Geometric Bayesian Update

post by SquirrelInHell · 2016-04-09T07:24:14.774Z · LW · GW · Legacy · 9 comments

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I have moved this post to my blog:


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comment by Good_Burning_Plastic · 2016-04-09T08:36:34.716Z · LW(p) · GW(p)

There's no reason why P(E|H) and P(E|~H) must sum to 1, but I can't move the lower right corner without the whole diagram rescaling.

Replies from: SquirrelInHell, Petter
comment by SquirrelInHell · 2016-04-09T08:50:52.072Z · LW(p) · GW(p)

Of course you are right, but it would just be a linear transformation of the whole diagram, so it doesn't change anything in the result. I've built the diagram starting from a square, so I can't change this easily... just imagine the whole thing scaling on the X axis, OK?

Edit: since two people asked for this, I remade the diagram and now you can put in any values of P(E|H) and P(E|~H)

Replies from: AlexMennen
comment by AlexMennen · 2016-04-09T17:34:07.310Z · LW(p) · GW(p)

When I drag the dot for P(E|~H), it only changes P(E|~H), but when I drag the dot for P(E|H), it still keeps P(E|H)+P(E|~H) conserved, which is a little weird. I think it would be better if changing either of them did not affect the other.

comment by Petter · 2016-04-09T10:28:20.006Z · LW(p) · GW(p)

Agreed. The diagram strongly suggests that they do sum to one, so this geometrical method is more confusing than helpful.

comment by gjm · 2016-04-11T13:10:23.275Z · LW(p) · GW(p)

Two-word proof: Prin'f gurberz. (Nebhaq gur gevnatyr, plpyvpnyyl, jr unir: rivqrapr, cevbe bqqf, erpvcebpny cbfgrevbe bqqf.)

I think this would be clearer with only the triangle where all the action is happening, and without the stuff on the left whose only job is to put the whole thing into a rectangle. You can still have the prior odds and the evidence on perpendicular axes: make it a right-angled triangle and let what's now the right-hand edge of the rectangle turn into the diagonal.

Replies from: SquirrelInHell
comment by SquirrelInHell · 2016-04-12T01:34:39.103Z · LW(p) · GW(p)

the stuff on the left whose only job is to put the whole thing into a rectangle.

You are forgetting that it makes it possible to keep the scale of all numerical input/outputs consistent.

Replies from: gjm
comment by gjm · 2016-04-12T08:38:21.449Z · LW(p) · GW(p)

Point taken. (I personally prefer odds ratios strongly enough for this kind of thing that keeping the scale consistent doesn't bother me.) You could fix that, kinda, by fixing the side lengths of the "prior" and "posterior" side while allowing the length of the "evidence" side to vary, but that means introducing extra not-so-visible constraints so maybe it's a bit of a cheat.

comment by ike · 2016-04-10T01:19:30.409Z · LW(p) · GW(p)

Not a proof:

Gur boivbhf ohg oehgr sbepr jnl gb qb guvf jbhyq or svaqvat gur rdhngvba bs gur gjb yvarf sebz gur gjb evtug pbearef (obgu rnfl orpnhfr bs fvzvyne gevnatyrf naq/be xabja yratguf/pbbeqvangrf), svaq gur cbvag jurer gurl vagrefrpg, jevgr gur rdhngvba sbe gur yvar cnffvat guebhtu vg naq gur ybjre yrsg pbeare, fbyir sbe gung yvar uvggvat gur evtug yvar, gura fvzcyvsl. V unira'g qbar vg, ohg guvf fubhyq onfvpnyyl jbex qverpgyl.

Vf gurer fbzr oevyyvnag jnl bs fubjvat vg ol zrer fvzvynevgvrf? V qvqa'g frr bar vzzrqvngryl.

Replies from: SquirrelInHell
comment by SquirrelInHell · 2016-04-10T01:48:50.040Z · LW(p) · GW(p)

I've added a hint in the main post.