In my earlier put up on Overlapping Confidence Intervals I requested what we are able to study from the overlap, or lack thereof, between confidence intervals for 2 inhabitants means constructed utilizing unbiased samples.
To recap: if the person confidence intervals for teams A and B don’t overlap, there have to be a statistically vital distinction between the inhabitants means for the 2 teams.
In different phrases, the interval for the distinction of means won’t embrace zero.
If the person intervals do overlap, alternatively, something goes.
The interval for the distinction of means might or might not embrace zero.
Certainly, it’s even potential for the person intervals for each A and B to incorporate zero whereas the interval for his or her distinction doesn’t!
In Half I we discovered a approach to rephrase our statistical drawback about overlapping confidence intervals as a well-known geometry drawback involving proper triangles.
We then solved this geometry drawback utilizing the Pythagorean Theorem and Triangle Inequality.
To construct the connection between confidence intervals and proper triangles, nevertheless, we assumed that our two estimators had been uncorrelated.
Sadly this assumption fails in lots of fascinating real-world functions.
At present we’ll ask what occurs to our earlier conclusions about overlapping intervals if we permit for correlation.
A Phrase About Notation
Whereas I phrased my first put up about overlapping intervals by way of pattern and inhabitants means for 2 teams, the concept is normal.
Nothing substantive adjustments if we substitute the parameters ((mu_A, mu_B)) with ((alpha, beta)), the estimators ((bar{A}, bar{B})) with ((hat{alpha}, hat{beta})), and the usual errors (left(textual content{SE}(bar{A}), textual content{SE}(bar{B})proper)) with (left(textual content{SE}(hat{alpha}), textual content{SE}(hat{beta})proper)).
So long as the 2 estimators are uncorrelated and (roughly) usually distributed, the outcomes from Half I apply to the person intervals for (alpha) and (beta) versus the interval for the distinction (alpha-beta).
At present we’ll ask what occurs when the estimators are doubtlessly correlated.
To make it clear that our outcomes are normal, I’ll use the extra agnostic ((alpha,beta)) notation all through.
A Motivating Instance
Contemplate a randomized managed trial with two energetic therapies.
In our paper on pawn lending, for instance, my co-authors and I evaluate default charges between debtors assigned to the establishment pawn contract (management), a brand new structured contract (Remedy A), and a selection arm (Remedy B) wherein they had been free to decide on whichever contract they most well-liked.
To study the causal impact of the structured contract, we evaluate the imply default charges of debtors who obtained Remedy A in opposition to the corresponding fee within the management group.
Name this distinction of means (hat{alpha}).
Equally, to study the impact of selection, we make the analogous comparability between Remedy B and the management group.
Name this distinction of means (hat{beta}).
Now suppose you’re studying a paper that reviews each of those estimators and their commonplace errors.
To search out out which remedy is simpler, you have to evaluate (hat{alpha}) and (hat{beta}), however these estimators should be correlated as a result of they each contain the management group common:
[
begin{align*}
hat{alpha} &= text{(Treatment A mean)} – text{(Control Group mean)}
hat{beta} &= text{(Treatment B mean)} – text{(Control Group mean)}.
end{align*}
]
Granted, if you happen to had entry to the uncooked information, you could possibly simply resolve this drawback with out worrying concerning the correlation.
Within the distinction (hat{alpha} – hat{beta}), the management group imply cancels out
[
hat{alpha} – hat{beta} = text{(Treatment A mean)} – text{(Treatment B mean)}.
]
So if we had the uncooked information for therapies A and B we’d be again to a well-known unbiased samples comparability of means, as in Half I.
However if you happen to’re studying a paper that solely reviews ((hat{alpha}, hat{beta})) and their commonplace errors, you can not straight calculate the usual error for the distinction.
The frequent variation from the management group imply is baked into the way in which each (textual content{SE}(hat{alpha})) and (textual content{SE}(hat{beta})) are computed, though this variation is irrelevant for (textual content{SE}(hat{alpha} – hat{beta})).
Permitting Correlation
So how will we compute (textual content{SE}(hat{alpha} – hat{beta})) if the 2 estimators are correlated?
Recall that the commonplace error is the usual deviation of that estimator’s sampling distribution, and a normal deviation is merely the sq. root of the corresponding variance.
Utilizing the properties of variance and covariance,
[
text{Var}(hat{alpha} – hat{beta}) = text{Var}(hat{alpha}) + text{Var}(hat{beta}) – 2text{Cov}(hat{alpha}, hat{beta}).
]
Defining (rho equiv textual content{Corr}(hat{alpha},hat{beta})), it follows that
[
text{SE}(hat{alpha} – hat{beta})^2 = text{SE}(hat{alpha})^2 + text{SE}(hat{beta})^2 – 2rho cdot text{SE}(hat{alpha}) cdot text{SE}(hat{beta}).
]
If (rho = 0) this reduces to our system from Half I:
(textual content{SE}(hat{alpha} – hat{beta})^2 = textual content{SE}(hat{alpha})^2 + textual content{SE}(hat{beta})^2) so we are able to equate the LHS with the size of the hypotenuse of a proper triangle whose legs have lengths (textual content{SE}(hat{alpha})) and (textual content{SE}(hat{beta})).
If (rho neq 0), nevertheless, this connection to the Pythagorean Theorem not holds.
Nonetheless, there are nonetheless triangles hiding on this commonplace error system!
To disclose them, we want a extra normal theorem about triangles.
Contemplate a triangle whose sides have lengths (a,b) and (c).
Let (theta) be the angle between the perimeters whose lengths are (a) and (b).
Then by the Legislation of Cosines
[
c^2 = a^2 + b^2 – 2cos(theta) cdot ab
]
This equality holds for any triangle.
For a proper triangle whose legs have lengths (a) and (b), we have now (theta = 90°) so the Legislation of Cosines reduces to the Pythagorean Theorem
[
c^2 = a^2 + b^2.
]
When (theta neq 90°), the “correction time period” (-2cos(theta)cdot ab) reveals how the size of (c) differs from that of a proper triangle with legs of lengths (a) and (b).
When (theta < 90°) the cosine is constructive so the correction time period shortens (c); when (theta > 90°) the cosine is unfavourable so the correction time period lengthens (c).
Whatever the angle (theta), nevertheless, the Triangle Inequality nonetheless holds: (c < a + b) as a result of the shortest distance between two factors is a straight line.
From Geometry to Statistics
The cosine of an angle is at all times between unfavourable one and one.
Are you able to consider anything that shares this property?
That’s proper: correlation!
So let’s put the Legislation of Cosines and our commonplace error system from above side-by-side:
[
begin{align*}
c^2 &= a^2 + b^2 – 2cos(theta) cdot ab
text{SE}(hat{alpha} – hat{beta})^2 &= text{SE}(hat{alpha})^2 + text{SE}(hat{beta})^2 – 2rho cdot text{SE}(hat{alpha}) cdot text{SE}(hat{beta}).
end{align*}
]
The analogy is good.
We will view (textual content{SE}(hat{alpha})) and (textual content{SE}(hat{beta})) because the lengths of two sides of a triangle and (rho) because the cosine of the angle between these sides.
This makes (textual content{SE}(hat{alpha} – hat{beta})) the size of the third aspect, indicated in blue within the following diagram.
When (rho) is constructive, the usual error of the distinction is smaller than it could be beneath independence; if (rho) is unfavourable, the usual error of the distinction is bigger than it could be beneath independence.
The Grand Finale
Since we’ve equated (textual content{SE}(hat{alpha} – hat{beta})), (textual content{SE}(hat{alpha})) and (textual content{SE}(hat{beta})) with the perimeters of a triangle, the Triangle Inequality provides
[
text{SE}(hat{alpha} – hat{beta}) < text{SE}(hat{alpha}) + text{SE}(hat{beta})
]
assuming that (|rho| < 1).
And now we’re on acquainted floor.
Let (z) be the suitable quantile of a standard distribution, i.e. (z approx 2) for a 95% confidence interval.
Simply as we argued in Half I, the person confidence intervals for (alpha) and (beta) overlap exactly when ((hat{alpha} – hat{beta})/z < textual content{SE}(hat{alpha}) + textual content{SE}(hat{beta})) and there’s a vital distinction between the 2 parameters when ((hat{alpha} – hat{beta})/z > textual content{SE}(hat{alpha} – hat{beta})).
The situation for overlap and a big distinction is
[
text{SE}(hat{alpha} – hat{beta}) < frac{hat{alpha} – hat{beta}}{z} < text{SE}(hat{alpha}) + text{SE}(hat{beta})
]
which holds by the Triangle Inequality.
So we’re again to precisely the identical state of affairs we had been in when (rho = 0)!
So long as (rho neq -1, 1) the identical outcomes regarding confidence interval overlap apply when (hat{alpha}) and (hat{beta}) are correlated as when they’re uncorrelated:
- Overlap doesn’t inform us something about whether or not there’s a vital distinction, however
- an absence of overlap implies that there should be a big distinction.
