Moments of Binomial Distribution
By Bindeshwar Singh Kushwaha — PostNetwork Academy
Moment Definition
Let \( X \sim B(n, p) \) be a binomial random variable.
- The \( r^\text{th} \) raw moment about origin:
\( \mu_r’ = \mathbb{E}(X^r) = \sum_{x=0}^{n} x^r \cdot \mathbb{P}(X = x) \)
- First-order moment (mean):
\( \mu_1′ = \mathbb{E}(X) \)
- Binomial PMF:
\( \mathbb{P}(X = x) = \binom{n}{x} p^x q^{n-x}, \quad q = 1 – p \)
Mean Computation
Substitute PMF:
\( \mu_1′ = \sum_{x=1}^{n} x \binom{n}{x} p^x q^{n – x} \)
Using identity \( x \binom{n}{x} = n \binom{n-1}{x-1} \):
\( \mu_1′ = n p \sum_{r=0}^{n-1} \binom{n-1}{r} p^r q^{n-1 – r} \)
Recognize the binomial expansion:
\( \mu_1′ = np \)
Second Moment and Variance
Start with:
\( \mu_2′ = \sum_{x=0}^{n} x^2 \binom{n}{x} p^x q^{n – x} \)
Use identity:
\( x^2 = x(x – 1) + x \)
Split summation:
\( \mu_2′ = n(n – 1)p^2 + \mu_1′ = n(n – 1)p^2 + np \)
Variance:
\( \text{Var}(X) = \mu_2′ – (\mu_1′)^2 = np(1 – p) \)
Third Moment
Using:
\( x^3 = x(x – 1)(x – 2) + 3x(x – 1) + x \)
Apply identities and simplify:
\( \mu_3′ = n(n-1)(n-2)p^3 + 3n(n-1)p^2 + np \)
Third central moment:
\( \mu_3 = \mu_3′ – 3\mu_2 \mu_1′ + 2(\mu_1′)^3 = npq(1 – 2p) \)
Fourth Moment and Kurtosis
Using:
\( x^4 = x(x-1)(x-2)(x-3) + 6x(x-1)(x-2) + 7x(x-1) + x \)
Compute:
\( \mu_4′ = n(n-1)(n-2)(n-3)p^4 + 6n(n-1)(n-2)p^3 + 7n(n-1)p^2 + np \)
Central moment:
\( \mu_4 = \mu_4′ – 4\mu_3’\mu_1 + 6\mu_2′(\mu_1)^2 – 3(\mu_1)^4 \)
Skewness and Kurtosis
Skewness:
\( \gamma_1 = \frac{q – p}{\sqrt{npq}} \)
Kurtosis:
\( \gamma_2 = \frac{1 – 6pq}{npq} \)
Summary
- \( \mu_1 = np \), \( \mu_2 = npq \)
- \( \mu_3 = npq(1 – 2p) \)
- \( \mu_4 = npq[1 + 3(n-2)pq] \)
- \( \gamma_1 = \dfrac{q – p}{\sqrt{npq}} \)
- \( \gamma_2 = \dfrac{1 – 6pq}{npq} \)
Video
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