<p>While in GMM one uses theoretical analytical moments, in MSM one uses simulated theoretical moments instead. </p>

<p>For <strong>G</strong>MM,</p>

<blockquote>
  <p>[t]he method requires that a certain number of moment conditions were specified for the model. These moment conditions are functions of the model parameters and the data, such that their expectation is zero at the true values of the parameters. The GMM method then minimizes a certain norm of the sample averages of the moment conditions. (<a href="https://en.wikipedia.org/wiki/Generalized_method_of_moments" rel="nofollow noreferrer">Wikipedia</a>)</p>
</blockquote>

<p>Crucially,</p>

<blockquote>
  <p>In order to apply GMM, we need to have "moment conditions", i.e. we need to know a vector-valued function $g(Y,\theta)$ such that $m(\theta_{0})\equiv \mathbb{E}[\,g(Y_{t},\theta _{0})\,]=0$.</p>
</blockquote>

<p>For <strong>S</strong>MM, values of functions of the data and the parameters are too difficult to calculate analytically. Essentially, we cannot calculate what the moments of the data are given the parameters; for example, we cannot calculate $\mathbb{E}(XY)$. Instead we simulate a large sample of $X$s and $Y$s and calculate $\frac{1}{n}\sum_{i=1}^{n}X_i Y_i$ (simulated sample, not the actual sample) to replace the theoretical $\mathbb{E}(XY)$. The rest is the same.</p>

CC BY-SA 3.0
2018-02-26T19:35:44.887Z
Richard, thanks for the answer. I'm not sure I understand what you're trying to say. So, you're saying that in MSM we must use some estimator $\hat m ( . )$ for function $m$? If so, then why is this different  from the sample analogues also used in GMM?
An old man in the sea.
CC BY-SA 3.0
2018-02-26T19:48:15.913Z
@Anoldmaninthesea., because we do not use the sample, we use a simulated sample instead. At this stage sample analogues do not enter the picture yet. At this stage you have the theoretical analytic quantity in GMM and theoretical simulated quantity in SMM. But I am no expert in this method, so perhaps my answer is indeed not so clear. But I find it easier with an example. Say the population moment is 5, then the function will be moment-5=0. Whether you calculate the value 5 analytically or simulate it makes the difference between GMM and SMM.
Richard Hardy
CC BY-SA 3.0
2018-02-26T17:51:47.040Z
2018-02-26T19:50:21.587Z
2018-02-26T19:50:21.587Z
Richard Hardy
<p>In wikipedia, the main difference between GMM and <a href="https://en.wikipedia.org/wiki/Method_of_simulated_moments" rel="noreferrer">MSM(Method of simulated Moments)</a> is in the former the moment/orthogonality conditions can be evaluated analytically, while in the latter they cannot and we have to use sample analogues. </p>

<p>However, in <a href="https://en.wikipedia.org/wiki/Generalized_method_of_moments" rel="noreferrer">wikipedia</a> and other texts, with GMM we also use sample analogues.</p>

<p>So, what is the difference between both methods?</p>

<p>Any help would be appreciated.</p>

<econometrics><dsge>
Differences between GMM and MSM?
An old man in the sea.
# Differences between GMM and MSM?

(Asked by: username_0 on 2018-02-26)

In wikipedia, the main difference between GMM and [MSM(Method of simulated Moments)](https://en.wikipedia.org/wiki/Method_of_simulated_moments) is in the former the moment/orthogonality conditions can be evaluated analytically, while in the latter they cannot and we have to use sample analogues.

However, in [wikipedia](https://en.wikipedia.org/wiki/Generalized_method_of_moments) and other texts, with GMM we also use sample analogues.

So, what is the difference between both methods?

Any help would be appreciated.

**Tags:** econometrics, dsge

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## 1 answers

### Answer by username_1 on 2018-02-26. Score: 2

While in GMM one uses theoretical analytical moments, in MSM one uses simulated theoretical moments instead.

For **G**MM,

> [t]he method requires that a certain number of moment conditions were specified for the model. These moment conditions are functions of the model parameters and the data, such that their expectation is zero at the true values of the parameters. The GMM method then minimizes a certain norm of the sample averages of the moment conditions. ([Wikipedia](https://en.wikipedia.org/wiki/Generalized_method_of_moments))

Crucially,

> In order to apply GMM, we need to have "moment conditions", i.e. we need to know a vector-valued function $g(Y,\theta)$ such that $m(\theta\_{0})\equiv \mathbb{E}[\,g(Y\_{t},\theta \_{0})\,]=0$.

For **S**MM, values of functions of the data and the parameters are too difficult to calculate analytically. Essentially, we cannot calculate what the moments of the data are given the parameters; for example, we cannot calculate $\mathbb{E}(XY)$. Instead we simulate a large sample of $X$s and $Y$s and calculate $\frac{1}{n}\sum\_{i=1}^{n}X\_i Y\_i$ (simulated sample, not the actual sample) to replace the theoretical $\mathbb{E}(XY)$. The rest is the same.

> **Comments:**
> **username_0**: Richard, thanks for the answer. I'm not sure I understand what you're trying to say. So, you're saying that in MSM we must use some estimator $\hat m ( . )$ for function $m$? If so, then why is this different  from the sample analogues also used in GMM?  
> **username_1**: @Anoldmaninthesea., because we do not use the sample, we use a simulated sample instead. At this stage sample analogues do not enter the picture yet. At this stage you have the theoretical analytic quantity in GMM and theoretical simulated quantity in SMM. But I am no expert in this method, so perhaps my answer is indeed not so clear. But I find it easier with an example. Say the population moment is 5, then the function will be moment-5=0. Whether you calculate the value 5 analytically or simulate it makes the difference between GMM and SMM.  

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