View Full Version : Confidence Interval of Unrelated Events

01-08-2005, 03:41 PM
Suppose I have two independent characteristics, x and y.

I want to determine a confidence interval for x+y with alpha = a.

Is it correct to take CI(x,sqrt(a)) + CI(y,sqrt(a)) to get the CI(x+y,a)?

blank frank
01-08-2005, 11:33 PM
No. Think about the case of 2 standard normals at 0.05. The CI for either is -1.96 to 1.96. Add them together your way and you are adding two intervals at alpha=0.224. That gives you (-0.76,.76)+(-0.76,0.76)=(-1.52,1.52). That indicates a lower variance, when the variance should be higher. In fact, the variance of the sum should be 2 (the sum of the variances), giving a standard deviation of sqrt(2). So the answer should be -/+ 1.96*sqrt(2) = -/+2.77.

The way to get the answer is to square the margins of error, add them together, and take the square root. Sort of a Pythagorean thing. With the sum of two standard normals, that gives us sqrt(1.96^2+1.96^2) = 2.77, the correct answer.

01-09-2005, 12:06 PM
Frank, thank you for your response.

Now, I am a bit confused how to compute this if I had weighted events. Suppose I want to find 2A+5B.

Also, if I wanted to do 3 events, I could use the same methods?

blank frank
01-09-2005, 06:21 PM
The key is that variance exists so we can do math with standard deviations. The two rules to remember are Var(X+Y) = Var(X) + Var(Y) and Var(aX) = a^2*Var(X). So, square the standard deviations to get the variances, do the math with the variance, then take the square root to get back to the standard deviation, which you use to do the confidence intervals.

So Var(2A+5B) = Var(2A) + Var(5B) = 4*Var(A) + 25*Var(B). Again, you would take the square root of that to get the standard deviation for use in the confidence interval. This works the same with three variables: Var(A+B+C) = Var(A) + Var(B) + Var(C).

01-10-2005, 01:32 AM
Thanks for your help!