Ex 1.8 - Show that the mean value of Gaussian distribution is µ. (Pattern Recognition and ML by Christopher Bishop)

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Part 4:

Using ChatGPT by the formula for E[x,y] when x and y are discrete

If xn and xm are values from a gaussian distribution, can you derive the expression for E[xn.xm]

If $X_n$ and $X_m$ are values from independent Gaussian distributions, each with its own mean ($\mu_n$ and $\mu_m$) and variance ($\sigma_n^2$ and $\sigma_m^2$), and you want to find the expected value of their product $E[X_n \cdot X_m]$, it is important to note that the product of independent Gaussian variables does not follow a Gaussian distribution.

However, if you are interested in the covariance of $X_n$ and $X_m$, you can use the fact that the covariance of independent variables is zero:

$Cov[X_n, X_m] = E[X_n \cdot X_m] - E[X_n] \cdot E[X_m]$

Since $X_n$ and $X_m$ are independent, $Cov[X_n, X_m] = 0$, and you get:

$E[X_n \cdot X_m] = E[X_n] \cdot E[X_m]$

So, the expected value of the product of independent Gaussian variables is the product of their means:

$E[X_n \cdot X_m] = \mu_n \cdot \mu_m$

No, no. xn and xm are from the same distribution

I apologize for the confusion. If $X_n$ and $X_m$ are from the same Gaussian distribution, let's denote their mean as $\mu$ and their variance as $\sigma^2$. In this case, you're asking for the expected value $E[X_n \cdot X_m]$.

Since $X_n$ and $X_m$ are from the same distribution, we can simplify it as follows:

$E[X_n \cdot X_m] = E[X_n]^2 + Cov[X_n, X_m]$

Now, since $X_n$ and $X_m$ are from the same distribution, their covariance simplifies to the variance:

$E[X_n \cdot X_m] = \mu^2 + \sigma^2$

So, the expression for $E[X_n \cdot X_m]$ for two variables from the same Gaussian distribution is the square of the mean plus the variance of that distribution.

Ex 1.9 - Show that the mode of the multivariate Gaussian is given by µ. (Pattern Recognition and ML by Christopher Bishop)

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:

$f(\mathbf{x}; \boldsymbol{\mu}, \boldsymbol{\Sigma}) = \frac{1}{(2\pi)^{n/2} \det(\boldsymbol{\Sigma})^{1/2}} \exp\left(-\frac{1}{2} (\mathbf{x} - \boldsymbol{\mu})^T \boldsymbol{\Sigma}^{-1} (\mathbf{x} - \boldsymbol{\mu})\right)$

where:

• $\mathbf{x}$ is the vector of random variables.
• $\boldsymbol{\mu}$ is the mean vector.
• $\boldsymbol{\Sigma}$ is the covariance matrix.
• $n$ is the dimensionality of the multivariate distribution.
• $\det(\boldsymbol{\Sigma})$ is the determinant of the covariance matrix.

To find the mode, we need to maximize this PDF with respect to $\mathbf{x}$. Taking the natural logarithm (log) of the PDF simplifies the maximization problem, as the logarithm is a monotonic function:

$\ln f(\mathbf{x}; \boldsymbol{\mu}, \boldsymbol{\Sigma}) = -\frac{n}{2} \ln(2\pi) - \frac{1}{2} \ln(\det(\boldsymbol{\Sigma})) - \frac{1}{2} (\mathbf{x} - \boldsymbol{\mu})^T \boldsymbol{\Sigma}^{-1} (\mathbf{x} - \boldsymbol{\mu})$

To find the mode, we need to find the point $\mathbf{x}$ at which the derivative of the log PDF with respect to $\mathbf{x}$ is zero. This can be expressed as:

$\frac{\partial}{\partial \mathbf{x}} \ln f(\mathbf{x}; \boldsymbol{\mu}, \boldsymbol{\Sigma}) = 0$

Solving this equation will give us the mode $\mathbf{x}$. Taking the derivative, we get:

$-\boldsymbol{\Sigma}^{-1} (\mathbf{x} - \boldsymbol{\mu}) = 0$

Solving for $\mathbf{x}$, we find:

$\mathbf{x} = \boldsymbol{\mu}$

Therefore, the mode of the multivariate Gaussian distribution is given by the mean vector $\boldsymbol{\mu}$.

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.

What happens when we reverse the limits of a definite integral?

When you reverse the limits of a definite integral, the sign of the integral changes. Mathematically, if you have a definite integral of a function $f(x)$ from $a$ to $b$:

$\int_{a}^{b} f(x) \,dx$

and you reverse the limits to $b$ and $a$:

$\int_{b}^{a} f(x) \,dx$

the new integral is equal to the negative of the original integral:

$\int_{b}^{a} f(x) \,dx = -\int_{a}^{b} f(x) \,dx$

This is a consequence of the fact that the definite integral represents the signed area under the curve of the function between the given limits. Reversing the limits effectively changes the orientation of the interval, resulting in a change of sign for the area.

In summary, reversing the limits of a definite integral changes the sign of the integral.