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I’d say it’s less random than the headline makes it seem. It’s a random value with mean of 7500 and standard deviation of 3000 pulled from a normal distribution

I’d say it’s less random than the headline makes it seem. It’s a random value with mean of 7500 and standard deviation of 3000 pulled from a normal distribution. And, it’s multiplied by the daily volume. So, there may be some thought here. It’s adding about $7,500 per trading unit volume. Not saying it’s legal or anything.

searealist 3y ago
So basically, your claim is that a normal distribution is less random than a uniform distribution? Can’t say I agree.
- deleted 3y ago
  [deleted]
- stavros 3y ago
  As I understand it, the claim is that the volume being added is tiny compared to the real volume.
- sillysaurusx 3y ago
  Well, by definition it is. It’s why we have names like normal and uniform.
  - uxp8u61q 3y ago
    By definition it's not. It's just as random. These names just tell you a few key properties of the distribution.
    - nomel 3y ago
      > just as random Let’s do a lottery together. We each throw in some money. Since they’re both just as random, I’m sure you won’t mind if I use a normal distribution, and choose each of our positions on the number line. :) Your assertion is missing some needed qualifiers.
    - maxbond 3y ago
      You've probably just got different understandings of the term "random."
    - jefftk 3y ago
      Would you say that a function that always returned 0 or 1 would be more/less/equally random than a function that returned any int32 with equal probability? I would normally say the latter is more random, because you're getting more entropy per call. In the same sense, a function drawing from a normal distribution is less random than one drawing from a uniform distribution. (All of this depends on a computers view of the world, where values are in data types that have some number of bits. In a mathy view of the world where every value is effectively infinitely many bits you get just as much entropy sampling from a normal distribution as a uniform one.)
      - eterm 3y ago
        > a function drawing from a normal distribution is less random than one drawing from a uniform distribution That's some mathematical nonsense. A normal distribution is continuous from negative infinity to positive infinity, so even by your flawed judgement of what makes something "random", a sample from the normal distrubtion has far more possibilities. The standard normal distribution is almost the definition of random.
        jefftk 3y ago
        Again, it depends on whether you are looking at this from a pure math perspective or a "math as implemented on computers" perspective. I'm only claiming that this use of more/less random makes sense in a computational context. But in that context, there is more entropy in a float32 drawn from a uniform distribution than a float32 drawn from a normal distribution, no?
        jefftk 3y ago
        Actually, I'm not sure this is mostly about a mathematical vs computational perspective. Let's look at how entropy is defined. It's not just the number of possibilities, it's how likely they are. For example, consider two functions: A: 1% chance of 1, 1% chance of 2, 98% chance of 3 B: 50% chance of 1, 50% chance of 2 The entropy of a distribution, in bits, is: H = - sum over all possibilities of (p * log_2 p) So for A it is: H_A = -(2 * 0.01 * log_2(0.01) + 0.98 * log_2(0.98)) = 0.16 And for B it is: H_B = -(2 * 0.5 * log_2(0.5)) = 1 So you can see that A has lower entropy than B, because it's more concentrated. For continuous distributions this is defined with an integral instead of a sum, but it's the same idea: H = - integral from -inf to + inf of (p * log_2 p) Whether there's more entropy in a sample drawn from a uniform or normal distribution, then, depends on how wide your distributions are. Consider: C: uniform distribution from -1000 to +1000 D: normal distribution centered on 0 with standard deviation 10 E: uniform distribution from -1 to 1 We then have: H_C = - log_2(2000) / log_2(2) = 10.97 H_D = 1/2 (log_2(2pi * 10^2) + 1) = 5.15 H_E = 1 Note that continuous and discrete entropies mean different things, and their values can't be compared directly. That H_E and H_B are both 1 is not meaningful.
    - sillysaurusx 3y ago
      That’s a bit like saying that our bodies are random numbers because of quantum fluctuations. It may be true, but the distribution is the part that matters. It’s still ridiculous to use a normal distribution to calculate how much money should go into an insurance fund that needs to cover sudden, severe losses.
  - sampo 3y ago
    > Well, by definition it is. If given only mean and standard deviation, the normal distribution is the maximum entropy distribution i.e. maximally random distribution.
- laurent_du 3y ago
  If you order randomness by entropy, the uniform distribution is indeed, in some sense, "more random" than any other probability distribution with the same support.
- deleted 3y ago
  [deleted]
TedDoesntTalk 3y ago
> multiplied by the daily volume How does volume correlate to the value of the insurance fund? Isnt value equal to number of tokens * price per token?
- gruez 3y ago
  see: https://news.ycombinator.com/item?id=37805256 https://news.ycombinator.com/item?id=37805256 It basically takes the daily volume, multiplies it by 0.00075% (on average) and adds that to the insurance fund. FTX charges a % fee on every trade, so this is the equivalent of taking a fraction of their fee revenue and putting that into an insurance fund, which seems pretty reasonable.
  - jahewson 3y ago
    No it doesn’t add it to the insurance fund - there’s no transactions here, it just increments the fake insurance fund variable.
    - wolverine876 3y ago
      In another company, it wouldn't be impossible that some other function in some other code used that variable for an actual transaction into an actual fund.
      - maweaver 3y ago
        Except the value isn't saved anywhere, it's just added to the total. And running again would give a new value.
  - tomku 3y ago
    If you look closely, you may notice that the code that "adds to the insurance fund" doesn't _take_ that money from anywhere. It's just retrieving a number from the database, adding a value to it out of thin air and writing it back. This number was then presented to users as representing the balance of the insurance fund protecting their assets. There was an actual insurance fund and it had a tiny fraction (5% or less) of what this number in the database claimed. The random numbers generated by this code did not represent actual contributions to said insurance fund. They were not, in fact, taking a fraction of their fee revenue and putting it into an insurance fund. They were not doing anything even remotely reasonable, this was a flat-out lie.
protastus 3y ago
Also not saying it's illegal. The non-zero mean and relatively small standard deviation (0.4 sigma) make all the difference here. I'd say the ethics and legality are entirely determined by how this distribution was determined and if it's consistent with promises to customers.
out-of-ideas 3y ago
agreed- title vs what is written in twitter is a bit misleading than what the code actually does. at first glance it is far too easy to assume the random numbers will be around 7500 - but in reality folks may not understand stddev's, and that the random number produced in this method can still yield a negative (just with less likelihood)
ncallaway 3y ago
What on earth are you taking about? It’s literally a random number taken from normal distribution, used to calculate the figure. What _possible_ justification could there be for using that as an accounting practice for a fund other than fraud?

searealist 3y ago

So basically, your claim is that a normal distribution is less random than a uniform distribution? Can’t say I agree.

deleted 3y ago
[deleted]
stavros 3y ago
As I understand it, the claim is that the volume being added is tiny compared to the real volume.
sillysaurusx 3y ago
Well, by definition it is. It’s why we have names like normal and uniform.
- uxp8u61q 3y ago
  By definition it's not. It's just as random. These names just tell you a few key properties of the distribution.
  - nomel 3y ago
    > just as random Let’s do a lottery together. We each throw in some money. Since they’re both just as random, I’m sure you won’t mind if I use a normal distribution, and choose each of our positions on the number line. :) Your assertion is missing some needed qualifiers.
  - maxbond 3y ago
    You've probably just got different understandings of the term "random."
  - jefftk 3y ago
    Would you say that a function that always returned 0 or 1 would be more/less/equally random than a function that returned any int32 with equal probability? I would normally say the latter is more random, because you're getting more entropy per call. In the same sense, a function drawing from a normal distribution is less random than one drawing from a uniform distribution. (All of this depends on a computers view of the world, where values are in data types that have some number of bits. In a mathy view of the world where every value is effectively infinitely many bits you get just as much entropy sampling from a normal distribution as a uniform one.)
    - eterm 3y ago
      > a function drawing from a normal distribution is less random than one drawing from a uniform distribution That's some mathematical nonsense. A normal distribution is continuous from negative infinity to positive infinity, so even by your flawed judgement of what makes something "random", a sample from the normal distrubtion has far more possibilities. The standard normal distribution is almost the definition of random.
      - jefftk 3y ago
        Again, it depends on whether you are looking at this from a pure math perspective or a "math as implemented on computers" perspective. I'm only claiming that this use of more/less random makes sense in a computational context. But in that context, there is more entropy in a float32 drawn from a uniform distribution than a float32 drawn from a normal distribution, no?
        jefftk 3y ago
        Actually, I'm not sure this is mostly about a mathematical vs computational perspective. Let's look at how entropy is defined. It's not just the number of possibilities, it's how likely they are. For example, consider two functions: A: 1% chance of 1, 1% chance of 2, 98% chance of 3 B: 50% chance of 1, 50% chance of 2 The entropy of a distribution, in bits, is: H = - sum over all possibilities of (p * log_2 p) So for A it is: H_A = -(2 * 0.01 * log_2(0.01) + 0.98 * log_2(0.98)) = 0.16 And for B it is: H_B = -(2 * 0.5 * log_2(0.5)) = 1 So you can see that A has lower entropy than B, because it's more concentrated. For continuous distributions this is defined with an integral instead of a sum, but it's the same idea: H = - integral from -inf to + inf of (p * log_2 p) Whether there's more entropy in a sample drawn from a uniform or normal distribution, then, depends on how wide your distributions are. Consider: C: uniform distribution from -1000 to +1000 D: normal distribution centered on 0 with standard deviation 10 E: uniform distribution from -1 to 1 We then have: H_C = - log_2(2000) / log_2(2) = 10.97 H_D = 1/2 (log_2(2pi * 10^2) + 1) = 5.15 H_E = 1 Note that continuous and discrete entropies mean different things, and their values can't be compared directly. That H_E and H_B are both 1 is not meaningful.
  - sillysaurusx 3y ago
    That’s a bit like saying that our bodies are random numbers because of quantum fluctuations. It may be true, but the distribution is the part that matters. It’s still ridiculous to use a normal distribution to calculate how much money should go into an insurance fund that needs to cover sudden, severe losses.
- sampo 3y ago
  > Well, by definition it is. If given only mean and standard deviation, the normal distribution is the maximum entropy distribution i.e. maximally random distribution.
laurent_du 3y ago
If you order randomness by entropy, the uniform distribution is indeed, in some sense, "more random" than any other probability distribution with the same support.
deleted 3y ago
[deleted]

TedDoesntTalk 3y ago

> multiplied by the daily volume How does volume correlate to the value of the insurance fund? Isnt value equal to number of tokens * price per token?

gruez 3y ago
see: https://news.ycombinator.com/item?id=37805256 https://news.ycombinator.com/item?id=37805256 It basically takes the daily volume, multiplies it by 0.00075% (on average) and adds that to the insurance fund. FTX charges a % fee on every trade, so this is the equivalent of taking a fraction of their fee revenue and putting that into an insurance fund, which seems pretty reasonable.
- jahewson 3y ago
  No it doesn’t add it to the insurance fund - there’s no transactions here, it just increments the fake insurance fund variable.
  - wolverine876 3y ago
    In another company, it wouldn't be impossible that some other function in some other code used that variable for an actual transaction into an actual fund.
    - maweaver 3y ago
      Except the value isn't saved anywhere, it's just added to the total. And running again would give a new value.
- tomku 3y ago
  If you look closely, you may notice that the code that "adds to the insurance fund" doesn't _take_ that money from anywhere. It's just retrieving a number from the database, adding a value to it out of thin air and writing it back. This number was then presented to users as representing the balance of the insurance fund protecting their assets. There was an actual insurance fund and it had a tiny fraction (5% or less) of what this number in the database claimed. The random numbers generated by this code did not represent actual contributions to said insurance fund. They were not, in fact, taking a fraction of their fee revenue and putting it into an insurance fund. They were not doing anything even remotely reasonable, this was a flat-out lie.

protastus 3y ago

Also not saying it's illegal. The non-zero mean and relatively small standard deviation (0.4 sigma) make all the difference here. I'd say the ethics and legality are entirely determined by how this distribution was determined and if it's consistent with promises to customers.

out-of-ideas 3y ago

agreed- title vs what is written in twitter is a bit misleading than what the code actually does. at first glance it is far too easy to assume the random numbers will be around 7500 - but in reality folks may not understand stddev's, and that the random number produced in this method can still yield a negative (just with less likelihood)

ncallaway 3y ago

What on earth are you taking about? It’s literally a random number taken from normal distribution, used to calculate the figure. What _possible_ justification could there be for using that as an accounting practice for a fund other than fraud?