🧖 🎮 😅 Pendekatan sederhana dan cepat untuk fungsi statistik 🚵🏾 📙 🦎

Tugas. Ada kalkulator , tetapi tidak ada tabel statistik di tangan . Misalnya, Anda memerlukan tabel titik kritis distribusi Student untuk menghitung interval kepercayaan. Dapatkan komputer dengan Excel? Tidak atletis.

Akurasi yang bagus tidak diperlukan, Anda dapat menggunakan rumus perkiraan. Ide dari rumus di bawah ini adalah dengan mengubah argumen, semua distribusi dapat direduksi menjadi normal. Perkiraan harus memberikan kalkulasi fungsi distribusi kumulatif dan kalkulasi fungsi kebalikannya.

Mari kita mulai dengan distribusi normal.

Φ (z) = P = \frac{1}{2} [1 + e r f (\frac{z}{\sqrt{2}})]

$\Phi(z)=P=\frac{1}{2}\left[1+\mathrm{erf}\left(\frac{z}{\sqrt{2}}\right)\right]$

z = Φ^{- 1} (P) = \sqrt{2} \cdot {e r f}^{- 1} (2 P - 1)

$z=\Phi^{-1}(P)=\sqrt{2}\cdot\mathrm{erf}^{-1}(2P-1)$

Ini membutuhkan penghitungan fungsi $\mathrm{erf}(x)$ dan kebalikannya. Saya menggunakan perkiraan [1]:

e r f (x) = s i g n (x) \cdot \sqrt{1 - \exp (- x^{2} \cdot \frac{\frac{4}{π} + a x^{2}}{1 + a x^{2}})}

$\mathrm{erf}(x)=\mathrm{sign}(x)\cdot\sqrt{1-\exp\left(-x^{2}\cdot\frac{\frac{4}{\pi}+ax^{2}}{1+ax^{2}}\right)}$

{e r f}^{- 1} (x) = s i g n (x) \cdot \sqrt{- t_{2} + \sqrt{t_{2}^{2} - \frac{1}{a} \cdot \ln t_{1}}}

$\mathrm{erf}^{-1}(x)=\mathrm{sign}(x)\cdot\sqrt{-t_2 + \sqrt{t_2^{2}-\frac{1}{a}\cdot \ln t_1}}$

Dimana $t_1$ dan $t_2$ - variabel pembantu:

t_{1} = 1 - x^{2}, t_{2} = \frac{2}{π a} + \frac{\ln t_{1}}{2}

$t_1=1-x^{2},\:t_2=\frac{2}{\pi a}+\frac{\ln t_1}{2}$

dan konstanta $a=0.147$ ... Di bawah ini adalah kode dalam bahasa Oktaf.

function y = erfa(x)
  a  = 0.147;
  x2 = x**2; t = x2*(4/pi + a*x2)/(1 + a*x2);
  y  = sign(x)*sqrt(1 - exp(-t));
endfunction

function y = erfinva(x)
  a  = 0.147; 
  t1 = 1 - x**2; t2 = 2/pi/a + log(t1)/2;
  y  = sign(x)*sqrt(-t2 + sqrt(t2**2 - log(t1)/a));
endfunction

function y = normcdfa(x)
  y = 1/2*(1 + erfa(x/sqrt(2)));
endfunction

function y = norminva(x)
  y = sqrt(2)*erfinva(2*x - 1);
endfunction

Sekarang, ketika ada fungsi distribusi normal, kami memberikan argumen dan menghitung distribusi-t Student [2]:

F_{t} (x, n) = Φ (\sqrt{\frac{1}{t_{1}} \cdot \ln (1 + \frac{x^{2}}{n})})

$F_t(x,n)=\Phi\left(\sqrt{\frac{1}{t_1}\cdot\ln(1+\frac{x^{2}}{n})}\right)$

t = F_{t}^{- 1} (P, n) = \sqrt{n \cdot \exp (Φ^{- 1} (P)^{2} \cdot t_{1}) - n}

$t=F_t^{-1}(P,n)=\sqrt{n\cdot\exp\left(\Phi^{-1}(P)^{2}\cdot t_1\right)-n}$

dimana variabel pembantu $t_1$ ada

t_{1} = \frac{n - 1.5}{(n - 1)^{2}}

$t_1=\frac{n-1.5}{(n-1)^{2}}$

function y = tcdfa(x,n)
  t1 = (n - 1.5)/(n - 1)**2;
 y = normcdfa(sqrt(1/t1*log(1 + x**2/n)));
endfunction

function y = tinva(x,n)
  t1 = (n - 1.5)/(n - 1)**2;
  y  = sqrt(n*exp(t1*norminva(x)**2) - n);
endfunction

Ide menghitung distribusi kira-kira $\chi^{2}$ diwakili dengan jelas oleh rumus [3]:

σ^{2} = \frac{2}{9 n}, μ = 1 - σ^{2}

$\sigma^{2}=\frac{2}{9n},\:\mu=1-\sigma^{2}$

F_{χ^{2}} (x, n) = Φ (\frac{{(\frac{x}{n})}^{1 / 3} - μ}{σ})

$F_{\chi^{2}}(x,n)=\Phi\left(\frac{\left(\frac{x}{n}\right)^{1/3}-\mu}{\sigma}\right)$

χ^{2} = F_{χ^{2}}^{- 1} (P, n) = n \cdot {(Φ^{- 1} (P) \cdot σ + μ)}^{3}

$\chi^2=F_{\chi^2}^{-1}(P,n)=n\cdot\left(\Phi^{-1}(P)\cdot\sigma + \mu\right)^3$

function y = chi2cdfa(x,n)
  s2 = 2/9/n; mu = 1 - s2;
  y  = normcdfa(((x/n)**(1/3) - mu)/sqrt(s2));
endfunction

function y = chi2inva(x,n)
 s2 = 2/9/n; mu = 1 - s2;
  y = n*(norminva(x)*sqrt(s2) + mu)**3;
endfunction

Distribusi Fisher (untuk $n/k\geq3$ dan $n\geq3$ ) . $\chi^2$ [4], , .

σ^{2} = \frac{2}{9 n}, μ = 1 - σ^{2}

$\sigma^2=\frac{2}{9n},\:\mu=1-\sigma^2$

λ = \frac{2 n + k \cdot x / 3 + (k - 2)}{2 n + 4 k \cdot x / 3}

$\lambda=\frac{2n+k\cdot x/3+(k-2)}{2n+4k\cdot x/3}$

F_{f} (x; k, n) = Φ (\frac{{(λ \cdot x)}^{1 / 3} - μ}{σ})

$F_f(x;k,n)=\Phi\left(\frac{\left(\lambda\cdot x\right)^{1/3}-\mu}{\sigma}\right)$

, .

q = {(Φ^{- 1} (P) \cdot σ + μ)}^{3}

$q=\left(\Phi^{-1}(P)\cdot\sigma+\mu\right)^3$

b = 2 n + k - 2 - 4 / 3 \cdot k q

$b=2n+k-2-4/3\cdot kq$

D = b^{2} + 8 / 3 \cdot k n q

$D=b^2+8/3\cdot knq$

x = F_{f}^{- 1} (P; k, n) = \frac{- b + \sqrt{D}}{2 k / 3}

$x=F_f^{-1}(P;k,n)=\frac{-b+\sqrt{D}}{2k/3}$

function y = fcdfa(x,k,n)
  mu = 1-2/9/k; s = sqrt(2/9/k);
  lambda = (2*n + k*x/3 + k-2)/(2*n + 4*k*x/3);
  normcdfa(((lambda*x)**(1/3)-mu)/s)
endfunction

function y = finva(x,k,n)
  mu = 1-2/9/k; s = sqrt(2/9/k);
  q = (norminva(x)*s + mu)**3;
  b = 2*n + k-2 -4/3*k*q;
  d = b**2 + 8/3*k*n*q;
  y = (sqrt(d) - b)/(2*k/3);
endfunction

Sergei Winitzki. A handy approximation for the error function and its inverse. February 6, 2008.
Gleason J.R. A note on a proposed Student t approximation // Computational statistics & data analysis. – 2000. – Vol. 34. – №. 1. – Pp. 63-66.
Wilson E.B., Hilferty M.M. The distribution of chi-square // Proceedings of the National Academy of Sciences. – 1931. – Vol. 17. – №. 12. – Pp. 684-688.
Li B. and Martin E.B. An approximation to the F-distribution using the chi-square distribution. Computational statistics & data analysis. – 2002. Vol. 40. – №. 1. pp. 21-26.

Pendekatan sederhana dan cepat untuk fungsi statistik

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