Author Archives: david

An infinitary extension to the previous theorem about random variables

This is an attempt at an infinitary version of my previous post on finite strategies

Let \(S\) be some set with a sigma-algebra on it such that points are measurable. Let \(\mathcal{M}\) be be the set of measurable subsets. A strategy is a function s from \(\mathcal{M}\) to the set of probability measures on \((S, \mathcal{M})\) such that \(s(A)(A) = 1\).

A strategy s is subset consistent if whenever \(U, V, W\) are measureable sets with \(U \subseteq V \subseteq W\) we have \(s(W)(U) = S(W)(V) S(V)(U)\).

It’s reasonably easy to see this reduces to the previous version if \(S\) is finite and the sigma-algebra is \(P(S)\). While random variables are replaced by measures, there’s not really much difference between them.

Now fix s as a subset consistent strategy.

We can convert this to the version we had before to get the relations \(\prec\) and \(\sim\) on \(S\). The finiteness of \(S\) was not used to derive any of the properties of these.

Further when we pass to the quotient we get a totally ordered set \(T\) and a partition \(\{L_t : t \in T\}\) such that if \(x \in L_s, y \in L_t\) then \(x \prec y\) iff \(s < t\). Everything we did after that point was reliant on the finiteness of \(S\), so we'll have to try harder here. Theorem: \(T\) is the reverse of a well-ordered relation. i.e. every set has a maximal element. Proof: Note that we need only show that every countable set has a maximal element, as given a set with no maximal element we can construct a strictly increasing sequence in it \(x_1 < x_2 < \ldots\), which would be a countable subset with no maximal element. Suppose now we have some countable subset \(A \subseteq T\). Pick \(U \subseteq S\) such that \(U\) chooses an element from each of \(\{L_a : a \in A\}\). Then \(s(U)(\{x\}) \neq 0\) for some \(x \in U\), as otherwise \(s(U)(U) = 0\) (because \(U\) is countable). Then it can't be the case that \(x \prec y\) for any \(y \in U\). Let \(x \in L_t\). Then \(t\) must be a maximal element of \(A\). QED Theorem: \(L_t\) is countable. Proof: First note that for any countable \(A \subseteq L_t\) it must be the case that for all \(x \in L_t\), \(s(A)(\{x\}) > 0\), for if not then they all must be \(0\) (because otherwise any zero ones would be \(\prec\) the non-zero ones), and thus we would have \(s(A)(A) = 0\) because of countable additivity.

Now suppose \(L_t\) is uncountable. We can find at least \(\aleph_1\) distinct elements in it. Let \((x_a : a < \omega_1)\) be a sequence of length \(\omega_1\) consisting of distinct elements in \(L_t\). Let \(X_a = \{x_b : b \leq a\}\) and consider the sequence \(p_a = s(X_a)(\{x_0\})\). Claim: If \(c < d\) then \(p_c > p_d\).
Proof: By subset compatibility, \(p_d = s(X_d)(X_c) s(X_c)(\{x_0\}) \leq (1 – s(X_d)(\{x_d\})) p_c < p_c\) So we have constructed a strictly decreasing sequence of real numbers of length \(\omega_1\). But this is impossible, for standard set theoretic reasons. QED This gives us a theorem very akin to our finitary version: Theorem: Let \(s\) be a subset consistent strategy on S. There is a reverse well ordered set \(T\), a partition \(\{L_t : t \in T\}\) and weight function \(w : S : \to (0, \infty)\) such that for countable \(U \subseteq S\), \(s(U)\) is determined as follows: Let \(L_t\) be the largest \(t\) such that \(U' L_t \cap U \neq \emptyset\). Then \(s(U)(V) = \frac{\sum\limits_{x \in U' \cap V} w(x)}{\sum\limits_{x in U'} w(x)}\). Proof: We've essentially done all the work for this. Define \(w(x) = s(L_t)(\{x\})\) where \(x \in L_t\). We need the countability of \(U\) only to get that \(s(U)(V) = \sum\limits_{x \in V} s(U)(\{x\})\), after which everything proceeds as in the finite case. QED So the behaviour for countable sets is extremely analagous to the finite case. How do things work for uncountable sets? I currently have no idea at all. I don't currently have any examples where the behaviour for uncountable sets is different, but I'd be moderately surprised if there weren't some. If anything, I'd expect it to be possible to build an example out of any probability measure.

A theorem about certain families of random variables

This is a bit of pure mathematics. I was originally interested in it for decision theoretic reasons, but it turned out to be the wrong structure for what I was looking for. Still, I found the result interesting enough in its own right that I thought I would write it up.

Let \(S\) be some non-empty finite set. Let \(\mathcal{C}\) be the set of non-empty subsets of \(S\). Let \(\mathcal{R}\) be the set of random variables taking values in \(S\). A strategy is a function \(s : \mathcal{C} \to \mathcal{R}\) such that \(P(s(U) \in U) = 1\).

A strategy is said to be subset consistent if when \(U \subseteq V\) and \(x \in U\) then \(P(s(U) = x) = P(s(V) = x) P(s(U) \in V)\). (The intuition here is “It doesn’t matter what order we add pieces of information in”).

It turns out that there’s a very nice characterisation of all subset-consistent decision strategies.

Theorem:

Let \(L_1, \ldots, L_n\) be a partition of \(S\) into non-empty sets. Let \(w : S \to (0, \infty)\). Define a strategy as follows:

Given \(U \in \mathcal{C}\), define \(U’ = L_i \cap U\), where i is the largest such that this set is non-empty. Now pick \(x \in U’\) with probability proportional to \(w(x)\).

This strategy is subset-consistent.

Proof:

Note that \(s(U) = s(U’)\).

Let \(V \subseteq U\), and let \(L_i\) be as above. If \(L_i \cap V = \emptyset\) then \(P(s(U) \in V) = 0\), and in particular for \(x \in V\) we have \(P(s(U) = x) = 0\), so the result is proved in this case as both sides of the equation are 0.

In the case where \(V \cap L_i \neq 0\), we have \(V’ = V \cap L_i\), and \(s(V)\) is chosen from \(V’\). Note that \(s(U) \in V\) iff \(s(U) \in V’\).

\[\begin{align*}
P(s(U) = x) &= \frac{w(x)}{\sum\limits_{y \in U’} w(y)} \\
&= \frac{w(x)}{\sum\limits_{y \in V’} w(y)} \frac{\sum\limits_{y \in V’} w(y)}{\sum\limits_{y \in U’} w(y)} \\
&= P(s(V’) = x) P(s(U) \in V’) \\
&= p(s(V) = x) P(s(U) \in V)\\
\end{align*}\]

As desired.

QED

We will spend the rest of this proving that in fact every subset-consistent strategy arises this way.

From now on let \(s\) be some fixed strategy.

We’ll need some notational convenience. Let \(\rho_U(x) = P(s(U) = x)\). Let \(\tau(x,y) = \rho_{\{x,y\}}(x)\) if \(x \neq y\) and \(\tau(x,x) = \frac{1}{2}\).

The following lemma is practically just definition shuffling, but is surprisingly useful:

Lemma: Let \(x, y \in U\). Then \(\rho_U(x) = (\rho_U(x) + \rho_U(y)) \tau(x, y)\).

Proof: If \(x = y\) then this just boils down to \(\rho_U(x) = 2 \rho_U(x) \frac{1}{2}\). So suppose \(x \neq y\).

Let \(V = \{x, y\}\). By subset consistency we have \(\rho_U(x) = \rho_V(x) P(s(U) \in V)\). But \(\rho_V(x) = \tau(x,y)\) by definition, and \(s(U) \in V\) iff \(s(U) = x\) or \(s(U) = y\), so \(P(s(U) \in V) = \rho_U(x) + \rho_U(y)\). Combining these gives the desired result.

QED

Say \(x \prec y\) if \(\tau(x,y) = 0\). Say \(x \sim y\) if neither \(x \prec y\) nor \(y \prec x\). Note that because \(\tau(x,y) = 1 – \tau(y, x)\) it’s not possible to have both \(x \prec y\) and \(y \prec x\).

Lemma: If \(x, y \in U\) with \(x \prec y\) then \(\rho_U(x) = 0\).

Proof: \(\rho_U(x) = (\rho_U(x) + \rho_U(y)) \tau(x, y) = 0\).

This has a partial converse:

Lemma: If \(x, y \in U\) with \(\rho_U(x) = 0\) and \(rho_U(y) \neq 0\) then \(x \prec y\).

Proof:

By the preceding lemma we have \(\tau(x,y) = \frac{\rho_U(x)}{\rho_U(x) + \rho_U(y)} = \frac{0}{0 + \rho_U(y)} = 0\). This is well defined because \(\rho_U(y) \neq 0\).

QED

Lemma: If \(x \prec y\) and \(y \sim z\) then \(x \prec z\). Similarly if \(x \sim z\) then \(z \prec y\).

Proof:

We’ll only prove the first version. The second will follow similarly.

Let \(V = \{x,y,z\}\). Then because \(y \prec z\) we have \(\rho_U(y) = 0\). Then \(\rho_U(x) = \rho_U(x) \tau(x, y)\). But because \(x \sim y\) we have that \(tau(x, y)\) is neither \(0\) nor \(1\), so the only possible solution to this is that \(\rho_U(x) = 0\).

QED

Lemma: If \(x \prec y\) and \(y \prec z\) then \(x \prec z\).

Proof: Again consider \(V = \{x,y,z\}\). We must have \(\rho_V(x) = \rho_V(y) = 0\) because \(x \prec y\) and \(y \prec z\) respectively. Therefore \(\rho_V(x) \neq 0\). Therefore by a previous lemma we have \(x \prec z\).

Lemma: \(\sim\) is an equivalence relationship.

We have \(\tau(x,x) = \frac{1}{2} \neq 0, 1\) by definition, so \(\sim\) is reflexive.

It is symmetric because \(\tau(x, y) = 1 – \tau(y, x)\), so if \(\tau(x,y) \neq 0, 1\) then \(\tau(y,x) \neq 0, 1\).

We now must show transitivity. Suppose \(x \sim y\) and \(y \sim z\). If it is not the case that \(x \sim z\) then we must have either \(x \prec z\) or \(z \prec x\). Suppose without loss of generality that \(x \prec z\). Then by a previous lemma we must have \(y \prec z\), contradicting our assumption that \(y \sim z\).

QED

Theorem: There exists a partition of \(S\) into finitely many sets \(L_1, \ldots, L_n\) such that for any \(U \in \mathcal{C}\) we have \(P(s(U) \in U \cap L_i) = 1\) where \(i\) is maximal such that this set is non-empty.

Proof:

We will take as our partition the set of \(\sim\) equivalence classes. These equivalence classes are given a total ordering by \(\prec\), because if \(w \sim x\) and \(y \sim z\) then \(x \prec y\) implies \(w \prec z\). There are finitely many of them, so we can put them in an order \(L_1, \ldots, L_n\) such that if \(x \in L_i\) and \(y \in L_j\) with \(i < j\) then \(x \prec y\). If \(L_i\) is maximal such that \(L_i \cap U \neq \emptyset\), then for any \(x \in L_j\) with \(j < i\) we must have \(\rho_U(x) = 0\), because for some \(y \in U\) we have \(y \in L_i\) and thus \(x \prec y\). Therefore \(P(s(U) \not\in L_i \cap U) = 0\) as desired. QED We now need only define our weight function. Define \(w(x) = \rho_{L_i}(x)\) where \(x \in L_i\). Note that \(w(x) \neq 0\) because if it were 0 then we'd have \(x \prec y\) for some \(y \in L_i\), contradicting that \(x \in L_i\). Theorem: Let \(s(U)\) is chosen by choosing from \(U'\) with probability proportional to \(w(x)\). Proof: We have \(P(s(U) = x) = P(s(U') = x)\) by the previous theorem. Now, \(U' \subseteq L_i\). Therefore by subset consistency we must have \(P(s(U') = x) = \frac{P(s(L_i) = x)}{P(s(L_i) \in U'\)} = \frac{w(x)}{\sum\limits_{y \in U'} w(y) }\). This was the desired result. QED We have now shown the converse to our original theorem: Every subset-consistent strategy can be constructed from a partition and a weight function. As an additional note, I've thought a bit about how this works if you lift the restriction that \(S\) must be finite. A lot of this carries through naturally, but some of it doesn't. I think if you replace the subset consistency hypothesis with the natural infinitary equivalent that if \(U \subseteq V \subseteq W\) then \(P(s(W) \in U) = P(s(W) \in V) P(s(V) \in U)\) and add the additional restriction that for every \(U\) there exists an \(x \in U\) with \(P(s(U) = x) \neq 0\) then a lot of this carries over naturally with the partition replaced by one whose reverse is a well-ordered set and each partition is countable. But I haven’t worked through the details yet, and I’m not sure I’m interested in doing so.

An amusing conceit for a sci-fi story

This story is pretty obviously inspired by The Road Not Taken, though I think it’s interestingly different in significant ways.

Interstellar travel (of sorts) is, it turns out, quite easy. It’s so easy in fact that entire species of animals have evolved to be able to do it naturally, including humans.

The ability is called “worldwalking”, and it allows you to move yourself and a reasonable amount of additional mass (say, on the order of about 100 people with associated gear as an upper limit) to other planets. The connection between these other planets doesn’t seem to map to positions in real-space terribly well, if it maps at all, and the planets you can move between are typically quite similar (so any planet you move to will probably have a broadly similar range of temperature, gravity, atmosphere, etc).

This talent is distributed unevenly, but typically in species that have it, more than half of the population can do it too some degree (carrying themselves between worlds) and some can do it to a much larger degree.

The difficulty of moving between planets varies. There’s effectively some sort of potential function which moving up it is hard and moving down it is easy. This potential changes gradually over time. There’s also some sort of distance factor. Imagining the placement of worlds as being on some sort of 2D mountainous land doesn’t put you too badly wrong.

At some point in the last 50k years or so, earth’s potential reached a peak where it was basically a worldwalking everest – quite easy to leave (there’s no equivalent of going “splat” when you jump off the peak), almost impossible to climb.

This created a very strong selection pressure against the worldwalking talent. If you ever worldwalked away (which was very easy to do even for extremely low levels of ability) you would basically never come back. So anything with non-trivial worldwalking talent very rapidly removed themselves from the gene pool as soon as their ability manifested. As a result the talent is basically entirely absent on earth.

Since peaking a while back, the potential level of earth has been gradually going down. It’s recently reached the point where travel to it has become reasonably feasible – as recently as 100 years ago, probably fewer than 1% of world-walkers could make it here and they probably couldn’t bring more than one or two people with them. The rate of decline seems to have increased recently and it’s got to the point where most world-walkers could reach earth if they had to and a reasonably large percentage can bring about a dozen people with them.

All of a sudden it’s become impossible to dismiss world walkers as random crazies and earth-bound humanity has basically been forced back into contact with humanity at large.

The rest of humanity is quite like us, by and large – most of them can probably interbreed with us just fine. They’re technologically much less advanced than we are, though they’re not primitives. They’ve likely got quite a decent grasp of basic science, maths, philosophy etc. but they’ve never been particularly driven to create a high functioning technological society because world walking was just easier. Sure we could build roads and trains and stuff, but why bother when we can worldwalk? We could build large cities, but there are entire virgin planets only a few hundred days travel away, so why?

They’re also incredibly racially diverse. There are millions of inhabited planets, each with their own subtly different selection pressures (different animal and plant species, more UV, more solar radiation, higher gravity, less water, etc) and there’s been a lot of time for those to take effect, but there’s also been enough migration that you can see ancestries from a wide variety of different locations wherever you go.

There’s also a lot of them. They’ve never experienced any population pressure or mass die offs, and they’ve been reproducing merrily for as long as we have. A million planets with maybe a million people apiece? Sounds plausible, though the true number is basically unknowable, but even given that we’re already outnumbered 100:1.

So we’ve just met our cousins. They’re more primitive than us, they look different and scary, and they have a large pool of natural resources they’ve never learned to exploit and a vast amount of living space for us to move in to.

This probably isn’t going to go well, is it?

(though a potential saving grace is that there is so much space and resources available that we’re likely not going to step on anyone’s toes too soon).

Objections to the VNM utility theorem, part 2

This is another post about reasons why I don’t agree with the VNM theorem. I’m still focusing on the actual axioms of it rather than taking it in the broader context that I object to it because, well, frankly it’s much easier.

Here I would like to object to the idea that preferences over lotteries should be complete. That is, for any pair of lotteries \(A, B\) you should be able to say that \(A \leq B\) or \(B \leq A\). I’m not even going to use lotteries to do it, I’m going to use concrete events.

The point I would like to make is that there are multiple relevant notions of indifference between outcomes. One is “I genuinely don’t care about the difference between these outcomes” and another is “I do not have sufficient information to decide which of these outcomes I prefer”. Although your observed behaviour in these cases may be the same (pick an option at random), they differ in important ways. In particular if you treat them as the same you lose transitivity of preference.

Let me pick a worked example. Suppose you’re running a school, and you’ve got some money to spend, and you want to spend it on improving the quality of education for the kids. You’ve decided you can spend it on two things: Hiring more teachers and students lunches. Less the latter example sound trivial, imagine you’re in some high poverty area where a lot of kids can’t afford to eat properly. It’s pretty well established that if you’re starving all day in school then you’re going to perform badly.

We basically have two numbers here completely describing the problem space we care about (we could introduce additional ones, but they wouldn’t change the problem): Number of children who are adequately fed and number of teachers who are employed.

I’ll even make this easier for you and give us a utility function. All we care about is maximizing one number: Say the number of students who get at least one B grade. This should be a poster child for how well subjected expected utility works as a metric.

Increasing either of these numbers while leaving the other one constant will result in a strictly better scenario (assuming the number of teachers is capped to something sensible so we’re not going to find ourselves in a stupid situation where we have twice as many teachers as students). Adding more teachers is good, increasing the number of students who are well fed is good.

What’s complicated is comparing the cases where one number goes up and the other goes down.

It’s not always complicated. Consider two scenarios: Scenario 1 is that we have 0 students well fed and 2 teachers. Scenario 2 is that we have 20 students well fed and 1 teacher. Scenario 2 is obviously better (supposing we have 50 students in our entire student body. If the student body were much larger than one teacher could handle then that might be different).

Now consider Scenario 3: We have 0 students well fed and 9 teachers. This is obviously worse than Scenario 1.

Now consider the transition from scenario 2 to 3: Take lunch away from students, one at a time. At what point does this flip and become no better than scenario 1? Is it really better all the way down to the last student looking forlornly at you as you take their lunch away? Probably not.

But there isn’t really some concrete point at which it flips. There’s one side on which we regard scenario 1 to be obviously worse and one on which we regard it to be obviously better, and a biggish region in between where we just don’t know.

Why don’t we know? We have a utility function! Surely all we need to do is work out the number of utilons a fed student gives us, the number a teacher gives us, add them up and whichever gets the highest score wins! Right?

Obviously I’m going to tell you that this is wrong.

The reason is that our utility function is defined by an outcome, not by the state of the world. Neither teachers nor hungry students contribute to it. What contributes are results. And those results are hard to predict.

We can make predictions about the extreme values relatively easily, but in the middle there’s honestly no way to tell which one will give better results without trying it and seeing. Sure, we could put on our rationality hats, do all the reading, run complicated simulations etc, but this will take far longer than we have to make the decision and will produce a less accurate decision than just trying it and seeing. In reality we will end up just picking an option semi-arbitrarily.

But this is not the same as being indifferent to the choice.

To see why the difference between ignorance and indifference is important, suppose there are at least two scenarios in the grey area between scenarios 2 and 3. Call them A and B, with A having fewer fed students than B. We do not know whether these are better than scenario 1. We do however know that B is a better than A – more students are fed. If we were to treat this ignorance as indifference then transitivity would force us to conclude that we were indifferent between A and B, but we’re not: We have a clear preference.

Note that this failure of totality of the preference relation occurred even though we are living in the desired conclusion of the VNM theorem. The problem in this case was not that we don’t have a utility function, or even that we want to maximize something other than its expected value. The problem is that the lotteries are hidden from us – we do not know what the probabilities involved are, and finding that out would take more than the available time and resources available to us.

You can argue that a way out of this is to use a Bayesian prior distribution for this value with our states as inputs. This is a valid way to do it, but without more information than we have those numbers will be pretty damn near guesses and are not much less arbitrary than using our own intuition. Moreover, this has simply become a normative claim rather than the descriptive one that the VNM professes to make.

On vegetarians who eat chicken

This is a chicken. What does it look like to you? Does it look like a vegetable? No. It looks like an animal. So how can you be a vegetarian who eats chicken? You’re lying to me when you say you’re a vegetarian, aren’t you?

Thus spake just about everyone ever presented with someone with a dietary preference more complicated than “I eat all the things” or “I’m vegetarian”. I’m sure I’ve said similar things in the past.

This post is not more of that. It’s a post about how calling yourself “vegetarian but I eat chicken” is entirely OK.

I am not vegetarian. If you ask me if I’m vegetarian I will say “kindof”. I might elaborate, but I probably won’t unless it clearly makes sense in context to do so. But if you say “Who here is vegetarian?” I will likely put my hand up.

Am I lying to you? Well, yes, technically I suppose. My dietary preferences are much more nuanced than “vegetarian”. But I’ll eat vegetarian food, and I probably won’t eat the non-vegetarian food, or will have a preference about it which goes sufficiently against the meat-eating members of the group’s preference that I will either a) End up eating the vegetarian option anyway or b) Annoy them by making them eat something they’d prefer not to eat. Does anyone really think that life would be improved by my going “Well, actually, I’m not vegetarian but I do have some very precise constraints about what sort of meat I think it is appropriate to eat. Here, let me tell you all about them”. I didn’t think so.

All communication is this sort of trade off. Many of the things we say are inaccurate, or at the very least imprecise.

In many cases you are simply much better off answering the question “Are you vegetarian?” with “Yes” than you are “No, but life will be much easier for all concerned if you just treat me like one”. Many people will still use the latter, and that’s fine, but many people will choose the former, and that’s fine too. It is no more “lying” than a thousand other social constructs we use every day.

Another reason why people will sometimes do this is that people can be real assholes about food. This isn’t something I’ve personally experienced much when I’ve been vegetarian (hypothesis: Because I’m male), but it’s something a lot of vegetarians encounter.

It is vastly more irritating when you do not neatly pigeon hole. People are mostly used to vegetarians now. Support for vegetarianism is pretty widespread – it might get you some funny looks, it might start a discussion, but it’s pretty normal and boring at this stage so it’s unlikely to create a big deal.

Try telling someone you can’t eat something specific. For me it was dairy – I spent a period of time with a dairy intolerance (yes, this is a thing that can go away). A literal quote from someone on hearing this: “I would sooner the sweet embrace of death than go without cheese”. Sure, it was funny, and it’s among the higher quality responses I received, but imagine how tiresome this gets when you get this sort of response every single fucking time you have to explain your dietary requirements to people.

Why is this? I think it’s because you’re in an unfamiliar category. You’re strange and different, and people don’t know how to react to you. Moreover, food is something that people seem to feel very strongly about because it’s such a big part of their life, so by refusing to eat things they love it feels like you are judging them for the fact that they do. So you’re a weird and unfamiliar thing which is attacking their way of life. What do they do? They attack back of course!

Sure, it’s entirely possible that they are a lovely person who would never attack you like this, but if you don’t know them very well then you have little way of knowing that, and if you do know them very well then they probably already know about your dietary preferences so what’s the issue?

So here are your options: You can apply a label which people are familiar with and might be a bit dickish about, but are probably basically going to be fine with, or you can give the more accurate truth which has a non-zero chance of getting you put on the spot and asked to defend your life choices. You encounter this every time you engage in a common social activity. Which one are you more likely to choose?

So this is where the motivation to tell people you are vegetarian when you’re not comes from.

Know where the motivation to tell people you are vegetarian but you eat chicken comes from? Well, it might come from the fact that you just really like chicken. I’m sure for many people it does. But another place it comes from? Consideration. You are giving people options to make their lives easier. It sure is nice when they use that as a reason to judge you, isn’t it?

Finally, from a language point of view, the construct “I am vegetarian but I eat chicken” is totally OK. People use it, you understand what it means, what’s the problem? It’s functionally equivalent to “I keep a mostly vegetarian diet but I eat chicken” or “I’m like a vegetarian but I eat chicken”, so it gets shortened. You wouldn’t object to the sentence “I am a vegetarian, with some exceptions”.

There are two ways to parse the sentence: One is “(I am a vegetarian) and (I eat chicken)”, which is a logical contradiction and so probably not what was meant. Another is “I am a (vegetarian with the exception that I eat chicken)”. One of these is the obviously intended parse of the sentence. The other is the parse you are insisting on out of a misguided sense of linguistic prescriptivism which you are using to police someone else’s life choices. Please stop it.

Edit: An update in response to some feedback from friends who are strict vegetarians.

An important thing to note here is that this is not the same thing as people who say “I’m vegetarian” whilst chowing down on a steak. Descriptive linguistics is a powerful tool, and with great power comes great responsibility. One of these responsibilities is this: Don’t ruin words for people who need them. One of the ways you can ruin the word “vegetarian” is by making people think that all vegetarians eat chicken.

Social lies are OK, but when you tell them you have a responsibility to exhibit behaviour consistent with what you are describing it as. If you describe yourself as “I’m vegetarian but I eat chicken”, this is fine. If anything it reinforces the idea that eating chicken is an exceptional behaviour for people who describe themselves as vegetarian. If you simply say “I’m vegetarian” whilst eating chicken, you are reinforcing the idea that vegetarians eat chicken and you are making other peoples’ lives worse. If you say “I’m vegetarian” and go on to eat vegetarian food, that’s fine. You’ve not hurt peoples’ perception of the world, and you probably have a perfectly good reason for having preferred to eat vegetarian right now even if the reality is more complicated.

There is a large gray area in the middle here as to what’s acceptable behaviour, and I’m not going to try to take a stand on where the dividing line is. All I’m saying is that there is a wide range of acceptable behaviour, and that the way people react to some of it is very unhelpful.