The problem with Black Lives Matter

[Manfred Belheim]

You're making a poorly supported claim to counter what you consider a poorly supported claim.

Right. Let me spell this out explicitly. A graph was posted that shows "Whites killed by blacks" and "Blacks killed by whites", showing that the numbers of the former have been consistently 2 to 3 times higher than the latter across a period of around 15 years. A factor of 2 or 3 difference when you have a sample of hundreds of individual measurements is not something you would expect to see from pure chance, so that's what makes it statistically significant. Whether or not these statistics are completely false is not relevant to what I'm saying. Whether or not they do not correctly convey "the bigger picture" is not relevant to what I'm saying. Why are these things not relevant? Because I am responding to two other comments specifically about this graph. Neither comment questioned the veracity of the statistics, they just made arguments which do not logically follow. To wit:

There are a lot more white people about, of course they get killed more often.

I'm no brain genius but aren't there like, a lot more white people in the United States. The fact that those lines are so close together doesn't say what I suspect you think it does.

My interpretation of these statements is that they are arguing from a purely numerical basis that we should expect to see a higher rate of whites killed by blacks because there are many more whites in the country ergo it makes sense that there would be more such victims.

Do you disagree with how I have interpreted these statements?

So my response, also from a purely numerical basis, is to say that if you actually work through the not particularly complicated maths, you will see that this is actually NOT what you should expect, even if when you first think about it it intuitively feels as though it should be.

So no, I'm not making a poorly supported claim to counter what I consider to be another poorly supported claim. I'm making an objectively correct claim to counter an objectively false claim. It's only poorly supported in the sense that I didn't demonstrate the maths, but I left that as an exercise for the reader because it's not that hard.

Of course it remains a possibility that I have misinterpreted both of those statements. If I have I invite you to correct me on what they actually meant.

Statistically significant how?

See above.

Using what methods?

Basic algebra.

Because I think you're making a false (and unevidenced) claim to statistical significance because as you admit you haven't looked at the wider context.

It's not false, it's objectively correct. I've not evidenced it because I didn't feel I needed to, it's simple to work out. The wider context is not relevant to anything I said so I have no reason not to admit I haven't looked at it.

Try google on the source included in the image. Its the top result.

No because it's not relevant and I'm not interested enough to do that.

 

[Senethro]

Thats a lot of words to fail to show your actual working. Numbers please. Statistically significant how? Actual numbers please. Gimme some goddamn p-values or something.

 

[Manfred Belheim]

Thats a lot of words to fail to show your actual working. Numbers please. Statistically significant how? Actual numbers please. Gimme some goddamn p-values or something.

It's got nothing to do with p-values. The reasoning of what you'd expect to see is basic algebra. The errors you'd expect to see in a sample of a certain number of events is Poissonian statistics, but even that part isn't really relevant because the quoted arguments were only about the expectation in the first place, not how closely the statistics agree with that expectation. Their expectations were wrong.

Seriously, just at least have a go at working it out yourself. It's not that hard, I presume you're out of high school, and I'm not your mother.

Edit: Also can I just point out that it's a bit silly to ask me for "numbers" when I've just said it was algebra.

 

[Senethro]

You're the one making the claim, show your working.

 

[Manfred Belheim]

You're the one making the claim, show your working.

I'm not making a claim about what deity I had over to dinner last night, I'm making a claim about how very simple maths works out if you bother to do it. The onus is actually on the people making the claim that "oh well there's more white people so there should be more white victims of interracial crime" to actually think about it and realise that that doesn't add up. If someone flips a coin 10 times in a row and gets heads, then they say that this means they're more likely to get tails the next time, and I point out that's not how probability works, would you say to me "you're the one making the claim, show your working"? I would hope not, because I'd hope that you'd realise that asking for proof of something like that would just make you look foolish as well. Saying that A x B gives the same number as B x A, regardless of how much bigger one is than the other, shouldn't require proof to anyone who isn't a child.

Plus we both know that if I did do your homework for you, you'd just move the goalposts anyway, and start talking about how maths doesn't matter when we're talking about people's lives, or that it's irrelevant to the bigger picture, or something like that. You don't actually want to be convinced, you're just here to score points and if it becomes apparent that there are no points to be scored on this tack you'll just switch to another one.

 

[Senethro]

Hey if you want to back down on your claim about statistical significance thats fine. All you've done is (analogy) say that 6 is three times greater than 2 while ignoring the sample size is 70. Is it any wonder I ask to see your working?

 

[Manfred Belheim]

Hey if you want to back down on your claim about statistical significance thats fine. All you've done is (analogy) say that 6 is three times greater than 2 while ignoring the sample size is 70. Is it any wonder I ask to see your working?

I'm not backing down, I'm just not allowing you to goad me into typing a load of stuff that we both know you'll just dismiss out of hand, because that's how you operate. I mean I already gave you a big clue on how to work it out yourself in my last post, but the fact that you're still talking about sample sizes and things suggests you still haven't even understood what my point was anyway.

But remember it's not just about you, other people can read the thread. I'd be likewise as happy to "back down" if you kept demanding to I justify my comment that the past history of a flipped coin doesn't influence its future history, safe in the knowledge that most other observers won't need that level of pandering

 

[Senethro]

In the future please consider everything a goad, and never respond to it.

 

[Arwon]

If there's six times as many white Americans than black Americans, but only three times as many white folk are killed by black folk than the reverse, then that suggests that any given black American is at last two times as likely to be killed by a white American than the reverse.

This gap strikes me as much lower than what you'd expect by pure random distribution. Much less what you'd expect given the differences in overall homicide rates by the black and white US populations.

 

[Manfred Belheim]

In the future please consider everything a goad, and never respond to it.

Takes two to tango Mr Octopus.

 

[Manfred Belheim]

If there's six times as many white Americans than black Americans, but only three times as many white folk are killed by black folk than the reverse, then that suggests that any given black American is at last two times as likely to be killed by a white American than the reverse.

This gap strikes me as much lower than what you'd expect by pure random distribution.

Okay well, you see more reasonable so...

Assume a total population of N. Assume the fraction of the population that Group A constitutes is X. Assume the fraction of the population that Group B constitutes is Y. Assume the fraction of the population that commits a murder in any given year is Z. Where X, Y and Z are (hopefully obviously) all between 0 and 1, and X + Y is hopefully equally obviously <= 1.

So the total number of murders committed by Group A in any year is N * X * Z. The total number of murders committed by Group B in any year is N * Y * Z.

The probability of a victim belonging to Group B, given random chance, is the same as the fraction that Group B makes up of the population. The same goes for Group A.

So the total number of members of Group B killed by members of Group A is N * X * Z * Y. The total number of members of Group A killed by members of Group B is N * Y * Z * X.

NXZY = NYZX, i.e. exactly the same number, regardless of whether X >> Y or vice versa.


You don't even need to bother about the total population or what fraction of people are going to commit a murder. You can say that the probability of any random murderer belonging to any group is proportional to the fraction they make up of the total population. The probability of any random victim belonging to any group is also proportional to the fraction they make up of the total population. The probability of any particular murder involving two people belonging to any two groups is therefore proportional to the product of those two fractions. The probability does not depend on which one is the murderer and which one is the victim. Therefore you'd expect to see exactly equal numbers of both cases, all other things being equal.

 

[metalhead]

The problem with your analysis is you assume an equal encounter rate across the entire population, i.e. that since 70% of Americans are white, that 70% of the people a hypothetical murderer sees each day are white, and therefore 70% of random murder victims are going to be white.

Aggregated across the entire population, this may hold true, but across subgroups, it likely does not. For large portions of the country, probably 100% or close to it of the people a given person will see are white. White murderers in those places will hardly ever have the opportunity to randomly kill a person of color. Conversely, during an average day, a black person living in a black neighborhood will still see a significant number of white people pretty much wherever he goes. We're all but unavoidable.

To actually figure a reasonable mathematical approximation of a given racial group's opportunity for interracial murder, you'd need to be able to figure an average encounter rate for one race with another. A simple proportion isn't sufficient because it doesn't capture the reality of day-to-day interactions between people of different races.

 

[Manfred Belheim]

The problem with your analysis is you assume an equal encounter rate across the entire population, i.e. that since 70% of Americans are white, that 70% of the people a hypothetical murderer sees each day are white, and therefore 70% of random murder victims are going to be white.

Aaand there's the goalpost moving. No. I'm just responding to the original two comments, which equally didn't take anything like that into account and seemed to be coming at it from a purely numerical basis in just the same way. As I've already said, if my interpretation of those comments was wrong I'm happy to be corrected. As one of those comments was yours, you'd be in a good position to do that I would imagine.

Edit: But also...

Aggregated across the entire population, this may hold true, but across subgroups, it likely does not. For large portions of the country, probably 100% or close to it of the people a given person will see are white. White murderers in those places will hardly ever have the opportunity to randomly kill a person of color. Conversely, during an average day, a black person living in a black neighborhood will still see a significant number of white people pretty much wherever he goes. We're all but unavoidable.

Go back up to my example and set X (white) = 0.999 and Y (black) = 0.0001. In other words, a person would have to meet, on average, 10,000 people in a day to meet just 1 black person. That more than matches what you just said. The maths still holds up. This is what I mean, you're making intuitive arguments about how you feel it should work, but if you go through the numbers you see how it actually works, and that it's actually completely symmetrical, no matter the difference in the size of the two groups. So I agree that there are lots of other factors involved (murder not generally being spontaneous and random for one thing), but the actual example you just chose is not one of those factors at all.

 

[Lexicus]

The problem with your analysis is you assume an equal encounter rate across the entire population, i.e. that since 70% of Americans are white, that 70% of the people a hypothetical murderer sees each day are white, and therefore 70% of random murder victims are going to be white.

Hmm, let's add the assumption that murderers choose their victims at random from the entire population, too.

 

[Manfred Belheim]

Hmm, let's add the assumption that murderers choose their victims at random from the entire population, too.

Hmm, let's.... ignore most of what's actually been said and just be snarky!! Yaaaaaaay!

You realise that was the implicit assumption in both the comments I was responding to right (and indeed Arwon's comment on this page)? That was not my assumption. My argument was with the flawed conclusions people were drawing from them.

Jesus H Bandicoot, how can you people be so clever when you choose to be, and yet so utterly, utterly <snip> when you just want to "win". It's like being on a board with a bunch of 13 year olds half the time.

Moderator Action: Warning included in post #104. ~ Arakhor
Please read the forum rules: http://forums.civfanatics.com/showthread.php?t=422889

 

[Arwon]

Okay well, you see more reasonable so...

Assume a total population of N. Assume the fraction of the population that Group A constitutes is X. Assume the fraction of the population that Group B constitutes is Y. Assume the fraction of the population that commits a murder in any given year is Z. Where X, Y and Z are (hopefully obviously) all between 0 and 1, and X + Y is hopefully equally obviously <= 1.

So the total number of murders committed by Group A in any year is N * X * Z. The total number of murders committed by Group B in any year is N * Y * Z.

The probability of a victim belonging to Group B, given random chance, is the same as the fraction that Group B makes up of the population. The same goes for Group A.

So the total number of members of Group B killed by members of Group A is N * X * Z * Y. The total number of members of Group A killed by members of Group B is N * Y * Z * X.

NXZY = NYZX, i.e. exactly the same number, regardless of whether X >> Y or vice versa.


You don't even need to bother about the total population or what fraction of people are going to commit a murder. You can say that the probability of any random murderer belonging to any group is proportional to the fraction they make up of the total population. The probability of any random victim belonging to any group is also proportional to the fraction they make up of the total population. The probability of any particular murder involving two people belonging to any two groups is therefore proportional to the product of those two fractions. The probability does not depend on which one is the murderer and which one is the victim. Therefore you'd expect to see exactly equal numbers of both cases, all other things being equal.

This seems to have very little to do with that chart containing a simple count of cross-racial homicides. That chart doesn't even contain a total count of all homicides in the US, we're not talking about all murders here. We're talking about the fairly small component which are cross-racial.

And what I said was about victims, not perpetrators. So Z doesn't seem to be important at any rate, even if it weren't unknowable from that chart (theoretically every killing counted on it could be done by two particularly prolific blokes).

Again, if there's six times as many white folk as black folk, but only twice as many white folk are being killed by black folk, then the rate at which black folk are being killed by white folk is over twice as high. CF:

 

[metalhead]

Aaand there's the goalpost moving. No. I'm just responding to the original two comments, which equally didn't take anything like that into account and seemed to be coming at it from a purely numerical basis in just the same way. As I've already said, if my interpretation of those comments was wrong I'm happy to be corrected. As one of those comments was yours, you'd be in a good position to do that I would imagine.

I didn't say anything other than "there are a lot more white people about, of course they get killed more often." It's not a goalpost move because that simple assumption still underlies my analysis. Don't make large logical leaps that turn out to be wrong and then accuse people of goalpost moving when they point out your chicanery.

Go back up to my example and set X (white) = 0.999 and Y (black) = 0.0001. In other words, a person would have to meet, on average, 10,000 people in a day to meet just 1 black person. That more than matches what you just said. The maths still holds up. This is what I mean, you're making intuitive arguments about how you feel it should work, but if you go through the numbers you see how it actually works, and that it's actually completely symmetrical, no matter the difference in the size of the two groups. So I agree that there are lots of other factors involved (murder not generally being spontaneous and random for one thing), but the actual example you just chose is not one of those factors at all.

No it doesn't. You're still assuming an encounter rate that is an exact match for each group's proportion of the population.

If black people are 10% of the population, but only 5% of the people an average white person sees each day are black, then the "random" murder rate - assuming the victim is truly selected at random - is only going to reflect the number of black people actually encountered, not their share of the population as a whole. Therefore, when accounting for random interracial murders, what matters is how frequently people encounter people of another race, not simply their proportion of the population as a whole.

 

[Lexicus]

You realise that was the implicit assumption in both the comments I was responding to right (and indeed Arwon's comment on this page)? That was not my assumption. My argument was with the flawed conclusions people were drawing from them.

Uh, I think they are in fact your assumptions. You're projecting them onto the things other people are saying because, as Arwon says, you either didn't see or were unable to comprehend the chart they were talking about.

 

[Manfred Belheim]

This seems to have very little to do with that chart containing a simple count of cross-racial homicides. That chart doesn't even contain a total count of all homicides in the US.

Well... what does that matter? Nothing I said depends on what the total number of homicides is at all. I even explicitly said that you don't need to even consider this in the second example I gave...

And what I said was about victims, not perpetrators. So Z doesn't seem to be important.

Well for every victim there's a perpetrator. Z determines both.

Perpetrators are of course also unknowable from that chart - theoretically every killing counted on it could be done by two particularly prolific blokes.

More goalpost moving. You were looking at it purely numerically. Now you're postulating two mass serial killers.

Again, if there's six times as many white folk as black folk, but only twice as many white folk are being killed by black folk, then the rate at which black folk are being killed by white folk is several times higher than the reverse.

That doesn't appear to make any sense. At the risk of setting off irony alarms... care to show your working there?

 

[Timsup2nothin]

It means letting bygones be bygones and treating people based on how they act, not on the color of their skin.

I'm more than willing to do this, as evidenced by the disdain with which I routinely respond to your posts.