Pre-read (1): Pre-read (2): Pre-read (3): Ex 1.9.A: Show that the mode (i.e. the maximum) of the Gaussian distribution (1.46) is given by µ. Ex 1.9.B: Similarly, show that the mode of the multivariate Gaussian (1.52) is given by µ. Solution hint from the book:
Solution by ChatGPT
The mode of a multivariate Gaussian distribution is the point at which the probability density function (PDF) is maximized. For a multivariate Gaussian distribution with mean vector μ and covariance matrix Σ, the PDF is given by:
where:
- is the vector of random variables.
- is the mean vector.
- is the covariance matrix.
- is the dimensionality of the multivariate distribution.
- is the determinant of the covariance matrix.
To find the mode, we need to maximize this PDF with respect to . Taking the natural logarithm (log) of the PDF simplifies the maximization problem, as the logarithm is a monotonic function:
To find the mode, we need to find the point at which the derivative of the log PDF with respect to is zero. This can be expressed as:
Solving this equation will give us the mode . Taking the derivative, we get:
Solving for , we find:
Therefore, the mode of the multivariate Gaussian distribution is given by the mean vector .
Solution by Bard
Here's a detailed proof demonstrating that the mode of a multivariate Gaussian distribution is indeed its mean:
1. Probability Density Function (PDF) of Multivariate Gaussian:
-
The PDF of a multivariate Gaussian distribution with mean vector μ and covariance matrix Σ is given by:
f(x) = (1 / ((2π)^n/2 * |Σ|^1/2)) * exp(-1/2 * (x - μ)^T * Σ^(-1) * (x - μ))
2. Finding the Mode:
- The mode is the point where the PDF has its maximum value.
- To find the mode, we need to find the maximum of f(x) with respect to x.
3. Taking the Gradient:
-
We differentiate f(x) with respect to x:
∇f(x) = -Σ^(-1) * (x - μ) * f(x)
4. Setting the Gradient to Zero:
-
At the maximum, the gradient will be zero:
∇f(x) = 0 ==> -Σ^(-1) * (x - μ) = 0
5. Solving for x:
-
Since Σ^(-1) is invertible, we can multiply both sides by its inverse:
x - μ = 0
6. Mode x = μ:
- This leads to the solution x = μ.
7. Second Derivative Test (Hessian):
- To confirm that this is a maximum, we can examine the Hessian matrix, which is the matrix of second partial derivatives.
- The Hessian of f(x) is -Σ^(-1), which is negative definite (all eigenvalues are negative).
- This guarantees that f(x) has a maximum at x = μ.
Therefore, we have proven that the mode of the multivariate Gaussian distribution is equal to its mean vector μ.
No comments:
Post a Comment