbb{E}(\mathbf{z}\mathbf{x}')]^{-1}\mathbb{E}(\mathbf{z}y)$, as desired.

Remarks: Condition 1 is useful to get the moment condition but the reduced form model with $\phi$ is crucial for the rank condition. Both conditions are usual.

At this point it should be clear why do we need this. In the one hand, without z OLS estimator of the true model will produce inconsisten estimators not only for $b_6$ but for all $b_i$. In the other hand (and somewhat related), our parameters are uniquely identifies so we are certain that we are estimating the true causal relationship as stated in our true model.

In regard to testing, condition 2 (z and exercise are partially correlated) can be tested directly and you should always report that step contrary to the comment in a previous answer. There is a huge literature in relation to this step, specially the weak-instrument literature.

Second condition cannot be directly tested nonetheless. Sometimes you might invoke economic theory to justify or provide alternative hypotheses thats support the use of z.

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The question doesn't make much sense to me as stated. If the problem says that exercise is endogenous (correlated with error term), you can't assume the opposite in the solution. Plus, one usually speaks about reduced vs. structural form in the context of IV estimation. If exercise is endogenous, you need an instrument for it (variable that predicts exercise, but doesn't affect health otherwise) to obtain causal effects. For example, if some people in your sample randomly won gym membership coupons, that could be a valid instrument.

Identification assumptions would then be

  1. coupon really does predict exercise

  2. coupon is orthogonal to $u$

What is called structural form would be two equations, one your original model, the other regression of exercise on coupon and other explanatory variables from the original model (the first stage). Reduced form would be when you substitute first stage into the main equation, so you regress health on age, weight, ... , work and coupon (but not exercise, as that has been substituted out). Reduced form is sometimes used for explaining properties of IV estimation, but AFAIK it's not much used in practice.

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