Statistics/Chi-squared distribution
The probability density function (pdf) of the chi-squared (or χ2) distribution as used in statistics is
Statistics/Chi-squared distribution is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
- , where
Here, denotes the Gamma_function.
The use of the gamma function in the equation below reflects the chi-squared distribution's origin as a special case of the gamma distribution.
In probability theory and statistics, the gamma distribution is a two-parameter family of continuous probability distributions. The exponential distribution, Erlang distribution, and chi-square distribution are special cases of the gamma distribution.
The probability density function (pdf) of the gamma distribution is given by the formula
where Γ(k) is the Gamma_function, with shape parameter k and a scale parameter θ.
The cumulative probability distribution of the gamma distribution is the area under the curve of the distribution, which indicates the increasing probability of the x value of a single random point within the gamma distribution being less than or equal to the x value of the cumulative probability distribution. The gamma cumulative probability distribution function can be calculated as
where is the lower incomplete gamma function.
The lower incomplete gamma function can be calculated as
and so, for the chi-squared cumulative probability distribution and substituting chi-square k into s as k/2 and chi-squared x into x as x / 2,
Because series formulas may be subject to accumulated errors from rounding in the frequently used region where x and k are under 10 and near one another, you may instead use a mathematics function library, if available for your programming task, to calculate gamma and incomplete gamma.
- Task
- Calculate and show the values of the χ2(x; k) for k = 1 through 5 inclusive and x integer from 0 and through 10 inclusive.
- Create a function to calculate the cumulative probability function for the χ2 distribution. This will need to be reasonably accurate (at least 6 digit accuracy) for k = 3.
- Calculate and show the p values of statistical samples which result in a χ2(k = 3) value of 1, 2, 4, 8, 16, and 32. (Statistical p values can be calculated for the purpose of this task as approximately 1 - P(x), with P(x) the cumulative probability function at x for χ2.)
- The following is a chart for 4 airports:
| Airport | On Time | Delayed | Totals |
|---|---|---|---|
| Dallas/Fort Worth | 77 | 23 | 100 |
| Honolulu | 88 | 12 | 100 |
| LaGuardia | 79 | 21 | 100 |
| Orlando | 81 | 19 | 100 |
| All Totals | 325 | 75 | 400 |
| Expected per 100 | 81.25 | 18.75 | 100 |
- χ2 on a 2D table is calculated as the sum of the squares of the differences from expected divided by the expected numbers for each entry on the table. The k for the chi-squared distribution is to be calculated as df, the degrees of freedom for the table, which is a 2D parameter, (count of airports - 1) * (count of measures per airport - 1), which here is (4 - 1 )(2 - 1) = 3.
- Calculate the Chi-squared statistic for the above table and find its p value using your function for the cumulative probability for χ2 with k = 3 from the previous task.
- Stretch task
- Show how you could make a plot of the curves for the probability distribution function χ2(x; k) for k = 0, 1, 2, and 3.
Julia
""" Rosetta Code task rosettacode.org/wiki/Statistics/Chi-squared_distribution """
""" gamma function to 7 places """
function gamma(x)
p = [ 0.99999999999980993, 676.5203681218851, -1259.1392167224028,
771.32342877765313, -176.61502916214059, 12.507343278686905,
-0.13857109526572012, 9.9843695780195716e-6, 1.5056327351493116e-7 ]
if x < 0.5
return π / (sinpi(x) * gamma(1.0 - x))
else
x -= 1.0
t = p[1]
for i in 1:8
t += p[i+1] / (x + i)
end
end
w = x + 7.5
return sqrt(2.0 * π) * w^(x+0.5) * exp(-w) * t
end
""" Chi-squared function, the probability distribution function (pdf) for chi-squared """
function χ2(x, k)
return x > 0 ? x^(k/2 - 1) * exp(-x/2) / (2^(k/2) * gamma(k / 2)) : 0
end
""" lower incomplete gamma by series formula with gamma """
function gamma_cdf(k, x)
return x^k * exp(-x) * sum(x^m / gamma(k + m + 1) for m in 0:100)
end
""" Cumulative probability function (cdf) for chi-squared """
function cdf_χ2(x, k)
return x <= 0 || k <= 0 ? 0.0 : gamma_cdf(k / 2, x / 2)
end
""" Cumulative probability function (cdf) for chi-squared """
function cdf_χ2(x, k)
return x <= 0 || k <= 0 ? 0.0 : gamma_cdf(k / 2, x / 2)
end
println("x χ2 k = 1 k = 2 k = 3 k = 4 k = 5")
println("-"^93)
for x in 0:10
print(lpad(x, 2))
for k in 1:5
s = string(χ2(x, k))
print(lpad(s[1:min(end, 13)], 18), k % 5 == 0 ? "\n" : "")
end
end
println("\nχ2 x cdf for χ2 P value (df=3)\n", "-"^36)
for p in [1, 2, 4, 8, 16, 32]
cdf = round(cdf_χ2(p, 3), digits=10)
println(lpad(p, 2), " ", cdf, " ", round(1.0 - cdf, digits=10))
end
airportdata = [ 77 23 ;
88 12;
79 21;
81 19 ]
expected_data = [ 81.25 18.75 ;
81.25 18.75 ;
81.25 18.75 ;
81.25 18.75 ; ]
dtotal = sum((airportdata[i] - expected_data[i])^2/ expected_data[i] for i in 1:length(airportdata))
println("\nFor the airport data, diff total is $dtotal, χ2 is ", χ2(dtotal, 3), ", p value ", cdf_χ2(dtotal, 3))
using Plots
x = 0.0:0.01:10
y = [map(p -> χ2(p, k), x) for k in 0:3]
plot(x, y, yaxis=[-0.1, 0.5], labels=[0 1 2 3])
- Output:
Graph of Chi-Squared for k values 0 through 3
x χ2 k = 1 k = 2 k = 3 k = 4 k = 5 --------------------------------------------------------------------------------------------- 0 0 0 0 0 0 1 0.24197072451 0.30326532985 0.24197072451 0.15163266492 0.08065690817 2 0.10377687435 0.18393972058 0.20755374871 0.18393972058 0.13836916580 3 0.05139344326 0.11156508007 0.15418032980 0.16734762011 0.15418032980 4 0.02699548325 0.06766764161 0.10798193302 0.13533528323 0.14397591070 5 0.01464498256 0.04104249931 0.07322491280 0.10260624827 0.12204152134 6 0.00810869555 0.02489353418 0.04865217332 0.07468060255 0.09730434665 7 0.00455334292 0.01509869171 0.03187340045 0.05284542098 0.07437126772 8 0.00258337316 0.00915781944 0.02066698535 0.03663127777 0.05511196094 9 0.00147728280 0.00555449826 0.01329554523 0.02499524221 0.03988663570 10 0.00085003666 0.00336897349 0.00850036660 0.01684486749 0.02833455534 χ2 x cdf for χ2 P value (df=3) ------------------------------------ 1 0.1987480431 0.8012519569 2 0.4275932955 0.5724067045 4 0.7385358701 0.2614641299 8 0.9539882943 0.0460117057 16 0.9988660157 0.0011339843 32 0.9999994767 5.233e-7 For the airport data, diff total is 4.512820512820512, χ2 is 0.08875392598443503, p value 0.7888504263193064
Phix
with javascript_semantics constant p = { 0.99999999999980993, 676.5203681218851, -1259.1392167224028, 771.32342877765313, -176.61502916214059, 12.507343278686905, -0.13857109526572012, 9.9843695780195716e-6, 1.5056327351493116e-7 } function gamma(atom z) if z<0.5 then return PI / (sin(PI*z)*gamma(1-z)) end if z -= 1; atom x := p[1]; for i=1 to length(p)-1 do x += p[i+1]/(z+i) end for atom t = z + length(p) - 1.5; return sqrt(2*PI) * power(t,z+0.5) * exp(-t) * x end function function pdf(atom x, k) if (x <= 0) then return 0 end if // this should be in task description return exp(-x/2) * power(x,k/2-1) / (power(2,k/2) * gamma(k/2)) end function function cpdf(atom x, k) x = x / 2 // need to do this to agree with Wikipedia formula for k = 2 atom t = exp(-x) * power(x,k/2) * gamma(k/2) / gamma(k), s = 0, m = 0, tol = 1e-15 // say while true do atom term = power(x,m) / gamma(k/2 + m + 1) s = s + term if abs(term) < tol then exit end if m = m + 1 end while return t * s end function printf(1," Values of the ?2 probablility distribution function\n") printf(1," x/k 1 2 3 4 5\n") for x=0 to 10 do printf(1,"%2d ", x) for k=1 to 5 do printf(1,"%f ", pdf(x, k)) end for printf(1,"\n") end for printf(1,"\n------Checking cpdf formula works for k = 2-------\n") printf(1," Values of the ?2 cumulative probability distribution function for k = 2\n") printf(1," x\n") for x=0 to 10 do printf(1,"%2d %f\n", {x, cpdf(x, 2)}) end for printf(1,"\n Values of the ?2 cumulative pdf using simple formula for k = 2\n") printf(1," x\n") for x=0 to 10 do printf(1,"%2d %f\n", {x, 1 - exp(-x/2)}) end for
- Output:
Same output (so probably not very helpful)
Values of the ?2 probablility distribution function
x/k 1 2 3 4 5
0 0.000000 0.000000 0.000000 0.000000 0.000000
1 0.241971 0.303265 0.241971 0.151633 0.080657
2 0.103777 0.183940 0.207554 0.183940 0.138369
3 0.051393 0.111565 0.154180 0.167348 0.154180
4 0.026995 0.067668 0.107982 0.135335 0.143976
5 0.014645 0.041042 0.073225 0.102606 0.122042
6 0.008109 0.024894 0.048652 0.074681 0.097304
7 0.004553 0.015099 0.031873 0.052845 0.074371
8 0.002583 0.009158 0.020667 0.036631 0.055112
9 0.001477 0.005554 0.013296 0.024995 0.039887
10 0.000850 0.003369 0.008500 0.016845 0.028335
------Checking cpdf formula works for k = 2-------
Values of the ?2 cumulative probability distribution function for k = 2
x
0 0.000000
1 0.393469
2 0.632121
3 0.776870
4 0.864665
5 0.917915
6 0.950213
7 0.969803
8 0.981684
9 0.988891
10 0.993262
Values of the ?2 cumulative pdf using simple formula for k = 2
x
0 0.000000
1 0.393469
2 0.632121
3 0.776870
4 0.864665
5 0.917915
6 0.950213
7 0.969803
8 0.981684
9 0.988891
10 0.993262
Python
''' rosettacode.org/wiki/Statistics/Chi-squared_distribution#Python '''
from math import gamma, exp
from matplotlib.pyplot import plot, legend, ylim
def χ2(x, k):
''' Chi-squared function, the probability distribution function (pdf) for chi-squared '''
return x**(k/2 - 1) * exp(-x/2) / (2**(k/2) * gamma(k / 2)) if x > 0 and k > 0 else 0.0
def gamma_cdf(k, x):
''' lower incomplete gamma by series formula with gamma '''
return x**k * exp(-x) * sum(x**m / gamma(k + m + 1) for m in range(100))
def cdf_χ2(x, k):
''' Cumulative probability function (cdf) for chi-squared '''
return gamma_cdf(k / 2, x / 2) if x > 0 and k > 0 else 0.0
print('x χ2 k = 1 k = 2 k = 3 k = 4 k = 5')
print('-' * 93)
for x in range(11):
print(f'{x:2}', end='')
for k in range(1, 6):
print(f'{χ2(x, k):16.8}', end='\n' if k % 5 == 0 else '')
print('\nχ2 x P value (df=3)\n----------------------')
for p in [1, 2, 4, 8, 16, 32]:
print(f'{p:2}', ' ', 1.0 - cdf_χ2(p, 3))
AIRPORT_DATA = [[77, 23], [88, 12], [79, 21], [81, 19]]
EXPECTED = [[81.25, 18.75],
[81.25, 18.75],
[81.25, 18.75],
[81.25, 18.75]]
DTOTAL = sum((d[pos] - EXPECTED[i][pos])**2 / EXPECTED[i][pos]
for i, d in enumerate(AIRPORT_DATA) for pos in [0, 1])
print(
f'\nFor the airport data, diff total is {DTOTAL}, χ2 is {χ2(DTOTAL, 3)}, p value {cdf_χ2(DTOTAL, 3)}')
X = [x * 0.001 for x in range(10000)]
for k in range(5):
plot(X, [χ2(p, k) for p in X])
legend([0, 1, 2, 3, 4])
ylim(-0.02, 0.5)
- Output:
x χ2 k = 1 k = 2 k = 3 k = 4 k = 5 --------------------------------------------------------------------------------------------- 0 0.0 0.0 0.0 0.0 0.0 1 0.24197072 0.30326533 0.24197072 0.15163266 0.080656908 2 0.10377687 0.18393972 0.20755375 0.18393972 0.13836917 3 0.051393443 0.11156508 0.15418033 0.16734762 0.15418033 4 0.026995483 0.067667642 0.10798193 0.13533528 0.14397591 5 0.014644983 0.041042499 0.073224913 0.10260625 0.12204152 6 0.0081086956 0.024893534 0.048652173 0.074680603 0.097304347 7 0.0045533429 0.015098692 0.0318734 0.052845421 0.074371268 8 0.0025833732 0.0091578194 0.020666985 0.036631278 0.055111961 9 0.0014772828 0.0055544983 0.013295545 0.024995242 0.039886636 10 0.00085003666 0.0033689735 0.0085003666 0.016844867 0.028334555 χ2 x P value (df=3) ---------------------- 1 0.8012519569012009 2 0.5724067044708798 4 0.26146412994911117 8 0.04601170568923141 16 0.0011339842897852837 32 5.233466447984725e-07 For the airport data, diff total is 4.512820512820513, χ2 is 0.088753925984435, p value 0.7888504263193064
Wren
import "./math" for Math
import "./fmt" for Fmt
class Chi2 {
static pdf(x, k) {
if (x <= 0) return 0
return (-x/2).exp * x.pow(k/2-1) / (2.pow(k/2) * Math.gamma(k/2))
}
static cpdf(x, k) {
var t = (-x/2).exp * (x/2).pow(k/2)
var s = 0
var m = 0
var tol = 1e-15 // say
while (true) {
var term = (x/2).pow(m) / Math.gamma(k/2 + m + 1)
s = s + term
if (term.abs < tol) break
m = m + 1
}
return t * s
}
}
System.print(" Values of the χ2 probability distribution function")
System.print(" x/k 1 2 3 4 5")
for (x in 0..10) {
Fmt.write("$2d ", x)
for (k in 1..5) {
Fmt.write("$f ", Chi2.pdf(x, k))
}
System.print()
}
System.print("\n Values for χ2 with 3 degrees of freedom")
System.print("χ2 cum cpdf p-value")
for (x in [1, 2, 4, 8, 16, 32]) {
var cpdf = Chi2.cpdf(x, 3)
Fmt.print("$2d $f $f", x, cpdf, 1-cpdf)
}
var airport = [[77, 23], [88, 12], [79, 21], [81, 19]]
var expected = [81.25, 18.75]
var dsum = 0
for (i in 0...airport.count) {
for (j in 0...airport[0].count) {
dsum = dsum + (airport[i][j] - expected[j]).pow(2) / expected[j]
}
}
var dof = (airport.count - 1) / (airport[0].count - 1)
System.print("\nFor airport data table: ")
Fmt.print(" diff sum : $f", dsum)
Fmt.print(" d.o.f. : $d", dof)
Fmt.print(" χ2 value : $f", Chi2.pdf(dsum, 3))
Fmt.print(" p-value : $f", Chi2.cpdf(dsum, 3))- Output:
Values of the χ2 probability distribution function
x/k 1 2 3 4 5
0 0.000000 0.000000 0.000000 0.000000 0.000000
1 0.241971 0.303265 0.241971 0.151633 0.080657
2 0.103777 0.183940 0.207554 0.183940 0.138369
3 0.051393 0.111565 0.154180 0.167348 0.154180
4 0.026995 0.067668 0.107982 0.135335 0.143976
5 0.014645 0.041042 0.073225 0.102606 0.122042
6 0.008109 0.024894 0.048652 0.074681 0.097304
7 0.004553 0.015099 0.031873 0.052845 0.074371
8 0.002583 0.009158 0.020667 0.036631 0.055112
9 0.001477 0.005554 0.013296 0.024995 0.039887
10 0.000850 0.003369 0.008500 0.016845 0.028335
Values for χ2 with 3 degrees of freedom
χ2 cum cpdf p-value
1 0.198748 0.801252
2 0.427593 0.572407
4 0.738536 0.261464
8 0.953988 0.046012
16 0.998866 0.001134
32 0.999999 0.000001
For airport data table:
diff sum : 4.512821
d.o.f. : 3
χ2 value : 0.088754
p-value : 0.788850