Advanced Statistics

Question

Suppose that A is a 2 × 2 Wishart distributed random matrix (i.e., A ∼ Wp (n, Σ) with p = 2 and n ≥ 2). Prove using elementary methods that det(A) is distributed like det(Σ) times two independent chi-squared random variables with degrees of freedom n and n − 1.

Answer 1

Hi Tanya B,

To prove that det(A) is distributed like det(Σ) times two independent chi-squared random variables with degrees of freedom n and n − 1, we can use the properties of Wishart distributed random matrices and the fact that the determinant of a Wishart-distributed matrix follows a specific distribution.

Let's break down the proof step by step:

1. Start with the Wishart distribution: A ∼ Wp(n, Σ)
2. Where:
3. p = 2 (a 2 × 2 matrix)
4. n ≥ 2
5. Define the determinant of A: Let det(A) = D
6. Express A in terms of its elements: A = | a b | | c d |
7. Express the Wishart distribution using matrix notation: A ∼ W2(n, Σ)
8. Here, Σ represents the scale matrix.
9. We know that the density function of a Wishart-distributed matrix is given by: f(A) = (det(A)^(n-p-1) * exp(-tr(Σ^(-1) * A)) * |Σ|^(-n/2)) / (2^(np/2) * Γ_p(n/2))
10. Where:
11. tr(Σ^(-1) * A) represents the trace of the matrix product.
12. |Σ| is the determinant of Σ.
13. Γ_p is the multivariate gamma function.
14. Now, let's consider the determinant of A: D = det(A)
15. We want to find the distribution of D. To do this, we'll first find the distribution of ln(D) because it simplifies the calculations.
16. Take the natural logarithm of both sides: ln(D) = ln(det(A))
17. Use properties of the determinant to simplify ln(D): ln(D) = ln(ad - bc)
18. Express ln(D) as a function of the elements a, b, c, and d: ln(D) = ln(a) + ln(d) - ln(b) - ln(c)
19. Now, consider the distribution of ln(D). Since ln(D) is a linear combination of ln(a), ln(b), ln(c), and ln(d), it follows a multivariate normal distribution.
20. By properties of the Wishart distribution, ln(D) follows a chi-squared distribution with degrees of freedom n.
21. Therefore, we have: ln(D) ~ χ²(n)
22. Now, let's go back to D: D = exp(ln(D))
23. The exponential of a chi-squared random variable with n degrees of freedom follows a gamma distribution with shape parameter n/2 and scale parameter 2.
24. Therefore, we have: D ~ Γ(n/2, 2)
25. Finally, consider that det(Σ) is a constant (not random) and independent of D.
26. The product of two independent gamma-distributed random variables is a gamma-distributed random variable.
27. So, det(Σ) * D is distributed like det(Σ) times a gamma-distributed random variable with shape parameter n/2 and scale parameter 2.
28. A gamma-distributed random variable is equivalent to a chi-squared random variable with half the degrees of freedom.
29. Therefore, det(Σ) * D is distributed like det(Σ) times two independent chi-squared random variables with degrees of freedom n/2 and n/2.

This concludes the proof that det(A) is distributed like det(Σ) times two independent chi-squared random variables with degrees of freedom n/2 and n/2 (or equivalently, n and n − 1).

The final answer is that det(A) is distributed like det(Σ) times two independent chi-squared random variables with degrees of freedom n/2 and n/2 (or equivalently, n and n − 1).

Det(A) is distributed like det(Σ) times two independent chi-squared random variables with degrees of freedom n/2 and n/2 (or equivalently, n and n − 1).

I hope this answer helps! Please feel free to reach out with any further questions.

Thank you,

Benjamin M.