Establishing the “L1L2 AUC” equals `1/2 + arctan(1/sqrt(π - 3)) / π`

(`≅ 0.8854404657887897`

) used a few nifty lemmas. One of which I am calling “the sign tilting lemma.”

The sign tilting lemma is:

For X, Y independent mean zero normal random variables with known variances

`s`

and_{x}^{2}`s`

, what is_{y}^{2}`P[(X + Y ≥ 0) = (X ≥ 0)]`

?

It turns out it is: `1/2 + arctan(s`

._{x} / s_{y}) / π

This is solving: how much does knowing a fraction of the variance tell you about the sign of a sum?

Let’s check if this is right in a few cases.

- For
`s`

large we expect_{x}/ s_{y}`P[(X + Y ≥ 0) = (X ≥ 0)]`

near 1 (`X`

dominates`X + Y`

, making`X`

and`X + Y`

highly correlated). Our formula agrees with this. - For
`s`

near zero we expect_{x}/ s_{y}`P[(X + Y ≥ 0) = (X ≥ 0)]`

near 1/2 (`Y`

dominates`X + Y`

, making`X`

and`X + Y`

nearly independent). Our formula agrees with this. - For
`s`

we expect_{x}/ s_{y}= 1`P[(X + Y ≥ 0) = (X ≥ 0)] = 3/4`

. Our formula agrees with this.

The reason we have that `s`

implies _{x} / s_{y} = 1`P[(X + Y ≥ 0) = (X ≥ 0)] = 3/4`

is as follows. We only get inequality when `X`

and `Y`

have different signs (chance 1/2 of that by symmetry of `X`

to `Y`

) and `|Y| > |X|`

(an independent chance 1/2 of that, again by a symmetry argument). So the mismatch chance is 1/4, meaning the match chance is 3/4. In this case the following diagram shows that 3/4 of the angles between `(0, 0)`

and `(x, y)`

are in the shaded regions where `sign(x) = sign(x + y)`

And this completes the `s`

case, as in this case _{x} / s_{y} = 1`(x, y)`

is spherically symmetric (generates all angles uniformly at random).

For the general case we convert the question of if `s`

(for _{x} u ≥ s_{y} v`u`

, `v`

independent normal mean zero variance 1 random variables) to checking if the point `(s`

is above or below the line _{x} u, s_{y} v)`x + y = 0`

. This in turn is equivalent to checking if the point `(u, v)`

is above or below the line `s`

(notice the swap of coefficients, as this is the algebra for checking if a point is above or below a line!). This swap is a bit tricky to visualize, it is a bit of how geometry and algebra differ in viewpoint._{y} x + s_{x} y = 0

The advantage of the last check being: we are again in spherically symmetric situation with all angles generated uniformly. We can illustrate the geometry as follows. We put on the x-axis `u = X / s`

, and on the y-axis _{x}`v = Y / s`

to get the uniform distribution on angles. In this notation we use _{y}`a = 1 / s`

and _{x}`b = 1 / s`

. This _{y}*can be very confusing*, as we get one reversal from how we check above/below lines and another (cancelling reversal) by inverting `(u, v) → (X, Y)`

to `(X, Y) → (u, v)`

. In honest practice, we use the diagram to work out the answer *has* to have `arctan(s`

or _{x} / s_{y})`arctan(s`

in it, and pick the one that works (though in math one has to pretend to never have such difficulty or to use a dirty move to get out of it!)._{y} / s_{x})

Adding up the indicated areas of the shaded regions completes our argument for the lemma.

To conclude, we can rephrase or lemma as follow.

If a mean zero normal random variable determines a

`f`

fraction of the variance of a mean zero sum of independent normal random variables, then its sign matches the sign of the sum a`1/2 + arctan(sqrt(f / (1-f))) / π`

fraction of the time. That is: it gives a`arctan(sqrt(f / (1-f))) / π`

advantage in guessing the sign of the sum. For small`f`

this advantage is approximately`sqrt(f) / π`

.

Categories: Mathematics