How do you calculate the priceImpact on Uniswap v2 pools? I can't find documentation to help me. The price impact appears on the UI when you select two tokens you want to swap.
You can calculate the price impact using the constant product formula.
Imagine you have a pool with 2,000,000 USDC and 1,000 ETH. The price of 1 ETH is 2,000 USDC.
Using the constant product formula (x*y=k) where x is USDC and y is ETH, k is 2,000,000,000.
Now, I'm going to swap 10,000 USDC for ETH.
2,000,000 USDC + 10,000 USDC = 2,010,000 USDC
x * y = k => y = k/x = 2,000,000,000/2,010,000 = 995.02 ETH
We receive 1,000 ETH - 995.02 ETH = 4.98 ETH
Price Impact = 4.98 ETH / 995.02 ETH = 0.5%
This article covers price impact in more detail.
Related
I have looked at the questions already asked about how to calculate precision and recall. However, I am still confused. There are specific questions that I face during the interviews that I am not sure how to approach them.
Question-1:
A binary classifier correctly filters 90% of spam emails but misclassifies 5% of non-spam emails as spam. Give the classifier's:
- recall
- precision
- accuracy,
or explain why there's not enough information to know.
For the above example, my solution would be:
TP = 90%
FP = 5%
Recall = TP / TP + FN => cannot be calculated without knowing FN
Precision = TP / TP + FP => 0.90/0.90 + 0.05 = 0.94
Accuracy = TP + TN / FN + FP + TP + TN => cannot be calculated without knowing TN & FN
Question-2:
1 percent of Uber transactions are fraud
We have a model that:
When a transaction is fraud, 99% classifies as fraud
When a transaction is not fraud, 99% classifies not fraud
What is the precision and recall of this model?
For the above example, my solution would be:
Let's say we have 1,000 daily transactions
number of positive classes (number of frauds): 10
number of negative classes (number of non-frauds): 990
TP = 10 * 99% = 9.9
TN = 990 * 99% = 980.1
FN = 990 * 1% = 9.9
FP = 10 * 1% = 0.1
Then use these in the below formula.
recall = TP / (TP + FN)
precision = TP / (FP + TP)
Can you please help me clarify an approach for these kinds of questions?
Thank you all!
Suppose we have sinusoidal with frequency 100Hz and sampling frequency of 1000Hz. It means that our signal has 100 periods in a second and we are taking 1000 samples in a second. Therefore, in order to select a complete period I'll have to take fs/f=10 samples. Right?
What if the sampling period is not a multiple of the frequency of the signal (like 550Hz)? Do I have to find the minimum multiple M of f and fs, and than take M samples?
My goal is to select an integer number of periods in order to be able to replicate them without changes.
You have f periods a second, and fs samples a second.
If you take M samples, it would cover M/fs part of a second, or P = f * (M/fs) periods. You want this number to be integer.
So you need to take M = fs / gcd(f, fs) samples.
For your example P = 1000 / gcd(100, 1000) = 1000 / 100 = 10.
If you have 60 Hz frequency and 80 Hz sampling frequency, it gives P = 80 / gcd(60, 80) = 80 / 20 = 4 -- 4 samples will cover 4 * 1/80 = 1/20 part of a second, and that will be 3 periods.
If you have 113 Hz frequency and 512 Hz sampling frequency, you are out of luck, since gcd(113, 512) = 1 and you'll need 512 samples, covering the whole second and 113 periods.
In general, an arbitrary frequency will not have an integer number of periods. Irrational frequencies will never even repeat ever. So some means other than concatenation of buffers one period in length will be needed to synthesize exactly periodic waveforms of arbitrary frequencies. Approximation by interpolation for fractional phase offsets is one possibility.
I have found that during training my model vw shows very big (much more than my features count ) feature number count in it's log.
I have tried to reproduce it using some small example:
simple.test:
-1 | 1 2 3
1 | 3 4 5
then "vw simple.test" command says that it have used 8 features. +one feature is constant but what are the other ? And in my real exmaple difference between my features and features used in wv is abot x10 more.
....
Num weight bits = 18
learning rate = 0.5
initial_t = 0
power_t = 0.5
using no cache
Reading datafile = t
num sources = 1
average since example example current current current
loss last counter weight label predict features
finished run
number of examples = 2
weighted example sum = 2
weighted label sum = 3
average loss = 1.9179
best constant = 1.5
total feature number = 8 !!!!
total feature number displays a sum of feature counts from all observed examples. So it's 2*(3+1 constant)=8 in your case. The number of features in current example is displayed in current features column. Note that only 2^Nth example is printed on screen by default. In general observations can have unequal number of features.
I want to find the standard deviation:
Minimum = 5
Mean = 24
Maximum = 84
Overall score = 90
I just want to find out my grade by using the standard deviation
Thanks,
A standard deviation cannot in general be computed from just the min, max, and mean. This can be demonstrated with two sets of scores that have the same min, and max, and mean but different standard deviations:
1 2 4 5 : min=1 max=5 mean=3 stdev≈1.5811
1 3 3 5 : min=1 max=5 mean=3 stdev≈0.7071
Also, what does an 'overall score' of 90 mean if the maximum is 84?
I actually did a quick-and-dirty calculation of the type M Rad mentions. It involves assuming that the distribution is Gaussian or "normal." This does not apply to your situation but might help others asking the same question. (You can tell your distribution is not normal because the distance from mean to max and mean to min is not close). Even if it were normal, you would need something you don't mention: the number of samples (number of tests taken in your case).
Those readers who DO have a normal population can use the table below to give a rough estimate by dividing the difference of your measured minimum and your calculated mean by the expected value for your sample size. On average, it will be off by the given number of standard deviations. (I have no idea whether it is biased - change the code below and calculate the error without the abs to get a guess.)
Num Samples Expected distance Expected error
10 1.55 0.25
20 1.88 0.20
30 2.05 0.18
40 2.16 0.17
50 2.26 0.15
60 2.33 0.15
70 2.38 0.14
80 2.43 0.14
90 2.47 0.13
100 2.52 0.13
This experiment shows that the "rule of thumb" of dividing the range by 4 to get the standard deviation is in general incorrect -- even for normal populations. In my experiment it only holds for sample sizes between 20 and 40 (and then loosely). This rule may have been what the OP was thinking about.
You can modify the following python code to generate the table for different values (change max_sample_size) or more accuracy (change num_simulations) or get rid of the limitation to multiples of 10 (change the parameters to xrange in the for loop for idx)
#!/usr/bin/python
import random
# Return the distance of the minimum of samples from its mean
#
# Samples must have at least one entry
def min_dist_from_estd_mean(samples):
total = 0
sample_min = samples[0]
for sample in samples:
total += sample
sample_min = min(sample, sample_min)
estd_mean = total / len(samples)
return estd_mean - sample_min # Pos bec min cannot be greater than mean
num_simulations = 4095
max_sample_size = 100
# Calculate expected distances
sum_of_dists=[0]*(max_sample_size+1) # +1 so can index by sample size
for iternum in xrange(num_simulations):
samples=[random.normalvariate(0,1)]
while len(samples) <= max_sample_size:
sum_of_dists[len(samples)] += min_dist_from_estd_mean(samples)
samples.append(random.normalvariate(0,1))
expected_dist = [total/num_simulations for total in sum_of_dists]
# Calculate average error using that distance
sum_of_errors=[0]*len(sum_of_dists)
for iternum in xrange(num_simulations):
samples=[random.normalvariate(0,1)]
while len(samples) <= max_sample_size:
ave_dist = expected_dist[len(samples)]
if ave_dist > 0:
sum_of_errors[len(samples)] += \
abs(1 - (min_dist_from_estd_mean(samples)/ave_dist))
samples.append(random.normalvariate(0,1))
expected_error = [total/num_simulations for total in sum_of_errors]
cols=" {0:>15}{1:>20}{2:>20}"
print(cols.format("Num Samples","Expected distance","Expected error"))
cols=" {0:>15}{1:>20.2f}{2:>20.2f}"
for idx in xrange(10,len(expected_dist),10):
print(cols.format(idx, expected_dist[idx], expected_error[idx]))
Yo can obtain an estimate of the geometric mean, sometimes called the geometric mean of the extremes or GME, using the Min and the Max by calculating the GME= $\sqrt{ Min*Max }$. The SD can be then calculated using your arithmetic mean (AM) and the GME as:
SD= $$\frac{AM}{GME} * \sqrt{(AM)^2-(GME)^2 }$$
This approach works well for log-normal distributions or as long as the GME, GM or Median is smaller than the AM.
In principle you can make an estimate of standard deviation from the mean/min/max and the number of elements in the sample. The min and max of a sample are, if you assume normality, random variables whose statistics follow from mean/stddev/number of samples. So given the latter, one can compute (after slogging through the math or running a bunch of monte carlo scripts) a confidence interval for the former (like it is 80% probable that the stddev is between 20 and 40 or something like that).
That said, it probably isn't worth doing except in extreme situations.
Query optimizers typically use summaries of data distributions to estimate the sizes of the intermediate tables generated during query processing. One popular such summarization scheme is a histogram, whereby the input range is partitioned into buckets and a cumulative count is maintained of the number of tuples falling in each bucket. The distribution within a bucket is assumed to be uniform for the purposes of estimation.
The following shows one such histogram for a relation R on a discrete attribute a with domain [1..10]:
Bucket 1: range = [1..2] Cumulative tuple count = 6
Bucket 2: range = [3..8] Cumulative tuple count = 30
Bucket 3: range = [9..10] Cumulative tuple count = 10
What is the estimated size of the self-join operation R x R
A) 46
B) 218
C) 248
D) 1,036
E) 5,672
Answer given in solutions : B
How is the answer to be calculated?
The size of a self-join on attribute R is equal to the summation of the frequency of each value of attribute R.
Here the frequency is given in buckets, e.g. the first bucket has 2 values r with frequency = 6, so we can assume the frequency of each value in bucket one is frequency = 3, similarly for bucket two frequency of each = 30/6 = 5, and for bucket three frequency of each value = 10/2 = 5.
Therefore, the size is
Size = [(3^2)*2] + [(5^2)*6] + [(5^2)*2]
= 218
I've been trying to figure this one out myself (it's from the GRE Computer Science subject test preparation exam).
So far I haven't found an answer as to why the answer is 218, but I have found a connection between the numbers given and the correct answer.
It turns out that that sum of the square of the cumulative tuple counts divided by the number of discrete values in each bucket, you get 218. Less abstractly: 6²/2 + 30²/5 + 10²/2 = 218.
It's not an answer, but at least there's a connection =)