9.6-dars: Kutilma va dispersiya

DISKRET:   E[X] = sum( x × P(X = x) )         — qiymatlar ehtimol bilan vaznlangan
UZLUKSIZ:  E[X] = integral( x × f(x) dx )

Zar:   E[X] = (1 + 2 + 3 + 4 + 5 + 6) × 1/6 = 3.5
Lotereya: bilet 10 000 so'm; 1/100 000 ehtimol bilan 500 mln
  E[yutuq] = 500 000 000 × 0.00001 = 5 000 so'm  →  E[foyda] = -5 000

Var(X) = E[(X - E[X])^2] = E[X^2] - (E[X])^2
SD(X)  = sqrt(Var(X))                         — X bilan bir xil birlikda

Zar: E[X^2] = (1 + 4 + 9 + 16 + 25 + 36) / 6 = 91/6
     Var = 91/6 - 3.5^2 = 35/12 ≈ 2.92,  SD ≈ 1.71

E[aX + b]  = a × E[X] + b
E[X + Y]   = E[X] + E[Y]         ← HAR DOIM (mustaqillik shart emas!)
E[X1 + ... + Xn] = E[X1] + ... + E[Xn]

Misol: 100 mijoz, har biri o'rtacha 45 000 so'm sarflaydi
  E[jami tushum] = 100 × 45 000 = 4.5 mln

Var(aX + b)   = a^2 × Var(X)          ← b (siljish) dispersiyani o'zgartirmaydi
Var(X + Y)    = Var(X) + Var(Y)       ← FAQAT mustaqil (korrelyatsiyasiz) bo'lsa
Var(X + Y)    = Var(X) + Var(Y) + 2 × Cov(X, Y)   ← umumiy holat

n ta mustaqil, bir xil X ning o'rtachasi:
  Var(o'rtacha) = Var(X) / n      →   SD(o'rtacha) = SD(X) / sqrt(n)

import numpy as np
from scipy import stats

# DISKRET TAQSIMOTDAN
x = np.array([1, 2, 3, 4, 5, 6]); p = np.full(6, 1 / 6)
E = (x * p).sum()                       # 3.5
Var = (x ** 2 * p).sum() - E ** 2       # 2.9167
SD = np.sqrt(Var)

# SCIPY
X = stats.poisson(20)
X.mean(), X.var(), X.std()
X.expect(lambda k: k ** 2)              # E[X^2]

# XOSSALAR
# E[aX + b] = a E[X] + b;  Var(aX + b) = a^2 Var(X)
# E[X + Y] = E[X] + E[Y]  (har doim)
# Var(X + Y) = Var(X) + Var(Y) + 2 Cov(X, Y)

# NAMUNA O'RTACHASI
se = s.std(ddof=1) / np.sqrt(len(s))    # standart xato
QOIDA: kutilma + xavf (SD) · chiziqlilik — mustaqillikka bog'liq emas · Var — kvadrat bilan

E[X] — vaznli o'rtacha (sum x × p); uzoq muddatli o'rtacha
Var(X) = E[X^2] - E[X]^2; SD — asl birlikda (xavf)
E[X + Y] = E[X] + E[Y] — har doim
Var(X + Y) = Var(X) + Var(Y) — faqat mustaqil
Var(aX + b) = a^2 Var(X)
SE = SD / sqrt(n) — o'rtachaning noaniqligi

"""Kutilma va dispersiya: formula, scipy, simulyatsiya (real numpy/scipy)."""

from fractions import Fraction

import numpy as np
from scipy import stats


def main() -> None:
    x = range(1, 7)
    p = Fraction(1, 6)

    print("=== 1. Formula (aniq kasr) ===")
    E = sum(k * p for k in x)
    E2 = sum(k * k * p for k in x)
    print(f"  E[X] = {E}, E[X^2] = {E2}, Var = {E2 - E ** 2}")

    print("\n=== 2. scipy (randint 1..6) ===")
    X = stats.randint(1, 7)
    print(f"  mean {X.mean()}, var {X.var():.4f}, std {X.std():.4f}")

    print("\n=== 3. Simulyatsiya — katta sonlar qonuni ===")
    rng = np.random.default_rng(0)
    for n in [10, 1_000, 1_000_000]:
        s = rng.integers(1, 7, n)
        print(f"  n={n:>9}: o'rtacha {s.mean():.3f}, dispersiya {s.var():.3f}")

    print("\n=== 4. E[X] bitta tajribada chiqmaydi ===")
    print("  zar hech qachon 3.5 ko'rsatmaydi")
    print("  ⭐ Kutilma — uzoq muddatli o'rtacha")


if __name__ == "__main__":
    main()

=== 1. Formula (aniq kasr) ===
  E[X] = 7/2, E[X^2] = 91/6, Var = 35/12

=== 2. scipy (randint 1..6) ===
  mean 3.5, var 2.9167, std 1.7078

=== 3. Simulyatsiya — katta sonlar qonuni ===
  n=       10: o'rtacha 2.800, dispersiya 2.960
  n=     1000: o'rtacha 3.602, dispersiya 2.882
  n=  1000000: o'rtacha 3.502, dispersiya 2.917

=== 4. E[X] bitta tajribada chiqmaydi ===
  zar hech qachon 3.5 ko'rsatmaydi
  ⭐ Kutilma — uzoq muddatli o'rtacha

"""Kutilgan qiymat qarori: kutilma + xavf (real numpy)."""

import numpy as np


def main() -> None:
    kampaniyalar = {
        "A (kafolatli)": ([50], [1.0]),
        "B (xavfli)":    ([100, -20], [0.6, 0.4]),
        "C (juda xavfli)": ([300, -30], [0.25, 0.75]),
    }

    print("=== 1. Kutilma va standart og'ish (mln so'm) ===")
    for nom, (x, p) in kampaniyalar.items():
        x, p = np.array(x), np.array(p)
        E = (x * p).sum()
        SD = np.sqrt((x ** 2 * p).sum() - E ** 2)
        zarar = p[x < 0].sum()
        print(f"  {nom:<16} E = {E:6.1f}, SD = {SD:6.1f}, P(zarar) = {zarar:.0%}")

    print("\n=== 2. 50 marta takrorlansa (B) ===")
    rng = np.random.default_rng(0)
    natija = rng.choice([100, -20], size=(10_000, 50), p=[0.6, 0.4]).sum(axis=1)
    print(f"  jami o'rtacha: {natija.mean():.0f} (50 × 52 = 2600)")
    print(f"  zarar bilan tugash: {(natija < 0).mean():.2%}")

    print("\n=== 3. Bir marta (B) ===")
    print("  40% ehtimol bilan -20 — bir martalik qarorda xavf muhim")
    print("  ⭐ Kutilma + xavf — to'liq rasm")


if __name__ == "__main__":
    main()

=== 1. Kutilma va standart og'ish (mln so'm) ===
  A (kafolatli)    E =   50.0, SD =    0.0, P(zarar) = 0%
  B (xavfli)       E =   52.0, SD =   58.8, P(zarar) = 40%
  C (juda xavfli)  E =   52.5, SD =  142.9, P(zarar) = 75%

=== 2. 50 marta takrorlansa (B) ===
  jami o'rtacha: 2600 (50 × 52 = 2600)
  zarar bilan tugash: 0.00%

=== 3. Bir marta (B) ===
  40% ehtimol bilan -20 — bir martalik qarorda xavf muhim
  ⭐ Kutilma + xavf — to'liq rasm

"""Kutilmaning chiziqliligi: indikatorlar bilan hisoblash (real numpy)."""

import numpy as np


def main() -> None:
    rng = np.random.default_rng(1)

    print("=== 1. 10 ta zar yig'indisi ===")
    s = rng.integers(1, 7, (200_000, 10)).sum(axis=1)
    print(f"  nazariya: 10 × 3.5 = 35, simulyatsiya: {s.mean():.2f}")

    print("\n=== 2. Indikator: sovg'a almashish (n = 20) ===")
    # har kishi tasodifiy sovg'a oladi; o'z sovg'asini olganlar soni?
    n = 20
    mos = np.array([(rng.permutation(n) == np.arange(n)).sum() for _ in range(50_000)])
    print(f"  nazariya: n × (1/n) = 1, simulyatsiya: {mos.mean():.3f}")

    print("\n=== 3. Bog'liq o'zgaruvchilarda ham ishlaydi ===")
    x = rng.normal(10, 2, 100_000)
    y = 3 * x + rng.normal(0, 1, 100_000)     # y x ga kuchli bog'liq
    print(f"  E[X] + E[Y] = {x.mean() + y.mean():.2f}, E[X + Y] = {(x + y).mean():.2f}")

    print("\n=== 4. Lekin E[XY] ≠ E[X]E[Y] (bog'liq) ===")
    print(f"  E[XY] = {(x * y).mean():.1f}, E[X]E[Y] = {x.mean() * y.mean():.1f}")
    print("  ⭐ E[X + Y] = E[X] + E[Y] — har doim")


if __name__ == "__main__":
    main()

=== 1. 10 ta zar yig'indisi ===
  nazariya: 10 × 3.5 = 35, simulyatsiya: 34.99

=== 2. Indikator: sovg'a almashish (n = 20) ===
  nazariya: n × (1/n) = 1, simulyatsiya: 0.999

=== 3. Bog'liq o'zgaruvchilarda ham ishlaydi ===
  E[X] + E[Y] = 40.07, E[X + Y] = 40.07

=== 4. Lekin E[XY] ≠ E[X]E[Y] (bog'liq) ===
  E[XY] = 313.0, E[X]E[Y] = 301.0
  ⭐ E[X + Y] = E[X] + E[Y] — har doim

"""Var(aX), mustaqil/bog'liq yig'indi, SE = SD/sqrt(n) (real numpy)."""

import numpy as np


def main() -> None:
    rng = np.random.default_rng(2)
    x = rng.normal(100, 10, 500_000)

    print("=== 1. Masshtab va siljish ===")
    print(f"  Var(X) = {x.var():.1f}")
    print(f"  Var(2X) = {(2 * x).var():.1f}  (4 marta)")
    print(f"  Var(X + 50) = {(x + 50).var():.1f}  (o'zgarmaydi)")

    print("\n=== 2. Yig'indi: mustaqil va bog'liq ===")
    y_must = rng.normal(100, 10, 500_000)
    y_bog = x + rng.normal(0, 3, 500_000)
    print(f"  mustaqil: Var(X + Y) = {(x + y_must).var():.0f}  (≈ 100 + 100)")
    print(f"  bog'liq:  Var(X + Y) = {(x + y_bog).var():.0f}  (2Cov qo'shiladi)")

    print("\n=== 3. Namuna o'rtachasining tarqoqligi ===")
    for n in [25, 100, 400]:
        ortachalar = rng.normal(100, 10, (20_000, n)).mean(axis=1)
        print(f"  n={n:>3}: SD(o'rtacha) = {ortachalar.std():.3f}, nazariya 10/sqrt(n) = {10 / np.sqrt(n):.3f}")
    print("  ⭐ 4 marta ko'p ma'lumot — 2 marta aniqroq")


if __name__ == "__main__":
    main()

=== 1. Masshtab va siljish ===
  Var(X) = 100.0
  Var(2X) = 399.8  (4 marta)
  Var(X + 50) = 100.0  (o'zgarmaydi)

=== 2. Yig'indi: mustaqil va bog'liq ===
  mustaqil: Var(X + Y) = 200  (≈ 100 + 100)
  bog'liq:  Var(X + Y) = 409  (2Cov qo'shiladi)

=== 3. Namuna o'rtachasining tarqoqligi ===
  n= 25: SD(o'rtacha) = 2.004, nazariya 10/sqrt(n) = 2.000
  n=100: SD(o'rtacha) = 0.995, nazariya 10/sqrt(n) = 1.000
  n=400: SD(o'rtacha) = 0.506, nazariya 10/sqrt(n) = 0.500
  ⭐ 4 marta ko'p ma'lumot — 2 marta aniqroq

tanlov = max(variantlar, key=kutilma)                              # ⚠️
# kutilma, SD va P(zarar) ni birga ko'rsating                      # ✅

var_dollar = var_som / 12_000                                      # ⚠️
var_dollar = var_som / 12_000 ** 2                                 # ✅

var_portfel = var_a + var_b                                        # ⚠️
var_portfel = var_a + var_b + 2 * np.cov(a, b)[0, 1]               # ✅

ortacha_tezlik = masofa / vaqtlar.mean()     # E[1/T] emas         # ⚠️
ortacha_tezlik = (masofa / vaqtlar).mean()                         # ✅

xato = s.std()               # o'rtacha bahosi uchun               # ⚠️
xato = s.std(ddof=1) / np.sqrt(len(s))                             # ✅

var = np.var(namuna)         # ddof=0 — kichik namunada siljigan   # ⚠️
var = np.var(namuna, ddof=1)                                       # ✅

e_xy = x.mean() * y.mean()   # bog'liq bo'lsa xato                 # ⚠️
e_xy = (x * y).mean()                                              # ✅

1.  # zar E[X]?
2.  # E[2X + 3] (E[X] = 5)?
3.  # Var(2X + 3) (Var(X) = 4)?
4.  # E[X + Y] mustaqillik talab qiladimi?
5.  # Var(X + Y) mustaqil (4 va 9)?
6.  # SD(X + Y) shu holda?
7.  # SE: SD = 20, n = 100?
8.  # n ni 4 marta oshirsak SE?
9.  # E[X^2] - E[X]^2 = ?
10. # lotereya kutilgan foydasi manfiy — nega sotib olishadi?
11. # 10 ta zar yig'indisi kutilmasi?
12. # Var(-X) = ?

1.  var_ming = var_som / 1000

2.  var_jami = a.var() + b.var()   # a, b korrelyatsiyali

3.  xato = s.std()   # o'rtacha uchun

4.  e_xy = x.mean() * y.mean()   # bog'liq

5.  tanlov = max(kutilma)   # bir martalik katta qaror