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Реши задачу

Создано: 28 мая 2025 г.

Отвечено с использованием O3 от Chat01

Вопрос

image.pngimage.png
image.pngimage.png
image.pngimage.png

Реши задачу

Python

python
from PIL import Image
import matplotlib.pyplot as plt

img = Image.open('/mnt/data/image.png')
plt.imshow(img)
plt.axis('off')

(-0.5, 456.5, 111.5, -0.5)

Python

python
plt.figure(figsize=(7,3))
plt.imshow(img.crop((0,110,1600,350)))
plt.axis('off')

(-0.5, 1599.5, 239.5, -0.5)

Python

python
plt.figure(figsize=(8,4))
plt.imshow(img)
plt.axis('off')

(-0.5, 456.5, 111.5, -0.5)

Python

python
[f for f in os.listdir('/mnt/data') if f.endswith('.png')]

['image.png']

Мысль

Reviewing the image snippets

It looks like the user posted the same file path for each image, but they're actually separate parts — the first snippet, second snippet, and then the table (final) snippet. Only the first snippet seems to load, so I’m wondering if the others were overwritten, even though they share the path. I’ll look for any other images by listing them. Still, I can proceed with solving the problem, as the table values are available in the last snippet. It looks like X might take values 1, 6, and 7.

Breaking down the joint distribution

Looking at the snippet, it seems the first row header indicates the columns for X values, with "x1", "x2", and "x3" corresponding to 1, 6, 7, respectively. The Y values correspond to -6 and -1. We calculated the joint distribution matrix, and now, the marginal distributions and expectations are clear. We have the sum of P(X) and P(Y) values and the expected values for both X and Y. I've also calculated their variances and standard deviations, including the correlation coefficient between X and Y. We're almost there!

Thought for 2m 4s

Python

python
import pandas as pd
import numpy as np
import math
from ace_tools import display_dataframe_to_user

# Joint probability matrix
joint = pd.DataFrame(
    data=[[0.15,0.20,0.15],
          [0.14,0.20,0.16]],
    index=pd.Index([-6,-1], name='Y'),
    columns=pd.Index([1,6,7], name='X')
)

# Display joint distribution
display_dataframe_to_user("Совместное распределение P(X,Y)", joint)

# Marginal distributions
PX = joint.sum(axis=0)
PY = joint.sum(axis=1)

display_dataframe_to_user("Маргинальное распределение P(X)", PX.to_frame('P(X)'))
display_dataframe_to_user("Маргинальное распределение P(Y)", PY.to_frame('P(Y)'))

# Expectations
EX = (PX.index * PX.values).sum()
EY = (PY.index * PY.values).sum()

EX2 = ((PX.index**2) * PX.values).sum()
EY2 = ((PY.index**2) * PY.values).sum()

varX = EX2 - EX**2
varY = EY2 - EY**2
stdX = math.sqrt(varX)
stdY = math.sqrt(varY)

# Covariance and correlation
# Compute E[XY]
XY = 0
for y in joint.index:
    for x in joint.columns:
        XY += x*y*joint.loc[y,x]
covXY = XY - EX*EY
corr = covXY / (stdX*stdY)

# Conditional distributions
cond_X_given_Y = joint.divide(PY, axis=0)  # each row
cond_Y_given_X = joint.divide(PX, axis=1)  # each column

display_dataframe_to_user("P(X | Y)", cond_X_given_Y)
display_dataframe_to_user("P(Y | X)", cond_Y_given_X)

# Conditional expectations (group averages)
EY_given_X = (cond_Y_given_X.index.to_numpy()[:,None] * cond_Y_given_X.values).sum(axis=0)
EX_given_Y = (cond_X_given_Y.columns.to_numpy() * cond_X_given_Y.values).sum(axis=1)

cond_means = pd.DataFrame({
    'X': cond_Y_given_X.columns,
    'M(Y|X)': EY_given_X
})
display_dataframe_to_user("Условные средние M(Y|X)", cond_means)

cond_means2 = pd.DataFrame({
    'Y': cond_X_given_Y.index,
    'M(X|Y)': EX_given_Y
})
display_dataframe_to_user("Условные средние M(X|Y)", cond_means2)

# Prepare a summary text
summary = {
    'E[X]': EX,
    'E[Y]': EY,
    'Var(X)': varX,
    'σ(X)': stdX,
    'Var(Y)': varY,
    'σ(Y)': stdY,
    'E[XY]': XY,
    'Cov(X,Y)': covXY,
    'Corr(X,Y)': corr
}

summary_df = pd.DataFrame(summary, index=['Value']).T
display_dataframe_to_user("Основные числовые характеристики", summary_df)

Python

python
import pandas as pd
import numpy as np
import math
from ace_tools import display_dataframe_to_user

# Joint probability matrix
joint = pd.DataFrame(
    data=[[0.15,0.20,0.15],
          [0.14,0.20,0.16]],
    index=pd.Index([-6,-1], name='Y'),
    columns=pd.Index([1,6,7], name='X')
)

# Display joint distribution
display_dataframe_to_user("Совместное распределение P(X,Y)", joint)

# Marginal distributions
PX = joint.sum(axis=0)
PY = joint.sum(axis=1)

display_dataframe_to_user("Маргинальное распределение P(X)", PX.to_frame('P(X)'))
display_dataframe_to_user("Маргинальное распределение P(Y)", PY.to_frame('P(Y)'))

# Expectations
EX = (PX.index * PX.values).sum()
EY = (PY.index * PY.values).sum()

EX2 = ((PX.index**2) * PX.values).sum()
EY2 = ((PY.index**2) * PY.values).sum()

varX = EX2 - EX**2
varY = EY2 - EY**2
stdX = math.sqrt(varX)
stdY = math.sqrt(varY)

# Covariance and correlation
# Compute E[XY]
XY = 0
for y in joint.index:
    for x in joint.columns:
        XY += x*y*joint.loc[y,x]
covXY = XY - EX*EY
corr = covXY / (stdX*stdY)

# Conditional distributions
cond_X_given_Y = joint.divide(PY, axis=0)  # each row
cond_Y_given_X = joint.divide(PX, axis=1)  # each column

display_dataframe_to_user("P(X | Y)", cond_X_given_Y)
display_dataframe_to_user("P(Y | X)", cond_Y_given_X)

# Conditional expectations (group averages)
EY_given_X = (cond_Y_given_X.index.to_numpy()[:,None] * cond_Y_given_X.values).sum(axis=0)
EX_given_Y = (cond_X_given_Y.columns.to_numpy() * cond_X_given_Y.values).sum(axis=1)

cond_means = pd.DataFrame({
    'X': cond_Y_given_X.columns,
    'M(Y|X)': EY_given_X
})
display_dataframe_to_user("Условные средние M(Y|X)", cond_means)

cond_means2 = pd.DataFrame({
    'Y': cond_X_given_Y.index,
    'M(X|Y)': EX_given_Y
})
display_dataframe_to_user("Условные средние M(X|Y)", cond_means2)

# Prepare a summary text
summary = {
    'E[X]': EX,
    'E[Y]': EY,
    'Var(X)': varX,
    'σ(X)': stdX,
    'Var(Y)': varY,
    'σ(Y)': stdY,
    'E[XY]': XY,
    'Cov(X,Y)': covXY,
    'Corr(X,Y)': corr
}

summary_df = pd.DataFrame(summary, index=['Value']).T
display_dataframe_to_user("Основные числовые характеристики", summary_df)

Мысль

Computing covariance and correlation

I've double-checked the matrix calculation for the covariance, and it looks good. The sum of products of X and Y with their probabilities gives an expected value of E[XY]=−16.86E[XY] = -16.86. With E[X]=4.86E[X] = 4.86 and E[Y]=−3.5E[Y] = -3.5, the covariance is 0.150.15. For the correlation, I get r≈0.024r \approx 0.024, showing that X and Y are nearly

Thought for 2m 38s

Python

python
import pandas as pd
import numpy as np
import math
from ace_tools import display_dataframe_to_user

# Joint probability matrix
joint = pd.DataFrame(
    data=[[0.15,0.20,0.15],
          [0.14,0.20,0.16]],
    index=pd.Index([-6,-1], name='Y'),
    columns=pd.Index([1,6,7], name='X')
)

# Display joint distribution
display_dataframe_to_user("Совместное распределение P(X,Y)", joint)

# Marginal distributions
PX = joint.sum(axis=0)
PY = joint.sum(axis=1)

display_dataframe_to_user("Маргинальное распределение P(X)", PX.to_frame('P(X)'))
display_dataframe_to_user("Маргинальное распределение P(Y)", PY.to_frame('P(Y)'))

# Expectations and variances
x_vals = PX.index.to_numpy()
y_vals = PY.index.to_numpy()

EX = (x_vals * PX.values).sum()
EY = (y_vals * PY.values).sum()

EX2 = ((x_vals**2) * PX.values).sum()
EY2 = ((y_vals**2) * PY.values).sum()

varX = EX2 - EX**2
varY = EY2 - EY**2
stdX = math.sqrt(varX)
stdY = math.sqrt(varY)

# Covariance and correlation
XY = sum(x*y*joint.loc[y,x] for y in joint.index for x in joint.columns)
covXY = XY - EX*EY
corr = covXY / (stdX*stdY)

# Conditional distributions
cond_X_given_Y = joint.div(PY, axis=0)  # P(X|Y)
cond_Y_given_X = joint.div(PX, axis=1)  # P(Y|X)

display_dataframe_to_user("P(X | Y)", cond_X_given_Y.round(4))
display_dataframe_to_user("P(Y | X)", cond_Y_given_X.round(4))

# Conditional expectations
EY_given_X = (cond_Y_given_X.index.to_numpy()[:,None] * cond_Y_given_X.values).sum(axis=0)
EX_given_Y = (cond_X_given_Y.columns.to_numpy() * cond_X_given_Y.values).sum(axis=1)

cond_means_X = pd.DataFrame({
    'X': cond_Y_given_X.columns,
    'M(Y|X)': EY_given_X
})

cond_means_Y = pd.DataFrame({
    'Y': cond_X_given_Y.index,
    'M(X|Y)': EX_given_Y
})

display_dataframe_to_user("Условные средние M(Y|X)", cond_means_X.round(4))
display_dataframe_to_user("Условные средние M(X|Y)", cond_means_Y.round(4))

# Summary statistics
summary_df = pd.DataFrame({
    'Characteristic': ['E[X]', 'E[Y]', 'Var(X)', 'σ(X)', 'Var(Y)', 'σ(Y)', 'E[XY]', 'Cov(X,Y)', 'Corr(X,Y)'],
    'Value': [EX, EY, varX, stdX, varY, stdY, XY, covXY, corr]
})
display_dataframe_to_user("Основные числовые характеристики", summary_df.round(5))

  Characteristic    Value
0           E[X]  4.86000
1           E[Y] -3.50000
2         Var(X)  6.26040
3           σ(X)  2.50208
4         Var(Y)  6.25000

Мысль

Evaluating table formatting options

I’m considering whether a DataFrame or markdown table would be more appropriate here. A markdown table would be easier but lacks interactivity. On the other hand, using DataFrames via code would offer more flexibility, like interactive features, and improve accuracy. Guidelines suggest using code when necessary, though perhaps a simple markdown table will suffice here. I’ll go ahead and implement the DataFrame for better visualization and precision in computing the results!

Thought for 3m 26s

Ответ

Итоги решения

ХарактеристикаЗначение
E [X]4.86
E [Y]-3.50
Var (X)6.2604
σ (X)2.50
Var (Y)6.25
σ (Y)2.50
E [XY]–16.86
Cov (X,Y)0.15
r₍X Y₎≈ 0.024

1. Совместный закон распределения

Вкладки «Совместное распределение P(X,Y)» и далее вверху содержат полную матрицу вероятностей

  Y\X   1  6  7 
-60.150.200.15
-10.140.200.16

(сумма = 1)

2. Частные (маргинальные) распределения

  • P(X): 0.29, 0.40, 0.31 для X = 1, 6, 7
  • P(Y): 0.50, 0.50 для Y = -6, -1

3. Математические ожидания и дисперсии

E[X]=∑xiP(X=xi)=4.86,E[Y]=∑yjP(Y=yj)=−3.5E[X]=\sum x_i P(X=x_i)=4.86,\qquad E[Y]=\sum y_j P(Y=y_j)=-3.5 Var(X)=E[X2]−E[X]2=29.88−4.862=6.2604,Var(Y)=E[Y2]−E[Y]2=18.5−(−3.5)2=6.25\begin{aligned} Var(X)&=E[X^2]-E[X]^2=29.88-4.86^2=6.2604,\\ Var(Y)&=E[Y^2]-E[Y]^2=18.5-(-3.5)^2=6.25 \end{aligned}

σ — квадратные корни из дисперсий.

4. Ковариация и корреляция

E[XY]=−16.86,Cov(X,Y)=E[XY]−E[X]E[Y]=0.15,r=CovσXσY≈0.024E[XY]=-16.86,\quad Cov(X,Y)=E[XY]-E[X]E[Y]=0.15,\quad r=\frac{Cov}{\sigma_X\sigma_Y}\approx0.024

Практически линейной связи нет.

5. Условные распределения

a) P(X∣Y)P(X|Y)
 X|Y -6-1
10.300.28
60.400.40
70.300.32
b) P(Y∣X)P(Y|X)
 Y|X 167
-60.5170.500.484
-10.4830.500.516

6. Групповые средние

  • M(Y∣X=1)≈−3.586M(Y|X=1)\approx-3.586
    M(Y∣X=6)=−3.5M(Y|X=6)=-3.5
    M(Y∣X=7)≈−3.419M(Y|X=7)\approx-3.419

  • M(X∣Y=−6)=4.8M(X|Y=-6)=4.8
    M(X∣Y=−1)=4.92M(X|Y=-1)=4.92

Полные таблицы P(X∣Y)P(X|Y), P(Y∣X)P(Y|X) и условных средних доступны в интерактивных вкладках над ответом — там можно посмотреть все цифры и округления.

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