Two-Sample Confidence Interval for Normal Distributions Let's say I have two independent random samples $X_1, X_2, \dots, X_n$ and $Y_1, Y_2, \dots, Y_n$ from normal distributions with real, unknown means $\mu_x$ and $\mu_y$ and known standard deviations $\sigma_x$ and $\sigma_y$.
How would I go about deriving a $100(1 - \alpha)$% confidence interval for $\mu_x - \mu_y$?  This is straight forward (in my mind) assuming the standard deviations are equal, but what if they are unequal?
 A: Alright, you say known variances.  So it's an exercise on a point of theory, not a realistic problem.
And you actually assume the two sample sizes are equal.
Start by recalling something from the one-sample problem:
$$
\bar{X} = \frac{X_1+\cdots+X_n}{n} \sim N\left(\mu_X,\frac{\sigma^2_X}{n} \right)
$$
$$
\bar{Y} = \frac{Y_1+\cdots+Y_n}{n} \sim N\left(\mu_Y,\frac{\sigma^2_Y}{n} \right)
$$
You don't explicitly state that the two samples are independent.  If they are, they we have
$$
\bar X - \bar Y \sim N\left(\mu_X-\mu_Y,\frac{\sigma^2_X+\sigma^2_Y}{n}\right)
$$
(If we had unequal sample sizes $n$ and $m$, then the variance would be $\dfrac{\sigma^2_X}{n}+\dfrac{\sigma^2_Y}{m}$.)
So
$$
\frac{((\bar X-\mu_X) - (\bar Y-\mu_Y))\sqrt{n}}{\sqrt{\sigma^2_X+\sigma^2_Y}} \sim N(0,1).
$$
So the probability that
$$
-A < \frac{(\bar X-\mu_X) - (\bar Y-\mu_Y)}{\sqrt{ \frac{\sigma^2_X+\sigma^2_Y}{n} }} <A \tag{1}
$$
is the desired confidence when the number $A$ is suitably chosen.  Now do a bit of algebra to rearrange the inequalities $(1)$:
$$
\bar X - \bar Y - A\sqrt{\frac{\sigma^2_X+\sigma^2_Y}{n}} < \mu_X-\mu_Y < \bar X - \bar Y + A\sqrt{\frac{\sigma^2_X+\sigma^2_Y}{n}}
$$
That's the confidence interval.
A: When the standard deviations are unequal the inference problem of comparing two means is often called the Behrens - Fisher problem.  The pivotal quantity for testing or constructing confidence intervals is a "t-like" statistic gotten by taking the difference of the two sample means and dividing by the sample estimate of the standard error of the mean difference.  The standard error is a function of the two unknown standard deviations and and the sample sizes used. The estimate involves replacing the unknown variances with their sample estimates.  The distribution of the test statistic under the null hpothesis that the means are equal is sometimes called Welch's distribution.  It can be approximated by a t distribution whoses degrees of freedom are fractional (not necessarily an integer).  This approximation is called the Satterthwaite approximation. This Wikipedia link provides the detailed information:
Welch-Satterthwaite Approximation .
If the variances are unequal and known then the pivotal quantity to use is what is given by Mike and it will have a standard normal distribution.  In practice the variance are not known unless you have knowledge that they are the same as variances that have been previously estimated based on very large samples.
