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Damn. Just realized I can only view 100 responses with the free surveymonkey account. It will cost me $25 per month to get an account that will handle 1000 responses, $325 per year for an account with unlimited responses. That would definitely have been the cause of your problems.

It would interesting to see what % of people are voting strategically and who people would like to vote for if they weren't voting strategically.

You might want to try this: http://www.kwiksurveys.com/?gclid=CI6cgd7Cl6gCFSYZQgodyxMZCw

Its the first response for googling "free survey with unlimited responses."

Another solution is a Google Docs Spreadsheet.

I agree with Simon, there are a lot of interesting issues that the standard polls don't bring up. They tend to ask "who will you vote for" or "what party do you support" which are two different questions. In a riding where Lib and Cons. candidate running neck-and-neck, and NDP or Green supporter might vote Liberal.

Yeah, what I'd really like to see is a poll that asks who people want to vote for then use that to assign seats proportionately, then take that and compare it the electoral outcome. More or less measuring what the difference would be if we had a proportional electoral system versus the single winner FPTP. Not sure if anyone's done it for the past election(assuming there was some poll that asked for party preference and not simply who people were voting for) but it would interesting.

Simon - The impact of proportional v. FPTP depends a lot upon how votes are distributed geographically (compare e.g. the number of seats held by the Green party and the number of seats held by the Bloc Quebecois). To do the kind of study that you're thinking about, it would be necessary to have a reasonable-sized sample at something close to the individual constituency level. That would require a pretty big sample - and one far more representative than the WCI sample! Also, in most publicly available data sets (e.g. the Gallop data) geographic data is masked, as it's really hard to maintain confidentiality when there is any kind of detailed geographic information in the data set.

Simon, thanks for the suggestions. Another question to ask - would you trade your vote? E.g. I'll vote Green in my riding where it doesn't matter if you'll vote Liberal in your riding where it does.

Wendy - the polls also tend to have this leader fixation, which is one of these Canada-as-US-divided-by-10 things. In the US, presidential races are separate from congressional or senate races, so it matters which presidential candidate the individual voter favours. In Canada, the prime minister is not directly elected, so there is no need to focus on individual leaders.

Simon -- Ask and you shall receive. Here's some pretty clever work looking at how having proportional or MMP would affect the election, if it was held today:


Haha thanks Azim, yeah that was more or less what I was looking for.

F: Terrific technology. That was a nice tight survey. n>200?

Such a large proportion of well-educated people, many of them economists voting against the economist-lead Conservatives (or for other parties). My, my.

So do Canadian economists who love to chatter policy and make lame jokes about accountants carry any political influence?


I was surprised - although I probably shouldn't have been - by the overwhelming majority of male responders. Sigh.

Odd that Liberal support is higher than the Conservatives. I can't say the Liberals are any less guilty of proposing bad economic policies for political expediency. Is it a perception that the CPC are more ideological?

Frances, this is a great survey! Having been land-line-less for several years now, I have been wondering when someone was going to ask how representative these surveys are becoming...

While you're right about the PM not being elected directly, I would disagree on the leader fixation thing. Given the huge power afforded to the PM by our system (much greater than the US president), especially party discipline, the leader/party is really what matters. Honestly, how important is the local MP, especially at the federal level? I know I choose the party (and leader) to vote for, not the individual candidate, and all my friends seem to do the same. We never even know the name of local candidates (unless they say very stupid things of course - so they can have a negative impact on voting decisions, but not a positive one).

On the other hand, asking who people would like to vote for vs who they will vote for would indeed be interesting.

Andrew F - I was surprised by the high Liberal support also. The sample size is small so it could be just a few people who all know each other. Or perhaps it reflects relatively high number of Ontario respondents? Perhaps also readers of this blog tend to be more libertarian than Conservative, so are unimpressed by things boutique tax credits and law and order agendas?

Linda, yes, I was surprised by that too. But a lot of my followers on twitter are male too. You can usually tell which WCI posts appeal to women, though, by looking at facebook likes - so the dumb men one hit a lot of nerves!

westslope, thanks. N is now over 300.

N = 383 at the close of markets Wednesday :-). Impressive.

For the record, I strategically checked the Liberal Party. In the early 1990s, I held my nose on social issues and voted Preston Manning's Reform Party for two reasons: conservative fiscal policy, and reasonable constraints to total immigration flows. Jaws of friends slammed to the floor. (It was Ottawa....)

Well, I'm still a fiscal conservative but my social issues nose hurts these days if I hold it. So I'll be damned if I vote for the Harper Conservatives. In addition, Harper strays too far in the direction of sectarian fear-mongering for my tastes.

Harper wants to appeal to electors who are what Argentinians aptly call 'patriotudos'. While I can be just as big a 'gran boludo' as the next aging white male, patriot with big _______ I am not.

There. I just gave an irrational identity-politics motivation for strategic voting. Or maybe overtly partisan discussions are simply subtle forms of cheap-talk marketing?

Linda: If Frances goes for leadership of let's say the Liberal Party, I'll raise funds. But now the two of you have me worried that she might alienate the female vote. :-(

Interesting poll. In particular, I'm surprised your readers are so disproportionately Liberal (relative to the general population). The poll seems strong enough that one might think WCI has a political bent, but I haven't picked that up from reading it.

"I would disagree on the leader fixation thing. Given the huge power afforded to the PM by our system (much greater than the US president), especially party discipline, the leader/party is really what matters. Honestly, how important is the local MP, especially at the federal level? I know I choose the party (and leader) to vote for, not the individual candidate, and all my friends seem to do the same."

Given your user name I suspect you think this way because you live in Ottawa? Outside the NHQ, and particularly in some of the smaller communities I've lived in, people seem to care quite a bit about the local candidates. For instance, Yvon Godin's (NDP) stronghold on the Acadie Bathurst riding preceded Jack Layton's leadership, and seems connected to his ties to the local unions (the riding is mostly working class). I'm sure there are other similar examples out there. My view is that the farther you are from Ottawa, or the more disillusioned you are with the federal leaders or federal policy making, the more likely you are to vote for a local candidate rather than a party leader. Also, in smaller communities, people are more likely to personally know some of the local candidates, in which case voting for your neighbour or some other familiar face has more appeal than voting for some unknown leader.

Damn! I really ought to be able to answer this, but my brain has stopped working. Somebody help me with this please.

Let x% of people interested in economics be male.

Assume the same x% of people using the internet are male. (Just an assumption).

Let 86% of people who are both internet users and interested in economics be male.

What is x? Thanks.

Nick: It depends on the correlation between internet use and maleness. If the correlation is one, then x=86%. If the correlation is zero, i.e. the events are independent, then x = sqrt(86%) = 92.7%.

I meant, it depends on the correlation between internet use and interest in economics among males... I think that makes it right :-( Sorry for the premature answer.

Actually, forget it. Can't think. Crap! Good night.

K: whatever it is that my brain has got, your brain has it too! Because my thinking was exactly the same!

"Given the huge power afforded to the PM by our system (much greater than the US president), especially party discipline, the leader/party is really what matters. "

This is a misnomer. It is true that the Prime Minister possesses a great deal of power, in a system characterized by strong party discipline. However, you have to remember that most parties have internal factions that act as veto players against government actions. Mulroney, for instance, had his western caucus and his Quebec caucus, and had to keep both happy in order to survive. Similarly, Chretien's problems with the Martinite faction of his party are well known. Party leaders face significant obstacles to doing whatever they want - just not obstacles we can see.

Incidentally, Harper is uniquely powerful among Canada's PMs in part because he has assembled (purposely in my view) as small a coalition of interests as possible. Harper wants a small winning coalition, which will enable his party to sustain its power over the long haul, instead of imploding like every Conservative government of the 20th century (namely Borden/Meighen, Bennett, Diefenbaker and Mulroney).

Serves you right for posting at midnight (and again at 5am!)

Preface: let A be the event that a person answers the survey. Let B be the event that a person uses the internet and is interested in economics. Assume A implies B. If 86% of A are M, does that imply 86% of B are M? No! Perhaps the B who are not A have a different fraction of M; then so must B, unless (B and ~A) is empty (but you were surely not assuming that A<=>B with N(A) < 400, were you? :-)

But lets accept the premise. Let I be the event that a person is interested in the internet, E be the event that the person likes economics, and M be the event that the person is male. We are given:

1. P(M|EI) = 0.86
2. P(M|E) = P(M|I) = x

Can we deduce x? No. This is a lot easier to explain with a Venn diagram, but basically what we would like to know is the ratio P(EM)/P(E), or equivalently, P(IM)/P(I). The trouble is that knowing the ratio P(EIM)/P(IE) doesn't tell us very much about the ratios we would like to know, even given that they are equal. Even allowing that we know P(M) endogenously doesn't help.

Aha! So Phil's brain is working!

So, we need an additional assumption. Some sort of independence assumption.

For example: if 50% of economics lovers were male, and 50% of internet lovers were male, it does not automatically follow that 50% of people who love both are male. But it would do if these were (in some undefined sense) independent characteristics. See where I'm coming from?

Nick, I understand what you are driving at, but what I was trying to say is that there isn't a nice, parsimonious assumption we can add that will solve the problem.

I can't draw a diagram here, so I am going to have to work with text. Picture (or draw) the traditional Venn diagram of three overlapping circles named for our basic events, E, I and M. Now I want to label the disjoint regions of the diagram; it will be convenient to work with disjoint events so that their probabilities will add. In the centre is the intersection set EIM which I will label D. Then there are the three regions in which exactly two events intersect; I will label EM~E by A, IE~M by B, and EM~I by C. Finally, there are the regions containing only one event; call M~E~I F, I~E~M G, and E~I~M H.

Here is our information:
1. P(D) / (P(B)+P(D)) = 0.86
2. (P(A)+P(D)) / (P(A)+P(B)+P(D)+P(G)) - (P(C)+P(D)) / (P(B)+P(C)+P(D)+P(H)) = 0

Which we may supplement with:
3. P(A)+P(C)+P(D)+P(F) = 0.5 [roughly]

Which beautiful, concise axiom would you add to solve this system? If there is one, then I apologize, but I don't see it.

Of course I forgot:

4. P(A)+P(B)+P(C)+P(D)+P(E)+P(F)+P(G) = 1

That doesn't help much though, does it?

Phil: I don't see it either.

Hang on. Now I'm sort of seeing something, based on your venn diagrams. Make them rectangles, rather than circles.

A vertical rectangle for internet-lovers. Split it vertically between M and F regions. A horizontal rectangle for Economics lovers, split horizontally between M and F regions.

So we have an L-shaped venn diagram, with both the vertical and horizontal bits of the L overlapping, and split into two. That's the orthogonality bit.

Now look at the square corner of the L, where the two rectangles overlap. It's subdivided into a larger MM square, a smaller FF square (plus 2 rectangles MF and FM which don't make any sense!). The MM and FF squares answered this survey.

So MM/(MM+FF) = 0.86 where M+F=1

That still doesn't quite make sense.

Nick and Phil:

I spent way to long thinking about this problem yesterday, because I shared Nick's intuition about the relevance of the independence of I and E.  First came up with what Phil wrote yesterday (exactly, right down to variable names).  Spent lots of time thinking about potentially simplifying symmetries.  In the end I decided there were just too many independent variables.  Seven to be exact: the 3 unconditional probs of M,E and I; the three pairwise joint probs and and the joint prob of all three (Phil's Venn diagram). Nick first proposed two equations plus the presumed P(M)=0.5.  Then "independence" presumable of I and E, but perhaps also of I|M and E|M, is another two equations, so we have 5. Couldn't think of any others.  Choosing two more constraints would be equivalent to choosing P(I) and P(E), which is obviously not very revealing.  Frustrating, as I kept feeling like I almost had something, yet ended up with nothing.  I'm still not totally convinced that nothing useful can be said, mostly because 86% is such a big number that I think it might provide some useful inequality constraints...

If I really wanted to go on from here I might try the simplest fitting Gaussian copula.

Just my thoughts for now.

K: I don't think we can get any inequality constraints. Suppose, for example, that all female economics lovers hate the internet, and all female internet lovers hate economics. Then Frances would get 100% males responding to her survey, even if 50% of economics lovers were female, and 50% of internet lovers were female.

When we (or someone else) eventually figures out the sensible independence assumption, we are going to be kicking ourselves for missing it!

Nick: "I don't think we can get any inequality constraints."

No, you're right: not in general. You'd need a simplifying correlation model, something like the gaussian factor copulas typically used in CDO pricing models. It's a great way to smoothly fit unconditional probs plus low dimensional information on marginals. And it does place sensible constraints on the structure of the marginals. In our case, think of three unit normal variables each with a threshold corresponding to the unconditional prob of each of E, I, and M. Each of these normals, in turn, is composed of a normalized sum of one idiosyncratic normal and one or more common normal variables. The coefficients of the common factors determine the marginal probabilities. It sounds bad, but it's pretty straightforward, and imposes sensible looking results (unless you use it for actual CDOs). Might work here.

Might it work here? Let's back up a step.

A gaussian copula does indeed impose a strong (yet often plausible) structure on the resulting joint distribution. That is clearly seen in the case of CDOs by examining the amount of information involved. A basket of N names has {sum(k=0...N) N choose k} = 2^N distinct default events. The gaussian copula can be described with the N marginal inverse distribution functions plus N(N-1)/2 pairwise correlations; a much smaller number for large N.

A factor gaussian copula reduces the required information even more, by reducing the number of correlations to the number of factors. A 1-factor copula, for instance, would reduce the correlations needed here from 3 to 1. If we had all of the marginal distributions, we would be left with one free parameter(the correlation) to manipulate; we could infer the value of this parameter by iterative search with P(EIM) the objective function, just like a CDO. Then we could calculate P(IM) or P(EM) and hence P(IM)/P(I) or P(EM)/P(E) since we know P(I) and P(E) by assumption.

But we don't in fact know P(I) or P(E), and furthermore a 1-factor copula is implausible; we might be willing to assume that rho(E,M) = rho(I,M), but surely not that these equal rho(E,I)? Thus I am not yet persuaded.

Seems like over 80% of your readers that responded to the poll are immoral and/or irrational. ;-)

Or they just believe in the democratic system as a way to voice opinion and express their proportional fraction of input into the country's leadership. Which doesn't rule out irrational, but we like it that way. :-)

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