TITLE: Two-Sample Confidence Interval for Normal Distributions
QUESTION [5 upvotes]: 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?

REPLY [6 votes]: 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.