Two sample $t$ test for the difference in means Assuming we have two independent samples, with the same variation, the T statistics is being constructed this way:
$$t = \frac{ (X_1 - X_2) - (\mu_1 - \mu_2)  } {S_p},$$
Where by $\bar X$ is denoted the sample mean, by $\mu$ the population mean.  Also, $S_p^2$ is
$$S_p^2 = \frac{ (n_1 - 1) S_1^2 + (n_2 - 1) S_2^2} {n_1 + n_2 - 2},$$
Where $S^2$ is the unbiased estimator of the variation (also referred as the sample variation), n is the population size.
$t$ has $n_1 + n_2 - 2$  DOF. My question is can we replace $S_p$ by $S_1$ or $S_2$ and get a $t$ with $n_1 - 1$ or $n_2 - 1$ DOF ? As far as I am seeing this is quite legit, as we will again get $N(0,1) / \sqrt{\chi^2(n_1-1)/(n_1-1)}.$
My other question is how are we able to construct this $t$ statistic, if the numerator and denominator are not independent?
Sorry for not being able to use MathJax properly.
 A: (1) $t$ has $n_1 + n_2 - 2$ DOF. My question is whether we can replace $S_p$ by $S_1$ or $S_2$ and get a $t$ with $n_1 - 1$ or $n_2 - 1$ DOF ? 
Because you are doing a 'pooled' t test, you have already assumed
that the variances $\sigma_1$ and $\sigma_2$ of the two populations
are equal, so we can write $\sigma_1 = \sigma_2 = \sigma.$ 
In that case $S_1,\, S_2\,$ and $S_p$ are all legitimate estimates
of $\sigma.$ So you $could$ legitimately make the substitution you
suggest and get $t$-statistics with the degrees of freedom you
state. $However,$ that would be a foolish substitution because,
under your assumptions, $S_p$ is the 'best' of the three estimates (being based on the most data),
and it will give you more 'power' to detect a significant
difference in means if such a difference exists.
(2) For normal data, the numerator and denominatorof $t$ $are\,$ (stochastically) independent. In particular, the random
variables $\bar X_1$ and $S_1$ are independent. Admittedly,
this is counterintuitive because $S_1$ is $functionally$
dependent on $\bar X_1$. (That is $\bar X_1$ appears in the
formula for $S_1.$) Similarly $\bar X_2$ and $S_2$ are
independent random variables. Finally, putting things together,
the numerator and denominator of $t$ are independent.
This independence of sample mean and sample SD is true $only$
for normal data. (Not for uniform data, exponential data, or
any other kind.) So it is not surprising that the result
for normal data would seem strange to you. But the independence for normal
data can be proved in a variety of ways (e.g., using linear
algebra or using moment generating functions).
