fix math equations in lists

03. Probability/1. Probability Basics.md

@@ -84,24 +84,18 @@ Based on intuition, personal judgment, or experience.
 1. **Range**:
    Probability always lies between 0 and 1:  
 
-   $$
-   0 \leq P(A) \leq 1
-   $$
+   $$0 \leq P(A) \leq 1$$
 
 2. **Sum of Probabilities**:  
 
    The sum of probabilities of all possible outcomes equals 1:  
 
-   $$
-   \sum P(A_i) = 1
-   $$
+   $$\sum P(A_i) = 1$$
 
 3. **Complement Rule**:  
    The probability of an event not happening is:  
 
-   $$
-   P(\text{Not A}) = 1 - P(A)
-   $$
+   $$P(\text{Not A}) = 1 - P(A)$$
 
    **Example**: If $P(A) = 0.7$, then $P(\text{Not A}) = 0.3$.
 

04. Inferential Statistics/3. Hypothesis Testing.md

@@ -103,9 +103,7 @@ A company claims the average lifetime of its batteries is 300 hours. A random sa
 
 3. **Calculate the Test Statistic**:
 
-   $$
-   t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} = \frac{290 - 300}{\frac{20}{\sqrt{25}}} = \frac{-10}{4} = -2.5
-   $$
+   $$t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} = \frac{290 - 300}{\frac{20}{\sqrt{25}}} = \frac{-10}{4} = -2.5$$
 
 4. **Find the Critical Value or P-Value**:  
    From $t$-distribution tables, critical $t$-value at $\alpha = 0.05$ (two-tailed, $df = 24$) is approximately $\pm 2.064$.  

05. Regression Analysis/1. Simple Linear Regression.md

@@ -17,9 +17,7 @@
 
 4. **Equation of the Regression Line**:
 
-   $$
-   Y = \beta_0 + \beta_1X + \epsilon
-   $$
+   $$Y = \beta_0 + \beta_1X + \epsilon$$
 
    Where:
 
@@ -85,45 +83,33 @@ A researcher wants to study the relationship between study hours ($X$) and test
 
 1. **Equation of Regression Line**:
 
-   $$
-   Y = \beta_0 + \beta_1X
-   $$
+   $$Y = \beta_0 + \beta_1X$$
 
 2. **Calculate Coefficients**:
 
    Using the least squares method:
 
    - Slope ($\beta_1$):
 
-     $$
-     \beta_1 = \frac{\text{Cov}(X, Y)}{\text{Var}(X)}
-     $$
+     $$\beta_1 = \frac{\text{Cov}(X, Y)}{\text{Var}(X)}$$
 
    - Intercept ($\beta_0$):
 
-     $$
-     \beta_0 = \bar{Y} - \beta_1\bar{X}
-     $$
+     $$\beta_0 = \bar{Y} - \beta_1\bar{X}$$
 
    After calculations:
 
-   $$
-   \beta_1 = 3.5, \quad \beta_0 = 46
-   $$
+   $$\beta_1 = 3.5, \quad \beta_0 = 46$$
 
 3. **Regression Equation**:
 
-   $$
-   Y = 46 + 3.5X
-   $$
+   $$Y = 46 + 3.5X$$
 
 4. **Predicted Values**:
 
    For $X = 6$:
 
-   $$
-   Y = 46 + 3.5(6) = 67
-   $$
+   $$Y = 46 + 3.5(6) = 67$$
 
 ---
 

05. Regression Analysis/2. Multiple Regression.md

@@ -14,9 +14,7 @@
 
 3. **Regression Equation**:
 
-   $$
-   Y = \beta_0 + \beta_1X_1 + \beta_2X_2 + \dots + \beta_kX_k + \epsilon
-   $$
+   $$Y = \beta_0 + \beta_1X_1 + \beta_2X_2 + \dots + \beta_kX_k + \epsilon$$
 
    Where:
 
@@ -80,9 +78,7 @@ A company wants to predict employee salaries ($Y$) based on years of experience
 
 1. **Regression Equation**:  
 
-   $$
-   Y = \beta_0 + \beta_1X_1 + \beta_2X_2
-   $$
+   $$Y = \beta_0 + \beta_1X_1 + \beta_2X_2$$
 
 2. **Fit the Model**:
 
@@ -94,17 +90,13 @@ A company wants to predict employee salaries ($Y$) based on years of experience
 
 3. **Regression Equation**:
 
-   $$
-   Y = 40,000 + 5,000X_1 + 10,000X_2
-   $$
+   $$Y = 40,000 + 5,000X_1 + 10,000X_2$$
 
 4. **Prediction**:
 
    For an employee with 7 years of experience ($X_1 = 7$) and education level 2 ($X_2 = 2$):
 
-   $$
-   Y = 40,000 + 5,000(7) + 10,000(2) = 40,000 + 35,000 + 20,000 = 95,000
-   $$
+   $$Y = 40,000 + 5,000(7) + 10,000(2) = 40,000 + 35,000 + 20,000 = 95,000$$
 
 ---
 

05. Regression Analysis/3. Logistic Regression.md

@@ -17,16 +17,12 @@
 3. **Logistic Function (Sigmoid Function)**:  
    Converts the linear combination of inputs into probabilities between 0 and 1.
 
-   $$
-   P(Y=1 \mid X) = \frac{1}{1 + e^{-(\beta_0 + \beta_1X_1 + \beta_2X_2 + \dots + \beta_kX_k)}}
-   $$
+   $$P(Y=1 \mid X) = \frac{1}{1 + e^{-(\beta_0 + \beta_1X_1 + \beta_2X_2 + \dots + \beta_kX_k)}}$$
 
 4. **Log-Odds Transformation**:
    The logistic model estimates the log-odds of the probability:
 
-   $$
-   \text{Log-Odds} = \ln\left(\frac{P}{1-P}\right) = \beta_0 + \beta_1X_1 + \beta_2X_2 + \dots + \beta_kX_k
-   $$
+   $$\text{Log-Odds} = \ln\left(\frac{P}{1-P}\right) = \beta_0 + \beta_1X_1 + \beta_2X_2 + \dots + \beta_kX_k$$
 
 ---
 
@@ -97,34 +93,22 @@ A bank wants to predict whether a loan applicant will default ($Y = 1$) or not (
 
 1. **Logistic Model**:
 
-   $$
-   P(Y=1 \mid X_1, X_2) = \frac{1}{1 + e^{-(\beta_0 + \beta_1X_1 + \beta_2X_2)}}
-   $$
+   $$P(Y=1 \mid X_1, X_2) = \frac{1}{1 + e^{-(\beta_0 + \beta_1X_1 + \beta_2X_2)}}$$
 
 2. **Estimated Coefficients**:
 
-   $$
-   \beta_0 = -10
-   $$
+   $$\beta_0 = -10$$
 
-   $$
-   \beta_1 = 0.0001
-   $$
+   $$\beta_1 = 0.0001$$
 
-   $$
-   \beta_2 = 0.02
-   $$
+   $$\beta_2 = 0.02$$
 
 3. **Prediction for New Applicant**:
    Applicant with income $X_1 = 65,000$ and credit score $X_2 = 720$:  
 
-   $$
-   \text{Log-Odds} = -10 + 0.0001(65,000) + 0.02(720) = -10 + 6.5 + 14.4 = 10.9
-   $$
+   $$\text{Log-Odds} = -10 + 0.0001(65,000) + 0.02(720) = -10 + 6.5 + 14.4 = 10.9$$
 
-   $$
-   P(Y=1) = \frac{1}{1 + e^{-10.9}} \approx 0.999
-   $$
+   $$P(Y=1) = \frac{1}{1 + e^{-10.9}} \approx 0.999$$
 
    The applicant is very likely to default.
 
@@ -140,21 +124,15 @@ A bank wants to predict whether a loan applicant will default ($Y = 1$) or not (
 
 2. **Accuracy**:
 
-   $$
-   \text{Accuracy} = \frac{\text{TP} + \text{TN}}{\text{Total Observations}}
-   $$
+   $$\text{Accuracy} = \frac{\text{TP} + \text{TN}}{\text{Total Observations}}$$
 
 3. **Precision**:
 
-   $$
-   \text{Precision} = \frac{\text{TP}}{\text{TP} + \text{FP}}
-   $$
+   $$\text{Precision} = \frac{\text{TP}}{\text{TP} + \text{FP}}$$
 
 4. **Recall**:
 
-   $$
-   \text{Recall} = \frac{\text{TP}}{\text{TP} + \text{FN}}
-   $$
+   $$\text{Recall} = \frac{\text{TP}}{\text{TP} + \text{FN}}$$
 
 5. **AUC-ROC**:
    Measures the model's ability to distinguish between classes. A higher value indicates better performance.

06. Advanced Topics/1. ANOVA and MANOVA.md

@@ -24,9 +24,7 @@ ANOVA tests whether there are statistically significant differences between the
 4. **F-Test**:  
    The test statistic is the ratio of between-group variance to within-group variance.
 
-   $$
-   F = \frac{\text{Variance Between Groups}}{\text{Variance Within Groups}}
-   $$
+   $$F = \frac{\text{Variance Between Groups}}{\text{Variance Within Groups}}$$
 
 ---
 

06. Advanced Topics/2. Time Series Analysis.md

@@ -128,27 +128,19 @@ Forecast sales for Jan 2023.
 
 1. **Mean Absolute Error (MAE)**:
 
-   $$
-   MAE = \frac{1}{n} \sum_{i=1}^{n} |y_i - \hat{y}_i|
-   $$
+   $$MAE = \frac{1}{n} \sum_{i=1}^{n} |y_i - \hat{y}_i|$$
 
 2. **Mean Squared Error (MSE)**:
 
-   $$
-   MSE = \frac{1}{n} \sum_{i=1}^{n} (y_i - \hat{y}_i)^2
-   $$
+   $$MSE = \frac{1}{n} \sum_{i=1}^{n} (y_i - \hat{y}_i)^2$$
 
 3. **Root Mean Squared Error (RMSE)**:
 
-   $$
-   RMSE = \sqrt{\frac{1}{n} \sum_{i=1}^{n} (y_i - \hat{y}_i)^2}
-   $$
+   $$RMSE = \sqrt{\frac{1}{n} \sum_{i=1}^{n} (y_i - \hat{y}_i)^2}$$
 
 4. **Mean Absolute Percentage Error (MAPE)**:
 
-   $$
-   MAPE = \frac{1}{n} \sum_{i=1}^{n} \left|\frac{y_i - \hat{y}_i}{y_i}\right| \times 100
-   $$
+   $$MAPE = \frac{1}{n} \sum_{i=1}^{n} \left|\frac{y_i - \hat{y}_i}{y_i}\right| \times 100$$
 
 ---
 

06. Advanced Topics/3. Bayesian Statistics.md

@@ -18,9 +18,7 @@
 4. **Bayes’ Theorem**:  
    The mathematical formula for updating probabilities:
 
-   $$
-   P(A \mid B) = \frac{P(B \mid A) \cdot P(A)}{P(B)}
-   $$
+   $$P(A \mid B) = \frac{P(B \mid A) \cdot P(A)}{P(B)}$$
 
    Where:
 
@@ -72,19 +70,13 @@ The prevalence of the disease is 1% in the population. What is the probability t
 
 3. **Calculate Evidence ($P(B)$)**:
 
-   $$
-   P(B) = P(B \mid A) \cdot P(A) + P(B \mid \text{Not } A) \cdot P(\text{Not } A)
-   $$  
+   $$P(B) = P(B \mid A) \cdot P(A) + P(B \mid \text{Not } A) \cdot P(\text{Not } A)$$  
 
-   $$
-   P(B) = (0.95 \cdot 0.01) + (0.10 \cdot 0.99) = 0.0095 + 0.099 = 0.1085
-   $$
+   $$P(B) = (0.95 \cdot 0.01) + (0.10 \cdot 0.99) = 0.0095 + 0.099 = 0.1085$$
 
 4. **Apply Bayes’ Theorem**:  
 
-   $$
-   P(A \mid B) = \frac{P(B \mid A) \cdot P(A)}{P(B)} = \frac{0.95 \cdot 0.01}{0.1085} \approx 0.0876
-   $$
+   $$P(A \mid B) = \frac{P(B \mid A) \cdot P(A)}{P(B)} = \frac{0.95 \cdot 0.01}{0.1085} \approx 0.0876$$
 
 ### Interpretation: