Math Problem Statement

1

Solution

To address part 1 of the problem, we need to prove that the quantity (Xμ)TΣ1(Xμ)(X - \mu)^T \Sigma^{-1} (X - \mu) is distributed as a chi-squared distribution with pp degrees of freedom, given that XX is distributed as Np(μ,Σ)N_p(\mu, \Sigma), where μ\mu is the mean vector and Σ\Sigma is the covariance matrix.

Proof:

Step 1: Definition and Setup

  • Let XX be a random vector following a multivariate normal distribution: XNp(μ,Σ)X \sim N_p(\mu, \Sigma) This means: X has mean vector μ and covariance matrix Σ.X \text{ has mean vector } \mu \text{ and covariance matrix } \Sigma.

Step 2: Standardization

  • Define the standardized random vector ZZ as: Z=Σ12(Xμ)Z = \Sigma^{-\frac{1}{2}} (X - \mu) Here, ZZ is a random vector with a standard normal distribution Np(0,I)N_p(0, I), where II is the identity matrix of size pp. This is because: ZNp(0,I)Z \sim N_p(0, I) where ZZ has mean 00 and covariance matrix II.

Step 3: Expression of the Quadratic Form

  • Consider the quadratic form: (Xμ)TΣ1(Xμ)(X - \mu)^T \Sigma^{-1} (X - \mu) Substitute ZZ into the expression: [ (X - \mu)^T \Sigma^{-1} (X - \mu) = \left[\Sigma^{\frac{1}{2}}Z\right]^T \Sigma^{-1} \left[\Sigma^{\frac{1}{2}}Z\right] ] Simplifying: =ZTΣ12TΣ1Σ12Z=ZTZ= Z^T \Sigma^{\frac{1}{2}T} \Sigma^{-1} \Sigma^{\frac{1}{2}} Z = Z^T Z

Step 4: Distribution of the Quadratic Form

  • Since ZNp(0,I)Z \sim N_p(0, I), the sum of the squares of pp independent standard normal variables Z12+Z22++Zp2Z_1^2 + Z_2^2 + \cdots + Z_p^2 follows a chi-squared distribution with pp degrees of freedom: ZTZ=Z12+Z22++Zp2χp2Z^T Z = Z_1^2 + Z_2^2 + \cdots + Z_p^2 \sim \chi^2_p

Conclusion:

  • Therefore, (Xμ)TΣ1(Xμ)(X - \mu)^T \Sigma^{-1} (X - \mu) follows a chi-squared distribution with pp degrees of freedom: (Xμ)TΣ1(Xμ)χp2(X - \mu)^T \Sigma^{-1} (X - \mu) \sim \chi^2_p

This completes the proof.

Do you have any questions on this proof? Would you like to proceed to part 2 or any other section?


Related Questions:

  1. How does the interpretation of (Xμ)TΣ1(Xμ)(X - \mu)^T \Sigma^{-1} (X - \mu) change when the covariance matrix Σ\Sigma is not diagonal?
  2. What role does the covariance matrix Σ\Sigma play in shaping the distribution of the quadratic form?
  3. Can you provide a practical example where this chi-squared distribution would be useful in real-world data analysis?
  4. How would you modify this proof if XX followed a non-normal distribution?
  5. Why is it important that the random vector ZZ is standardized in this proof?

Tip:

When proving distribution properties, it's often useful to standardize the variables involved. This reduces the problem to working with standard distributions, making it easier to apply known results.

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Math Problem Analysis

Mathematical Concepts

Multivariate Normal Distribution
Chi-Squared Distribution
Quadratic Forms

Formulas

(X - \mu)^T \Sigma^{-1} (X - \mu)
Chi-squared distribution

Theorems

Distribution of Quadratic Forms in Multivariate Normal Distribution

Suitable Grade Level

Advanced Undergraduate