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I cannot answer on which one is preferred (they are actually really close so I don't think it matters much), but generally, major statistical software packages use Satterthwaite's method. SPSS and SAS both use it. In Stata, some commands like ttest would allow user to specify Welch's method, but Satterthwaite's is still the default.
And in literature, I have mostly seen Satterthwaite's formula being cited. Time to time it's referred to as Satterthwaite-Welch's degrees of freedom, but the formula cited is Satterthwaite's. I guess having published it one year earlier did matter.
The Welch-Satterthwaite d.f. can be shown to be a scaled weighted harmonic mean of the two degrees of freedom, with weights in proportion to the corresponding standard deviations.
The original expression reads:
$$\nu_{_W} = \frac{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)^2}{\frac{s_1^4}{n_1^2\nu_1}+\frac{s_2^4}{n_2^2\nu_2}}$$
Note that $r_i=s_i^2/n_i$ is the estimated variance of the $i^\text{th}$ sample mean or the square of the $i$-th standard error of the mean. Let $r=r_1/r_2$ (the ratio of the estimated variances of the sample means), so
\begin{align}
\nu_{_W} &= \frac{\left(r_1+r_2\right)^2}{\frac{r_1^2}{\nu_1}+\frac{r_2^2}{\nu_2}} \newline
\newline
&=\frac{\left(r_1+r_2\right)^2}{r_1^2+r_2^2}\frac{r_1^2+r_2^2}{\frac{r_1^2}{\nu_1}+\frac{r_2^2}{\nu_2}} \newline
\newline
&=\frac{\left(r+1\right)^2}{r^2+1}\frac{r_1^2+r_2^2}{\frac{r_1^2}{\nu_1}+\frac{r_2^2}{\nu_2}}
\end{align}
The first factor is $1+\text{sech}(\log(r))$, which increases from $1$ at $r=0$ to $2$ at $r=1$ and then decreases to $1$ at $r=\infty$; it's symmetric in $\log r$.
The second factor is a weighted harmonic mean:
$$H(\underline{x})=\frac{\sum_{i=1}^n w_i }{ \sum_{i=1}^n \frac{w_i}{x_i}}\,.$$
of the d.f., where $w_i=r_i^2$ are the relative weights to the two d.f.
Which is to say, when $r_1/r_2$ is very large, it converges to $\nu_1$. When $r_1/r_2$ is very close to $0$ it converges to $\nu_2$. When $r_1=r_2$ you get twice the harmonic mean of the d.f., and when $s_1^2=s_2^2$ you get the usual equal-variance t-test d.f., which is also the maximum possible value for $\nu_{_W}$ (given the sample sizes).
--
With an equal-variance t-test, if the assumptions hold, the square of the denominator is a constant times a chi-square random variate.
The square of the denominator of the Welch t-test isn't (a constant times) a chi-square; however, it's often not too bad an approximation. A relevant discussion can be found here.
A more textbook-style derivation can be found here.
Best Answer
Yes.
The Welch test uses the Satterthaite-Welch adjustment for the degrees of freedom:
$$ df'=\frac{\left(\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}\right)^2}{\frac{\left(\frac{s^2_1}{n_1}\right)^2}{n_1-1}+\frac{\left(\frac{s^2_2}{n_2}\right)^2}{n_2-1}} $$ As you can see, it's rather ugly (and in fact is approximated numerically), but it necessitates that $df'<df$. Here's a reference: Howell (2002, p. 214) states that, "$df'$ is bounded by the smaller of $n_1-1$ and $n_2-1$ at one extreme and by $n_1+n_2-2~df$ at the other".
Here are the 'official' references (note that the adjustment above--the one that is typically used--is derived in the second paper):
(Googling may yield ungated versions of these.)