I recognize that method of moments is not the best way to estimate $\nu$ for the t-distribution, but I am just wondering how this would be calculated since $E[X^n] = 0$ if $n$ is odd. Here we're assuming iid $X_i$, from a standard t-distribution.
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$\begingroup$ Are we talking about a standard t ($\mu=0,\sigma=1$)? Or are we estimating three parameters? $\endgroup$– Glen_bCommented Dec 14, 2015 at 0:25
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$\begingroup$ Generally you use the lowest collection of moments that let you solve for the parameter set. Since $E(X)$ doesn't contain $\nu$, you can't use it. So ... what do you think you should try next? $\endgroup$– Glen_bCommented Dec 14, 2015 at 1:31
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1$\begingroup$ I am studying for an exam. I was looking at method of moments for different distributions but couldn't find anything on the t-distribution online or in the text. Most people just say it isn't the best method for estimating and move on. $\endgroup$– conv3dCommented Dec 14, 2015 at 4:34
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Note that for a standard $t$ with $\nu>2$, $E[X]=0$, which is no help in estimating $\nu$.
For a standard $t_\nu$, you should be able to show that
$E[X^2] = Var[X] = \frac{\nu}{\nu-2}$.
So for method of moments we could equate sample and population second moments. If $m_2 = \frac{1}{n}\sum_i x^2_i$ is the sample second (raw) moment, we'd solve $m_2 = \frac{\hat{\nu}}{\hat{\nu}-2}$ for $\hat{\nu}$ (or we could arguably write something in terms of variance).
The explicit solution I'll leave for you.
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Edit: note that if you didn't know the location and scale, you'd use the sample mean to estimate $\mu$, and you'd have $\text{Var}(X)=\sigma^2\frac{\nu}{\nu-2}$ and so would need another moment -- in this case the fourth, which is also a function of $\sigma^2$ and $\nu$ -- to identify the two.
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$\begingroup$ I see. So $\frac{1}{n}\Sigma_i x_i^2$ = $\frac{\hat{\nu}}{\hat{\nu} - 2}$, or $\hat{\nu}$ = $2 - \frac{2}{n}\Sigma_i x_i^2$ $\endgroup$– conv3dCommented Dec 14, 2015 at 4:43
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$\begingroup$ @jchaykow I don't think you got that right. $\endgroup$– Glen_bCommented Dec 14, 2015 at 4:59
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$\begingroup$ Oops. $\hat{\nu}$ = $\frac{\frac{2}{n}\Sigma x_i^2}{\frac{1}{n}\Sigma x_i^2 - 1}$ $\endgroup$– conv3dCommented Dec 14, 2015 at 5:07
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$\begingroup$ @jchaykow So now all you have to worry about is whether I got the variance part right. I could have made that bit up. $\endgroup$– Glen_bCommented Dec 14, 2015 at 7:02
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$\begingroup$ I think that is correct because var(x) = E[X^2] + (E[X])^2. And E[X] is 0. $\endgroup$– conv3dCommented Dec 14, 2015 at 17:03