Let X ~ N(μ, σ2), then the fourth central moment of X is:
3 σ4
Central moments are statistical measures that describe the shape and characteristics of a probability distribution. The \( k \)-th central moment of a random variable \( X \) is defined as the expected value of \( (X - E[X])^k \), where \( E[X] \) is the mean of \( X \).
The question asks for the fourth central moment of a normal distribution \( X \sim N(\mu, \sigma^2) \). Here, \( \mu \) represents the mean and \( \sigma^2 \) represents the variance of the distribution. The standard deviation is \( \sigma \).
The central moments for a normal distribution \( N(\mu, \sigma^2) \) follow a specific pattern:
The double factorial \( (k-1)!! \) is the product of all positive odd integers up to \( k-1 \).
We need to find the fourth central moment, so we use the formula for an even order moment with \( k = 4 \):
\( \mu_4 = \sigma^4 (4-1)!! \)
First, let's calculate the double factorial \( (4-1)!! \):
\( (4-1)!! = 3!! \)
The double factorial of 3 (3!!) is the product of positive odd integers less than or equal to 3:
\( 3!! = 3 \times 1 = 3 \)
Now, substitute this value back into the formula for \( \mu_4 \):
\( \mu_4 = \sigma^4 \times 3 \)
\( \mu_4 = 3\sigma^4 \)
Thus, the fourth central moment of a normal distribution is \( 3\sigma^4 \).
| Moment Order (k) | Central Moment (\( \mu_k \)) | Formula |
|---|---|---|
| 1 | \( \mu_1 \) (Mean Deviation from Mean) | 0 |
| 2 | \( \mu_2 \) (Variance) | \( \sigma^2 \) |
| 3 | \( \mu_3 \) (Skewness related) | 0 |
| 4 | \( \mu_4 \) (Kurtosis related) | \( 3\sigma^4 \) |
| Moment | Definition | Normal Distribution Value \( N(\mu, \sigma^2) \) |
|---|---|---|
| Mean (First raw moment) | \( E[X] \) | \( \mu \) |
| Variance (Second central moment) | \( E[(X-\mu)^2] \) | \( \sigma^2 \) |
| Skewness (Standardized third central moment) | \( \gamma_1 = \mu_3 / \sigma^3 \) | 0 |
| Kurtosis (Standardized fourth central moment) | \( \beta_2 = \mu_4 / \sigma^4 \) | 3 |
| Excess Kurtosis | \( \gamma_2 = \beta_2 - 3 \) | 0 |
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