📖 Two Types of Moments
1. Raw Moments (μᵣ'): Moments taken about the origin (zero). Also called crude moments.
2. Central Moments (μᵣ): Moments taken about the mean. Also called moments about the mean.
Unit III · Lecture 20 of 40 · B.Sc. Data Science, 1st Year, Semester II
In statistics, moments are a set of numerical measures that describe the shape of a probability distribution or a dataset. They are essentially averages of powers of deviations — and they generalise the concepts of mean, variance, skewness, and kurtosis into a single unified framework.
If the mean tells you the "location" of a distribution, and the variance tells you its "spread," then moments give you the complete "fingerprint" — they capture location, spread, symmetry, and tail behaviour all in one systematic framework. The first few moments are enough to characterise most practical distributions.
1. Raw Moments (μᵣ'): Moments taken about the origin (zero). Also called crude moments.
2. Central Moments (μᵣ): Moments taken about the mean. Also called moments about the mean.
The concept of moments was introduced by Karl Pearson in the late 19th century. The term "moment" is borrowed from physics, where the "moment of force" describes the turning effect of a force about a point. Similarly, statistical moments describe the "turning tendencies" of a distribution about a point.
The r-th raw moment (moment about the origin, A = 0) is denoted by μᵣ' (mu-r-prime) and defined as:
μᵣ' = (1/n) × Σ xᵢʳ (Ungrouped data)
μᵣ' = (1/N) × Σ fᵢ xᵢʳ (Grouped data)
Where r = 1, 2, 3, 4, … is the order of the moment.
μ₁' = (1/n) Σxᵢ = the mean (x̄)
μ₂' = (1/n) Σxᵢ² = mean of squared values
μ₃' = (1/n) Σxᵢ³ = mean of cubed values
μ₄' = (1/n) Σxᵢ⁴ = mean of fourth powers
Raw moments are easy to compute but hard to interpret directly (except μ₁' = mean). They are primarily used as intermediate steps to compute central moments, which have clear statistical interpretations.
Think of raw moments as ingredients — you need them to cook the dish (central moments), but you don't eat them raw.
The r-th central moment (moment about the mean) is denoted by μᵣ (without prime) and defined as:
μᵣ = (1/n) × Σ (xᵢ − x̄)ʳ (Ungrouped data)
μᵣ = (1/N) × Σ fᵢ (xᵢ − x̄)ʳ (Grouped data)
Where r = 1, 2, 3, 4, … is the order of the moment.
μ₁ = (1/n) Σ(xᵢ − x̄) = 0 (Always zero — deviations from mean sum to zero)
μ₂ = (1/n) Σ(xᵢ − x̄)² = σ² = Variance (Measures spread)
μ₃ = (1/n) Σ(xᵢ − x̄)³ (Measures skewness — asymmetry of distribution)
μ₄ = (1/n) Σ(xᵢ − x̄)⁴ (Measures kurtosis — peakedness / tail heaviness)
Each central moment captures a different aspect of the distribution's shape:
Raw moments are easier to compute (no need to calculate deviations from the mean), but central moments are easier to interpret. We need conversion formulas to get central moments from raw moments.
μ₁ = 0 (always, by definition)
μ₂ = μ₂' − (μ₁')²
= (mean of x²) − (mean of x)²
μ₃ = μ₃' − 3μ₂'μ₁' + 2(μ₁')³
μ₄ = μ₄' − 4μ₃'μ₁' + 6μ₂'(μ₁')² − 3(μ₁')⁴
Let a = μ₁' = x̄ (the mean). Then:
μ₂ = μ₂' − a²
μ₃ = μ₃' − 3aμ₂' + 2a³
μ₄ = μ₄' − 4aμ₃' + 6a²μ₂' − 3a⁴
These formulas let you compute central moments without ever calculating individual deviations — a huge time-saver for hand calculations.
μ₁' = x̄ (the mean)
μ₂' = μ₂ + (μ₁')² = μ₂ + x̄²
μ₃' = μ₃ + 3x̄μ₂ + x̄³
μ₄' = μ₄ + 4x̄μ₃ + 6x̄²μ₂ + x̄⁴
After computing central moments from raw moments (or vice versa), you can verify your work by checking:
μ₂ = μ₂' − (μ₁')² ≥ 0 (variance is always non-negative)
μ₁ = 0 (first central moment is always zero)
If either condition fails, there is a calculation error.
Find the first four raw moments and central moments for the data: 2, 4, 6, 8, 10
n = 5
Σxᵢ = 2+4+6+8+10 = 30
Σxᵢ² = 4+16+36+64+100 = 220
Σxᵢ³ = 8+64+216+512+1000 = 1800
Σxᵢ⁴ = 16+256+1296+4096+10000 = 15664
μ₁' = 30/5 = 6
μ₂' = 220/5 = 44
μ₃' = 1800/5 = 360
μ₄' = 15664/5 = 3132.8
μ₁ = 0 (always)
μ₂ = μ₂' − (μ₁')² = 44 − 36 = 8 (= variance = σ²)
μ₃ = μ₃' − 3μ₂'μ₁' + 2(μ₁')³ = 360 − 3(44)(6) + 2(216) = 360 − 792 + 432 = 0
μ₄ = μ₄' − 4μ₃'μ₁' + 6μ₂'(μ₁')² − 3(μ₁')⁴
= 3132.8 − 4(360)(6) + 6(44)(36) − 3(1296)
= 3132.8 − 8640 + 9504 − 3888 = 108.8
μ₁ = 0 | μ₂ = 8 | μ₃ = 0 | μ₄ = 108.8
μ₃ = 0 confirms the data is perfectly symmetric about the mean (as expected for 2, 4, 6, 8, 10).
Find the first four raw and central moments for the following data:
| xᵢ | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| fᵢ | 2 | 4 | 6 | 4 | 2 |
N = 2+4+6+4+2 = 18
Σfᵢxᵢ = 2+8+18+16+10 = 54 → μ₁' = 54/18 = 3
Σfᵢxᵢ² = 2+16+54+64+50 = 186 → μ₂' = 186/18 = 10.333
Σfᵢxᵢ³ = 2+32+162+256+250 = 702 → μ₃' = 702/18 = 39
Σfᵢxᵢ⁴ = 2+64+486+1024+1250 = 2826 → μ₄' = 2826/18 = 157
μ₁ = 0
μ₂ = 10.333 − 9 = 1.333 (Variance)
μ₃ = 39 − 3(10.333)(3) + 2(27) = 39 − 93 + 54 = 0 (Symmetric!)
μ₄ = 157 − 4(39)(3) + 6(10.333)(9) − 3(81) = 157 − 468 + 558 − 243 = 4
μ₂ = 1.333 | μ₃ = 0 | μ₄ = 4
σ = √1.333 = 1.155 | Data is symmetric (μ₃ = 0) — confirmed by the mirror-like frequencies (2, 4, 6, 4, 2).
Karl Pearson's coefficient of skewness:
β₁ = μ₃² / μ₂³
γ₁ = √β₁ = μ₃ / μ₂^(3/2) (signed skewness)
• γ₁ = 0 → Symmetric
• γ₁ > 0 → Right-skewed (positive skew)
• γ₁ < 0 → Left-skewed (negative skew)
Karl Pearson's coefficient of kurtosis:
β₂ = μ₄ / μ₂²
Excess kurtosis: γ₂ = β₂ − 3
• β₂ = 3 (γ₂ = 0) → Mesokurtic (normal-like)
• β₂ > 3 (γ₂ > 0) → Leptokurtic (sharp peak, heavy tails)
• β₂ < 3 (γ₂ < 0) → Platykurtic (flat peak, light tails)
Skewness is like tilting a hat — it tells you which way the distribution leans. Kurtosis is like the hat's crown — it tells you whether the distribution is pointy (top hat), flat (beret), or normal (baseball cap).
Enter your data to compute all four raw moments, central moments, skewness, and kurtosis:
When data is arranged in class intervals (grouped data), we use the mid-point of each class to represent all observations in that class. This introduces a systematic error because the actual values are spread across the interval, not concentrated at the mid-point.
This error is called the grouping error or Sheppard's bias, and it affects the computed moments — especially the even-order moments (μ₂, μ₄, etc.).
Imagine a class of students whose heights range from 150 cm to 160 cm (class interval). Instead of recording each student's actual height, you record everyone as 155 cm (the mid-point). This underestimates the spread — the real variation within the class is lost. Sheppard's correction compensates for this lost variation.
William Fleetwood Sheppard (1863–1936) was a British mathematician and actuary who derived correction formulas for moments computed from grouped data. His corrections are applied when data is continuous and grouped into equal-width class intervals.
Let c be the class width (uniform for all intervals). Let μᵣ be the uncorrected central moment computed from grouped data using mid-points. The corrected moments (μᵣᶜ) are:
μ₁ᶜ = μ₁ = 0 (No correction needed — 1st central moment is always 0)
μ₂ᶜ = μ₂ − c²/12
μ₃ᶜ = μ₃ (No correction needed for odd-order moments)
μ₄ᶜ = μ₄ − (c²/2)μ₂ + (7c⁴/240)
Consider a single class interval of width c, from a to a+c, with mid-point m = a + c/2. If observations are uniformly distributed within this interval, the actual variance within the class is:
Variance within class = c²/12
This is the variance of a uniform distribution on [a, a+c]. When we replace all values by the mid-point, we lose this within-class variance. Sheppard's correction subtracts c²/12 from the computed μ₂ to compensate for the artificially inflated spread.
When you use mid-points, you are pretending every value in the class is exactly at the centre. But in reality, values are spread across the interval. This "pretending" adds extra variance of exactly c²/12 to each class. Sheppard's correction removes this artificial inflation.
Analogy: If you measure everyone in a room as being at the room's centre, you'll overestimate how tightly clustered they are around the building's average position. The correction accounts for the spread within each room.
The following grouped data gives the marks of 50 students. The class width is c = 10. Compute the corrected variance and SD using Sheppard's correction.
| Marks | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 |
|---|---|---|---|---|---|
| fᵢ | 5 | 10 | 20 | 10 | 5 |
Mid-points (xᵢ): 5, 15, 25, 35, 45
N = 50
Σfᵢxᵢ = 25+150+500+350+225 = 1250 → x̄ = 1250/50 = 25
Σfᵢxᵢ² = 125+2250+12500+12250+10125 = 37250 → μ₂' = 37250/50 = 745
Uncorrected μ₂ = μ₂' − x̄² = 745 − 625 = 120
Uncorrected σ = √120 = 10.95
Class width c = 10
Corrected μ₂ = μ₂ − c²/12 = 120 − 100/12 = 120 − 8.333 = 111.667
Corrected σ = √111.667 = 10.57
Uncorrected: σ² = 120.00 | σ = 10.95
Corrected: σ² = 111.67 | σ = 10.57
The correction reduced the variance by 8.33 (which equals c²/12). The corrected SD is a more accurate estimate of the true population SD.
For a grouped dataset with class width c = 5, the uncorrected central moments are:
μ₂ = 30, μ₃ = 15, μ₄ = 2800
Find the corrected moments and the corrected coefficients of skewness and kurtosis.
Sheppard's Corrections:
μ₁ᶜ = 0 (no correction)
μ₂ᶜ = μ₂ − c²/12 = 30 − 25/12 = 30 − 2.083 = 27.917
μ₃ᶜ = μ₃ = 15 (no correction for odd moments)
μ₄ᶜ = μ₄ − (c²/2)μ₂ + (7c⁴/240)
= 2800 − (25/2)(30) + (7 × 625/240)
= 2800 − 375 + 18.229
= 2443.229
β₁ = μ₃²/μ₂³ = 225/27000 = 0.00833
γ₁ = √0.00833 = 0.0913
β₂ = μ₄/μ₂² = 2800/900 = 3.111
γ₂ = 3.111 − 3 = 0.111
β₁ᶜ = (15)²/(27.917)³ = 225/21762 = 0.01034
γ₁ᶜ = √0.01034 = 0.1017
β₂ᶜ = 2443.229/(27.917)² = 2443.229/779.36 = 3.135
γ₂ᶜ = 3.135 − 3 = 0.135
1. Data is continuous (measured, not counted).
2. Data is grouped into equal-width class intervals.
3. The distribution is unimodal (single peak) and roughly symmetric.
4. Class frequencies taper off smoothly in both tails (no abrupt cutoffs).
5. There are no open-end classes (e.g., "50 and above").
1. Data is discrete (counted, like number of children).
2. Class intervals are unequal.
3. Distribution is highly skewed or bimodal.
4. There are open-end classes (e.g., "above 100").
5. Data is ungrouped (individual values are known).
Enter the uncorrected moments and class width to see the corrected values:
Q1: The first central moment (μ₁) is always:
Q2: The second central moment (μ₂) equals:
Q3: If μ₃ = 0, the distribution is:
Q4: Sheppard's correction for μ₂ is:
Q5: Sheppard's correction is NOT needed for:
Q6: The value c²/12 in Sheppard's correction represents:
Q7: Sheppard's correction should NOT be applied when:
Q8: If μ₂' = 50 and μ₁' = 6, then the central moment μ₂ equals:
1. Compute the first four raw moments and central moments for: 1, 3, 5, 7, 9
2. Given μ₁' = 10, μ₂' = 120, μ₃' = 1500, μ₄' = 20000, find all four central moments.
3. For a grouped dataset with class width c = 8, the uncorrected moments are: μ₂ = 64, μ₃ = 30, μ₄ = 5000. Apply Sheppard's correction to find the corrected moments.
4. The following frequency distribution gives the weights (in kg) of 40 students. Class width = 5 kg. Compute the corrected variance and SD.
Weight: 40–45, 45–50, 50–55, 55–60, 60–65
Frequency: 4, 8, 14, 10, 4
5. Explain why Sheppard's correction is not needed for discrete data. Give an example.
6. If the corrected μ₂ = 45.67 and c = 10, what was the uncorrected μ₂?
7. The coefficient of kurtosis (β₂) computed from grouped data is 2.8 before Sheppard's correction. After applying the correction, will β₂ increase, decrease, or stay the same? Explain your reasoning.
8. BONUS: Derive the formula for the variance of a uniform distribution on [a, a+c] and show that it equals c²/12. (Hint: Use the formula for the variance of a continuous uniform distribution.)
Questions? Let's discuss!
Lecture 21: Skewness — Measures & Interpretation
Lecture 22: Kurtosis — Measures & Interpretation
Lecture 23: Lorenz Curve & Measures of Inequality
Unit III · Lecture 20 · BDS113 Introduction to Statistics · B.Sc. Data Science, Semester II