How can I work out the standard deviation of a t-distribution? Given a t-distribution with a certain degrees of freedom, how can I work out what the standard deviation of that distribution is?
 A: You can get the population standard deviation by computing the variance via integration and then taking the square root. (It's not the only possible way to compute a variance but it's fairly routine integration for this problem.)
Assuming you have a t$_{(\mu,\sigma^2,\nu)}$ distribution, you can replace $\mu$ by 0 without changing the variance. You can reparameterize to get rid of $\sigma$ (the variance is $\sigma^2$ times that of a standard t). Note that $\nu$ is the degrees of freedom. The location $\mu$ will only be the mean when it exists and $\sigma$ is a scale parameter but $\sigma^2$ is not the variance.
So you're left with finding the variance of a standard t$_\nu$.
$$f(x) = \frac{\Gamma(\frac{\nu+1}{2})} {\sqrt{\nu\pi}\,\Gamma(\frac{\nu}{2})} \left(1+\frac{x^2}{\nu} \right)^{-\frac{\nu+1}{2}},$$
So 
$$\int_{-\infty}^\infty x^2f(x)dx = 2c\int_0^\infty x^2(1+\frac{x^2}{\nu})^{-\frac{\nu+1}{2}}dx$$ 
where $c$ is the constant term out the front. There are a number of ways to approach that.
You could write that as $2c\nu$ times an integral with $x^2/\nu$ in it, then add 1 and subtract 1, pull off the -1 term into a separate integral (easily done), so you end up with a different power (and so a different $\nu$, $\nu^*$ say). You then rescale $x$ to that new  $\nu^*$ and pull out the appropriate constant to leave something times $\int_0^\infty (1+\frac{x^2}{\nu^*})^{-\frac{\nu^*+1}{2}}dx$, supply the required constant to make that a pdf (multiply and divide by by that constant), cancel the integral of a pdf out, and you're left with constant terms being a function of $\nu$.
An alternative would be to recognize something like a beta of the second kind in the integral $\int_0^\infty x^2(1+\frac{x^2}{\nu})^{-\frac{\nu+1}{2}}dx$. Or there are other approaches to doing the integral.
The required result is given on the Wikipedia page for the distribution, though, so if you just need the result for the standard t, you can get it there -- $\text{Var}(X)=\frac{\nu}{\nu-2}$, so $\text{sd}(X)=\sqrt{\frac{\nu}{\nu-2}}$.
A: Here is the generic equation for standard deviation assuming you only have a sample of the whole population.
$$\sqrt{\frac{1}{n-1}\sum_{i=1}^{n} (x_i - \bar{x})^2}$$


*

*Get the mean of the distribution ($\bar{x}$)

*Subtract the mean from each number in the vector ($x_i$) and square the result $(x_i - \bar{x})^2$

*Sum the results and multiply by (1/total_number - 1) ($\frac{1}{n-1}$)

*Take the square root


If you have the entire population the equation is:
$$\sqrt{\frac{1}{n}\sum_{i=1}^{n} (x_i - \mu)^2}$$
