<p>It depends<span class="math-container">$\ldots$</span>
In your setting, the structural parameter <span class="math-container">$\psi$</span> is simply <span class="math-container">$1$</span> minus the value of the reduced form parameter <span class="math-container">$\beta$</span>:
<span class="math-container">$$
\psi = 1 - \beta.
$$</span>
In such case, the mean of the estimator <span class="math-container">$\psi_n$</span> will be simply the mean of <span class="math-container">$1 - \beta_n$</span> and they will have the same variance (as the variance does not change if we add a fixed number or change sign).</p>
<p>If we have a general (non-linear) transformation:
<span class="math-container">$$
\psi = g(\beta),
$$</span>
then the most common way to proceed (and to get asymptotic consistent standard errors) is  by using the delta-method, which is based on a first order Taylor expansion (<a href="https://en.wikipedia.org/wiki/Delta_method" rel="noreferrer">wiki</a>). In particular, if <span class="math-container">$g$</span> is <span class="math-container">$C^1$</span> and if:
<span class="math-container">$$
\sqrt{n}(\beta_n - \beta) \to^D {\cal N}(0, \sigma^2),
$$</span>
we will have that:
<span class="math-container">$$
\sqrt{n}(g(\beta_n) - g(\beta)) = \sqrt{n}(\psi_n - \psi) \to^D {\cal N}(0, \sigma^2 \cdot [g'(\beta)]^2).
$$</span>
So the asymptotic mean of <span class="math-container">$\sqrt{n}(\psi_n - \psi)$</span> will be equal to zero (which means that <span class="math-container">$\psi_n$</span> is consistent and the asymptotic variance is the asymptotic variance of <span class="math-container">$\beta_n$</span> multiplied by the derivative <span class="math-container">$g'(\beta)$</span> squared.</p>
<p>Notice that if <span class="math-container">$g$</span> is a simple translation then <span class="math-container">$g' = \pm 1$</span> so the square is <span class="math-container">$1$</span> and the variance will be equal to <span class="math-container">$\sigma^2$</span>.</p>
<p>To estimate <span class="math-container">$\sigma^2 \cdot [g'(\beta)]^2$</span> you can use the consistent plugin estimator:
<span class="math-container">$$
\hat \sigma^2_n \cdot [g'(\beta_n)]^2. 
$$</span>
where <span class="math-container">$\hat \sigma^2_n$</span> is a consistent estimator of <span class="math-container">$\sigma^2$</span>.</p>

CC BY-SA 4.0
2021-09-19T04:37:49.240Z
Thanks. Will try that in MATLAB. Lemme ask though if you've tried this in other econometric software - STATA, EVIEWS, OXMETRICS, R, PYTHON?
Emmanuel Ameyaw
CC BY-SA 4.0
2021-09-17T06:31:39.187Z
2021-09-17T06:31:39.187Z
tdm
<p>Let say I have some structural model <span class="math-container">$y_t = \alpha + (1+\psi)x_t$</span>. I estimate the model <span class="math-container">$y_t = \alpha + \beta x_t + e_t$</span> using OLS. I can back out the structural parameter <span class="math-container">$\psi$</span> by hand.</p>
<p>How about standard errors and t-statistics for <span class="math-container">$\psi$</span>? Can some econometric software do that? Or do I need to do that by hand? By hand I mean use the model <span class="math-container">$(y_t-x_t) = \alpha_1 + \psi_1 x_t + u_t$</span> to get standard errors for <span class="math-container">$\psi$</span>?</p>
<p>Update: I checked in stata, SE for <span class="math-container">$\psi_1$</span> and <span class="math-container">$\beta$</span> are the same. So I guess that is it. Thus, standard errors for reduced-form and structural parameters are the same?</p>

<structural-estimation>
How to get statistics for structural parameters after estimating a reduced form model
Emmanuel Ameyaw
# How to get statistics for structural parameters after estimating a reduced form model

(Asked by: username_0 on 2021-09-16)

Let say I have some structural model $y\_t = \alpha + (1+\psi)x\_t$. I estimate the model $y\_t = \alpha + \beta x\_t + e\_t$ using OLS. I can back out the structural parameter $\psi$ by hand.

How about standard errors and t-statistics for $\psi$? Can some econometric software do that? Or do I need to do that by hand? By hand I mean use the model $(y\_t-x\_t) = \alpha\_1 + \psi\_1 x\_t + u\_t$ to get standard errors for $\psi$?

Update: I checked in stata, SE for $\psi\_1$ and $\beta$ are the same. So I guess that is it. Thus, standard errors for reduced-form and structural parameters are the same?

**Tags:** structural-estimation


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## 1 answers

### Answer by username_1 on 2021-09-17. Score: 5 | Accepted answer

It depends$\ldots$
In your setting, the structural parameter $\psi$ is simply $1$ minus the value of the reduced form parameter $\beta$:
$$
\psi = 1 - \beta.
$$
In such case, the mean of the estimator $\psi\_n$ will be simply the mean of $1 - \beta\_n$ and they will have the same variance (as the variance does not change if we add a fixed number or change sign).

If we have a general (non-linear) transformation:
$$
\psi = g(\beta),
$$
then the most common way to proceed (and to get asymptotic consistent standard errors) is by using the delta-method, which is based on a first order Taylor expansion ([wiki](https://en.wikipedia.org/wiki/Delta_method)). In particular, if $g$ is $C^1$ and if:
$$
\sqrt{n}(\beta\_n - \beta) \to^D {\cal N}(0, \sigma^2),
$$
we will have that:
$$
\sqrt{n}(g(\beta\_n) - g(\beta)) = \sqrt{n}(\psi\_n - \psi) \to^D {\cal N}(0, \sigma^2 \cdot [g'(\beta)]^2).
$$
So the asymptotic mean of $\sqrt{n}(\psi\_n - \psi)$ will be equal to zero (which means that $\psi\_n$ is consistent and the asymptotic variance is the asymptotic variance of $\beta\_n$ multiplied by the derivative $g'(\beta)$ squared.

Notice that if $g$ is a simple translation then $g' = \pm 1$ so the square is $1$ and the variance will be equal to $\sigma^2$.

To estimate $\sigma^2 \cdot [g'(\beta)]^2$ you can use the consistent plugin estimator:
$$
\hat \sigma^2\_n \cdot [g'(\beta\_n)]^2.
$$
where $\hat \sigma^2\_n$ is a consistent estimator of $\sigma^2$.

> **Comments:**
> **username_0**: Thanks. Will try that in MATLAB. Lemme ask though if you've tried this in other econometric software - STATA, EVIEWS, OXMETRICS, R, PYTHON?  

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