why equal variance assumption is necessary in T-test So I generally understand the basis for the t-test: i.e. you take advantage of the fact that you can make $\bar{X}-\bar{Y}$ standard normal:
$$Z = \frac{((\bar{X}-\bar{Y})-(\mu_X-\mu_Y))}{\sigma \sqrt{\frac{1}{n}+\frac{1}{m}}}$$
where $\sigma$ is the variance shared by the normal random variables $X$ and $Y$.
Furthermore, the unbiased estimator for $\sigma^2$ is:
$$S_p^2 = \frac{1}{m+n-2}\left(\sum{(X_i-\bar{X})^2}+\sum{(Y_i-\bar{Y})^2}\right)$$
which can be represented as a $\chi^2$ random variable using the fact that:
$\;\;\;\;\;\;\;\;\: \dfrac{(m+n-2)S_p^2}{\sigma^2}\:\:$ is $\:\:\:\chi^2$
and then you combine these distribution functions to get a T distribution.
What I don't understand is why the variances have to be equal (in this case, not talking about the wilcox test or anything like that). For example, what if you just had $\sigma_X^2 = a \sigma_Y^2$? where you knew the constant $a$. I tried to go through the derivation for the T-distribution and didn't come across any problems, I just had to include the constant $a$ in different places. If someone could direct me towards a reference or help me understand why this is the case I'd be very appreciative. I tried searching for the answer with no luck.
Thanks!
 A: If the population variances are not known to be equal, then the appropriate statistic for
testing equality of population means or finding a confidence interval for their difference would be (in your notation)
$$T = \frac{(\bar X = \bar Y) - (\mu_X - \mu_Y)}
{\sqrt{\frac{S_X^2}{n}+\frac{S_Y^2}{m}}}.$$
Distribution theory. But attempts do derive the exact distribution of this
statistic in a useful form have been unsuccessful. This is known
as the 'Behrens-Fisher Problem'. Nevertheless, it has been shown
that the distribution is approximately Student's t with 
degrees of freedom between $\min(n-1, m-1)$ and $n + m - 2.$
The precise formula for df is given in many statistics texts, but
messier than I want to type here. (Google 'Welch-Satterthwaite
equation'.)
Comprehensive simulation studies have shown
that this approximate t distribution is very accurate for 
a wide variety of ratios $\sigma_X^2/\sigma_Y^2$ and choices
of $m$ and $n$.
In practice. A consensus seems to have developed among applied statisticians that it is best to use
'Welch' or 'separate variances' t procedures, instead of 'pooled t'
procedures, in all cases---regardless whether an F-test shows
that $\sigma_X^2 \ne \sigma_Y^2.$ Many software packages use
the Welch test as their default two-sample t test, doing the
pooled test only if one overrides the default. (Two examples
in my experience are R and Minitab.) 
The pooled test has the
advantage of being workable on a statistics exam where calculators
are available, but not statistical software. This may account
for its continued use in the classroom. Even so, many recent elementary texts (including the popular
elementary books by Moore) recommend using the Welch test with
$df = \min(n-1, m-1).$ This is a conservative choice (never
rejecting when the messier, more accurate df would fail to reject).
And if both sample sizes are above 30 or so, this is not much of
a compromise.
Example. Here is an example using fake normal data to illustrate output
of the Welch test in R. Sample 1 has 10 observations from
$Norm(\mu=100, \sigma=15)$ and Sample 2 has 13 observations
from $Norm(\mu=80, \sigma=20).$ 
Because the population means
differ we might hope for rejection, but the P-value is a
little above 5% for the particular sample generated. The
population standard deviations (hence variances) are large
and unequal. (Of course, other samples generated using the
same parameters might show different test outcomes.)
In particular, notice that $df \approx 18$ here, whereas a pooled 
t test would have $df = 10 + 13 - 2 = 21,$ Roughly, the df for the Welch
test is decreased below $n + m - 2$ more markedly as the two
sample variances differ more markedly.
 x1 = round(rnorm(10, 100, 15), 1);  sort(x1)
 ## 83.2  85.9  88.3  92.3  92.4  92.5  96.3 101.4 104.6 124.2
 x2 = round(rnorm(13, 80, 20), 1);  sort(x2)
 ##  47.7  51.1  56.2  61.3  65.8  66.6  81.9  83.5  88.1  97.8 100.6 104.2 136.0

 t.test(x1, x2, alte="two.sided")

 ##        Welch Two Sample t-test

 ## data:  x1 and x2 
 ## t = 2.0137, df = 17.854, p-value = 0.05936
 ## alternative hypothesis: true difference in means is not equal to 0 
 ## 95 percent confidence interval:
 ## -0.7048102 32.8017333 
 ## sample estimates:
 ## mean of x mean of y 
 ## 96.11000  80.06154 

 Values = c(x1, x2);  Group = c(rep(1,10),rep(2,13))
 stripchart(Values ~ Group, ylim=c(.5, 2.5))


Reference: The Wikipedia article on 'Welch t test' gives
formulas and additional examples. Unfortunately, it uses
the phrase 'unequal variances test' where I would prefer
'separate variances test' because population variances
need not be unequal in order for the test to be valid.
