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This is a very basic question that I'm pretty sure I understand but I wanted to double check. Given a regression model of: $$y_t = \mathbf{x_{t}^{\prime}}\beta + u_t$$

We can use one of the CLT varieties to get a limiting distribution for $\hat{\beta}$ depending on the assumptions regarding $u_t$. If we assume that $E(\mathbf{x_{t}}u_t|\mathbf{X})=0$, that is that $\mathbf{x_{t}}u_t$ form a martingale difference sequence then there is a CLT that says: $$\frac{1}{\sqrt{T}}\sum_{t=1}^{T}x_{t}u_{t} \to N(0, \Omega)$$The question is how to compute $\Omega$?

$$ \begin{align*} \Omega &= E[(\frac{1}{\sqrt{T}}\sum_{t=1}^{T}x_{t}u_{t})(\frac{1}{\sqrt{T}}\sum_{t=1}^{T}x_{t}u_{t})^\prime] \\ &= \frac{1}{T}E\big[(\mathbf{x_1}u_1 + \mathbf{x_2}u_2 + ... + \mathbf{x_T}u_T)\times (\mathbf{x_1}u_1 + \mathbf{x_2}u_2 + ... + \mathbf{x_T}u_T)^\prime\big] \\ &= \frac {1}{T}E\big[\mathbf{x_1}u_1\times(\mathbf{x_1}u_1 + \mathbf{x_2}u_2 + ... + \mathbf{x_T}u_T) + \mathbf{x_2}u_2\times(\mathbf{x_1}u_1 + \mathbf{x_2}u_2 + ... + \mathbf{x_T}u_T) + ... \big] \end{align*} $$ The martingale difference assumption tells us that the cross terms fall out so that: $$ \begin{align*} \Omega &= \frac{1}{T}E\big[T u_t^2 \mathbf{x_t}\mathbf{x_t^\prime} \big]\\ &= E\big[u_t^2 \mathbf{x_t}\mathbf{x_t^\prime} \big]\\ \end{align*} $$

which we can then estimate from the sample. The thing that confuses me is that the way the regular CLT is stated is that

$$\frac{1}{\sqrt{T}}\sum_{t=1}^{T}u_t \to N(0,\sigma^2)$$

where $E(u_t^2)=\sigma^2$ which makes it seem you just plug in the variance of the term under the sum sign into the normal and you are done whereas the example above suggests that what you are really doing to compute the variance is:

$$ \begin{align*} Variance(\frac{1}{\sqrt{T}}\sum_{t=1}^{T}u_t) &= \frac{1}{T}\sum_{t=1}^{T}Variance(u_t)\\ &= \frac{1}{T}T\sigma^2=\sigma^2 \end{align*} $$

and it happens to be the same simply because of independence of $u_t$. Is that the right way to think about computing the variance term for the different forms of the CLT? This becomes much more important for correlated variables as then the variance of the sum isn't the sum of the variances (a CLT exists for that also).

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  • $\begingroup$ anyone? this seems like a really simple question but for some reason i keep going in circles so any help would be appreciated $\endgroup$ – Alex Feb 4 at 18:07

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