News In Delaware and it is funtime for elections

[Gokul43201]

I think Ivan was simply referring to Black's theorem.

With the degree of polarization in the electorate today, I wouldn't be surprised if the best 1-dim approximation of voter philosophy shows a bimodal distribution. I believe the Palin effect on the McCain campaign (lost more independents/moderates than gains in extreme right wing) is argument against this, but I also believe we are more polarized today than we were 2 yrs ago.

 

[Office_Shredder]

It doesn't matter what the distribution is. What matters is that anybody whose more extreme than you will still vote for you rather than the person whose on the other side of the spectrum. This doesn't take into account voter participation rates which is what decides elections as much as anything else it seems

 

[Gokul43201]

It doesn't matter what the distribution is. What matters is that anybody whose more extreme than you will still vote for you rather than the person whose on the other side of the spectrum. This doesn't take into account voter participation rates which is what decides elections as much as anything else it seems

And it is essentially via the voter turnout that I think the distribution matters. If you model voter turnout as suppressed in proportion to the politico-philosophical "difference" between the candidate and the voter, then one can easily find bimodal distributions where a more extreme candidate may fare better. See, for example, the second hypothetical distribution in the figure below (x-axis is political philosophy, with the left end being the liberal end; arrows represent positions of 3 hypothetical candidates: D=Dem, R=Rep, T=Tea):

On the other hand, unless there is an (IMO unphysical) inverse relationship between ideological compatibility and voter enthusiasm, there is no way that T fares better than R in distribution #1.

 

[CRGreathouse]

Of course in the example Gokul gives above, the key point is the difference between the primary and the final election, because they have different electorates. The tea party candidate may swing the median Republican without capturing the median voter.

On the other hand, unless there is an (IMO unphysical) inverse relationship between ideological compatibility and voter enthusiasm, there is no way that T fares better than R in distribution #1.

If you're saying that the distribution must be bimodal for the Tea Party candidate to defeat the Republican (assuming a single axis on which the Tea candidate is to the right of the Republican), I disagree.

 

[Office_Shredder]

And it is essentially via the voter turnout that I think the distribution matters. If you model voter turnout as suppressed in proportion to the politico-philosophical "difference" between the candidate and the voter, then one can easily find bimodal distributions where a more extreme candidate may fare better. See, for example, the second hypothetical distribution in the figure below (x-axis is political philosophy, with the left end being the liberal end; arrows represent positions of 3 hypothetical candidates: D=Dem, R=Rep, T=Tea):

On the other hand, unless there is an (IMO unphysical) inverse relationship between ideological compatibility and voter enthusiasm, there is no way that T fares better than R in distribution #1.

The theorem being discussed only compares two candidates in an election, and why being more extremist tends to be a disadvantage in general elections.

 

[Gokul43201]

If you're saying that the distribution must be bimodal for the Tea Party candidate to defeat the Republican (assuming a single axis on which the Tea candidate is to the right of the Republican), I disagree.

No, I'm not saying that. There are two significant differences between that statement and what I am saying:

1. I am talking not about the outcome of the primary, but rather, the possible outcomes in the general election. My contention is that R would fare better against D than T would.

2. Assuming the positions are fixed as shown, I assert that T can not fare better than R could have (assuming also that turnouts do not act counter-intuitively, as described previously) for any symmetric bell-shaped distribution (or more generally, for any symmetric distribution which increases monotonically from x=extreme liberal to x=mid-point moderate).

The reason I went with a bimodal distribution is that I think that may be a more likely reflection of reality than say, a unimodal distribution with mean to the right of the position of T. I think we will probably end up with a House that is no more than 52% R and a Senate that is no more than 50% R, so my 2 example distributions were chosen to be symmetric.

 

[Gokul43201]

The theorem being discussed only compares two candidates in an election, and why being more extremist tends to be a disadvantage in general elections.

The theorem also applies only to unimodal distributions.

 

[lisab]

No, I'm not saying that. There are two significant differences between that statement and what I am saying:

1. I am talking not about the outcome of the primary, but rather, the possible outcomes in the general election. My contention is that R would fare better against D than T would.

2. Assuming the positions are fixed as shown, I assert that T can not fare better than R (assuming also that turnouts do not act counter-intuitively, as described previously) for any symmetric bell-shaped distribution (or more generally, for any symmetric distribution which increases monotonically from x=extreme liberal to x=mid-point moderate).

The reason I went with a bimodal distribution is that I think that may be a more likely reflection of reality than say, a unimodal distribution with mean to the right of the position of T. I think we will probably end up with a House that is no more than 52% R and a Senate that is no more than 50% R, so my 2 example distributions were chosen to be symmetric.

I like your bimodal distribution, it seems to fit my view of how things are in the country now.

But the analysis you draw from it assumes that the left and right distributions will vote in equal proportions, and that's clearly not so. In 2008, the left was fired up; in 2010 the right is. So, while I still don't think a T can beat a D, it may be closer that we think.

What's going to save the Ds is the independents - I don't see them going for a T at all. If you were to superimpose them over your distribution I think they'd be a narrow normal distribution.

 

[Gokul43201]

But the analysis you draw from it assumes that the left and right distributions will vote in equal proportions, and that's clearly not so.

No, I do not assume this. In fact, this is exactly what I talk about in the part about voter turnout. The distributions in the figure are only meant to reflect political views and are not scaled by voter enthusiasm. My argument about how a T might have a better chance than a R is based on taking that raw distribution and scaling it by some modulation for voter turnout.

 

[Office_Shredder]

The theorem also applies only to unimodal distributions.

No, it doesn't. You're just interpreting too much from the picture. Given any distribution of voters along a single dimension, if voters vote for the person amongst two candidates who is closer to them on the axis, then the optimal strategy for each candidate is to take a position exactly in the median of the distribution. It doesn't matter what the distribution looks like

 

[CRGreathouse]

1. I am talking not about the outcome of the primary, but rather, the possible outcomes in the general election. My contention is that R would fare better against D than T would.

That's fine, and is the conclusion of Black's theorem under our working assumptions. But this has nothing to do with the shape of the distribution -- it requires only that T is to the right of R.

2. Assuming the positions are fixed as shown, I assert that T can not fare better than R could have (assuming also that turnouts do not act counter-intuitively, as described previously) for any symmetric bell-shaped distribution (or more generally, for any symmetric distribution which increases monotonically from x=extreme liberal to x=mid-point moderate).

You don't need all of those extra assumptions!

 

[CRGreathouse]

No, it doesn't. You're just interpreting too much from the picture. Given any distribution of voters along a single dimension, if voters vote for the person amongst two candidates who is closer to them on the axis, then the optimal strategy for each candidate is to take a position exactly in the median of the distribution. It doesn't matter what the distribution looks like

You beat me to it. Yes, this is quite correct. This of course leads to law[/url]: both parties move toward that median voter.

 

[Gokul43201]

No, it doesn't. You're just interpreting too much from the picture. Given any distribution of voters along a single dimension, if voters vote for the person amongst two candidates who is closer to them on the axis, then the optimal strategy for each candidate is to take a position exactly in the median of the distribution. It doesn't matter what the distribution looks like

Oops, that is correct. Sorry, I got a little off-track in my previous response.

The theorem, however, assumes voter turnout is flat across the spectrum. My argument is based on the rationale that it is not, and specifically on the expectation that your likelihood of showing up at a polling station increases with the proximity of a candidate's position to your own.

 

[CRGreathouse]

The theorem, however, assumes voter turnout is flat across the spectrum. My argument is based on the rationale that it is not, and specifically on the expectation that your likelihood of showing up at a polling station increases with the proximity of a candidate's position to your own.

Under those assumptions, though, and with a distribution like your second, T could fare better against D than R.

 

[Gokul43201]

That's fine, and is the conclusion of Black's theorem under our working assumptions. But this has nothing to do with the shape of the distribution -- it requires only that T is to the right of R.
You don't need all of those extra assumptions!

For some reason, it seems that I am conveying that my arguments are based on Black's Theorem, while in fact, they are not. My argument is based on the principle that underlies BT as well as a distribution that accounts for voter enthusiasm (assumed to be uniform in BT).

 

[Gokul43201]

Under those assumptions, though, and with a distribution like your second, T could fare better against D than R.

This is EXACTLY what I'm saying. Have I been that inarticulate?

 

[CRGreathouse]

This is EXACTLY what I'm saying. Have I been that inarticulate?

Well, either you're inarticulate or I'm unable to properly interpret others.

 

[mheslep]

Nice discussion, but makes an over generous assumption about the nature of the domain space (ordinal) in my view. I believe that many voters entering the process this year see the issue as entrenched, self serving, business as usual politicians versus those unconnected with traditional party machines and media glad handing. On such a 'political & medial machine' graph, D&R are closely grouped, while the T candidates are way out there, as in that space political 'right' and 'left' have little (no?) meaning.

Most importantly, the M voter is not between the D & R points in such a view, it's close to T. On many issues, I grant the D,R, and T distribution is as Gokul indicates, with T to the right of R on most issues and the median between D & R. So, if a T candidate makes the (faulty) assumption that the electorate is operating with an M between D & R calculus and moves toward D&R on the political issues, simultaneously she would seen as moving back to the traditional 'machines and media' D&R on my graph, moving away from M, when in reality the dominant M issue was all along close to T, then support will likely vanish.

(above somewhat rambling I know, but got to go)

 

[airborne18]

I would have thought your earlier attempts would have slowed down, if not retired, your predilection for predictions and mass characterizations.

Lol. Oh I will admit when I am wrong. Like I said, so much for my future as a commentator. I should really leave the house more.

I can't wait to see the state polling now that she has won. Most of the tracking polls were Castle/Coons. The state Democrat party does have an advantage in numbers, but have the moderates all moved into the "against status quo". I think she needs the republican base, plus the independents and some democrats to move to her side.

I ruled out any chance of her beating coons, but that was before yesterday. So she could have a chance. She apparently has money for the election.

I hope she doesn't win, but I can't argue with that result. What is key is that it was a record turnout, which I would have thought favored the party machine. ( I know I was WRONG ).

Shep, You can rub this in for a while and I will accept my punishment.

I do apologize for being so cynical about the whole thing. Apparenlty there are rainbows, kittens, and puppies.

 

[Office_Shredder]

Nice discussion, but makes an over generous assumption about the nature of the domain space (ordinal)

I should have known you were an anti-Choice loon

Spoiler

I'm talking about the axiom of choice of course

 

[CRGreathouse]

Nice discussion, but makes an over generous assumption about the nature of the domain space (ordinal) in my view.

Presumably it's multidimensional, in which case Black' theorem fails (and Condorcet's paradox rules). Heck, it's even known to fail for one and a half dimensions -- a left-right axis and a quality axis (on which all voters agree). But it's interesting to consider the one-dimensional case because in Duvergerian states like the US it's often a good first approximation, and because it's the only case where we really have good properties to analyze.

 

[airborne18]

As punishment, well my shame, is that I changed my avatar to an image with me and Mike Castle. The one that lost.

This actually was from the other day at a Veterans event. No I don't smile when I actually have to leave the house. Was not a good day really. CNN was at the event and none of the coverage hit the air.

My wife is happy because it would have been of me not finishing sentences ( real bad day )..

You people think I mindlessly ramble here.. geez you ain't seen nuttin.

 

[talk2glenn]

Nice discussion, but makes an over generous assumption about the nature of the domain space (ordinal) in my view. I believe that many voters entering the process this year see the issue as entrenched, self serving, business as usual politicians versus those unconnected with traditional party machines and media glad handing. On such a 'political & medial machine' graph, D&R are closely grouped, while the T candidates are way out there, as in that space political 'right' and 'left' have little (no?) meaning.

Most importantly, the M voter is not between the D & R points in such a view, it's close to T. On many issues, I grant the D,R, and T distribution is as Gokul indicates, with T to the right of R on most issues and the median between D & R. So, if a T candidate makes the (faulty) assumption that the electorate is operating with an M between D & R calculus and moves toward D&R on the political issues, simultaneously she would seen as moving back to the traditional 'machines and media' D&R on my graph, moving away from M, when in reality the dominant M issue was all along close to T, then support will likely vanish.

(above somewhat rambling I know, but got to go)

This.

Everyone else is assuming an electorate evenly divided between liberal and conservative, with the winning median left of R and right of D.

Clearly, this is not the case. The electorate is both anti-incumbent and more conservative than that of 2 years ago. The D is also assumed to have moved leftward relative to the R and the original popular median.

The most electable candidate, generally, is closer to T than R and especially D this year. Locally, there may be differentiation, however.

 

[Gokul43201]

Everyone else is assuming an electorate evenly divided between liberal and conservative, with the winning median left of R and right of D.

Everyone else is not assuming this.

I, however, did, in my example cases. And my justification for modeling the distribution as symmetric (besides for simplicity) is based on my expectation that Conservatives will not control more that 55% of either the Senate or the House after this election - I consider that pretty close to an even split.

And let's not forget that this is all gross simplification, if for no other reason than that it's a simple 1-d model.

The most electable candidate, generally, is closer to T than R and especially D this year.

That's a bold prediction! Let's check back in a couple months.

 

[BobG]

Nice discussion, but makes an over generous assumption about the nature of the domain space (ordinal) in my view. I believe that many voters entering the process this year see the issue as entrenched, self serving, business as usual politicians versus those unconnected with traditional party machines and media glad handing. On such a 'political & medial machine' graph, D&R are closely grouped, while the T candidates are way out there, as in that space political 'right' and 'left' have little (no?) meaning.

Presumably it's multidimensional, in which case Black' theorem fails (and Condorcet's paradox rules). Heck, it's even known to fail for one and a half dimensions -- a left-right axis and a quality axis (on which all voters agree). But it's interesting to consider the one-dimensional case because in Duvergerian states like the US it's often a good first approximation, and because it's the only case where we really have good properties to analyze.

mshelp's post has some truth to it, especially in a multi-dimensional case. In a way, this is happening 2 years too early for Republicans. The mood has some similarities to the mood that propelled Jimmy Carter to victory. One of the favorite cartoons of the time was of Jimmy Carter and his huge smile saying, "Heck, I've never even seen Washigton."

Just don't put too much emphasis on it. The reality is that winning an upset victory in the primaries in a Republican state/district is likely to result in a win in the general election. An upset victory in a Democratic state/district is likely to result in a loss (O'Donnell is going to lose - heck, Castle would have been an underdog; just an underdog with a chance).

The real test comes in the toss up states/districts. Tea-party candidates win a few of these and they have some credibility.

 

[Gokul43201]

Just don't put too much emphasis on it. The reality is that winning an upset victory in the primaries in a Republican state/district is likely to result in a win in the general election. An upset victory in a Democratic state/district is likely to result in a loss (O'Donnell is going to lose - heck, Castle would have been an underdog; just an underdog with a chance).

Rasmussen agrees. They just downgraded (upgraded?) Delaware from "leans Dem" to "solid Dem" after the Primary results were announced.

http://www.rasmussenreports.com/pub...ections/election_2010_senate_balance_of_power

 

[BobG]

Just looking at a few races with tea partiers:

Colorado: Ken Buck is a toss up against the Democratic incumbent
Nevada: Sharron Angle is a toss up against the Democratic incumbent
Florida: Marco Rubio has a double digit lead against former Republican Crist and a Dem (turns out Crist is getting more votes from Dems than Republicans)
Alaska: Joe Miller has turned a Republican lock into a 6 point lead that's closing.
Kentucky, Utah: Tea party candidates in a Rep state will win easily.
Delaware: O'Donnell turned the race from leans Dem to a lock for Dems.

I'd say Tea Party candidates are 3-2 as far as doing better or worse than establishment candidates. But the three positives will be wins, while the two negatives probably won't turn out different than the election would have with an establishment Republican.

I'm surprised by Florida. I would have bet Rubio and Crist would split the Republican vote and make Meeks the favorite, with Crist second, and Rubio third. It's turned out exactly opposite with Crist and Meeks splitting the Democratic vote - so far. Normally, once a third party candidate falls behind, the votes go back to the major party candidates (Rubio & Meeks). Could Florida go even more bizarre than they have so far and voters abandon the third place major party candidate instead? If Meeks were out of the race, Crist would be a near lock. I don't think the same could be said about Crist dropping out.

 

[airborne18]

The Republicans are not a driving force in Delaware. Out of the gate the Democrats have a major advantage. Believe it or not Biden is still has clout ( don't ask me I have no clue why ). If his son, who is the Attorney General would have ran, even castle would have had a hard time.

A republican primary in Delaware is probably not a good indicator of the electorae in the state. Though the turnout was outragous, and I totally pegged it wrong.

However, I talked to several campaigns today, and they all echoed the same thing. They didn't like how the 9/12 Patriots and O'Donnell attacked Castle. Dirty politics is new to Delaware.

What they did is attack Castle, and then when he defended himself they portrayed him as going negative. It was very effective.

The Republicans must peal off 15% away from any Democrat to win the office. The Tea Party might get some angry moderates, but they also split the republican party. The state party might give her money because they have to, but they are not backing her at all.

I am not making predicitions. But if you look at Bidens last race for Senate, against O'Donnell as a matter of fact, you will see how the vote usually breaks for Democrats.
It is not even a close race at this point. She will keep whatever she got in the primary, but she will be hard pressed to increase it much. There are probably some Independents and Moderate Democrats mad at O'bama.

Also Delaware has a large Gay population, which she will never get. And most of the voters are in Wilmington, a very liberal inner city.

Coons was county manager, not sure what that is, never heard of one. But he has a large following in Wilmington. That city alone is enough to pull the vote for coons.

The tracking polls show that O'Donnell is not clinching the Republicans, and after the way it went I doubt she will.

Delaware voters generally will vote for the best for the state, and will break party lines. So O'Donnell can get more support. But she has to just to get in striking distance.

I think that is why you hear that nationally the Tea Party has too many races to back everyone, so they probably willl not be putting much more into Delaware. Though her website is not even campainging anymore, she is just asking for money.

 

[edward]

The Tea Party candidates got a lot of out of state support here in AZ. It came in the form of TV ads. Jesse Kelly tea party candidate and college drop out beat Jonathan Paton a former military attorney.

My republican relatives are still in shock. Will the real republican party please stand up??

 

[airborne18]

The Tea Party candidates got a lot of out of state support here in AZ. It came in the form of TV ads. Jesse Kelly tea party candidate and college drop out beat Jonathan Paton a former military attorney.

My republican relatives are still in shock. Will the real republican party please stand up??

The ironic part is that they only care about themselves, and through that they will end up handing Obama reelection. Just like when Perot handed the election to Clinton. In Delaware they handed the Senate back to the Democrats. It was a seat that was going to change party to the Republicans.

In Delaware it was not even about issues, in their agenda, they only had Castle on Cap and Trade. That was it. The 9/12 Patriots in DE are anti-abortion, so they had that on him.

O'Donnell is the same thing, mostly out of state money. I can't wait to see her next FEC report to see where it came from