Exercise 4.5 - Normalization constant for a multidimensional Gaussian

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Assume $μ = 0$ w.l.o.g. If the covariance matrix already takes a diagonal form:

Σ = (\begin{matrix} λ_{1}^{- 1} & \dots \\ \dots & \dots \\ \dots & λ_{d}^{- 1} \end{matrix}),

then:

\begin{aligned} \int \exp (- \frac{1}{2} 𝐱^{T} Σ 𝐱) d 𝐱 & = \int \exp (- \frac{1}{2} (\sum_{i = 1}^{d} \frac{x_{i}^{2}}{λ_{i}})) d 𝐱 \\ = \prod_{i = 1}^{d} \int \exp (- \frac{x_{i}^{2}}{2 \cdot λ_{i}}) d x_{i} \\ = {(2 π λ_{i})}^{- \frac{d}{2}} . \end{aligned}

Plugging in $| Σ | = \prod_{i = 1}^{d} λ_{i}^{- 1}$ yields the desired normalization constant. In the second equation, using the distribution law (though somewhat intimidating).

For the general case, we begin by diagonalizing $Σ$ into:

Σ = U^{T} Λ U,

where $Λ$ is a diagonal matrix with components $λ_{1}^{- 1} \dots λ_{d}^{- 1}$ and $U$ is a orthogonal matrix. The integral now becomes:

\int \exp (- \frac{1}{2} {(U 𝐱)}^{T} Λ (U 𝐱)) d 𝐱 .

Since $| U | = 1$ uniformly, we can directly rewrite the integral into:

\int \exp (- \frac{1}{2} 𝐮^{T} Λ 𝐮) d 𝐮 .

The rest is repeating the diagonal case.

solour_lfq

2021-03-24 13:42

Exercise 4.5 - Normalization constant for a multidimensional Gaussian

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