1. ICA Overview

PCA pursues variance maximization and decorrelation, while ICA pursues deeper independence and non-Gaussianity.

1. The starting point of the problem: Blind Source Separation (Blind Source Separation)

Suppose we have a linear model (exactly the same as your previous FA model):

$$X = Lz$$
  • $X \in \mathbb{R}^p$: Observed Signals.
  • $z \in \mathbb{R}^r$: Latent Sources that we cannot see.
  • $L$: Mixing Matrix.

Goal of ICA: Without knowing $L$ and $z$, find an unmixing matrix $W$ (that is, $L^{-1}$) based on only $X$ such that $Y = WX \approx z$.

2. Why can’t PCA work? (Connection Problem 2: Uncorrelated $\neq$ Independent)

You may ask: “I have PCA. PCA can uncorrelated data. Doesn’t this mean separation?”

Problem 2 gave us a slap in the face. In that question, we saw that two variables $x_1, x_2$ satisfy $\text{Cov}(x_1, x_2) = 0$ (uncorrelated), but their joint distribution is $f(x_1, x_2) \neq f(x_1)f(x_2)$ (not independent).

  • Limitations of PCA: PCA only handles second-order statistics (Second-order statistics), which is the covariance matrix. It rotates the axes so that $\text{Cov} = 0$. This is indeed equivalent to independence for Gaussian Distribution, but for the vast majority of non-Gaussian signals in the real world (such as human voices, image edges), “uncorrelated” is far from enough.
  • Advanced of ICA**: ICA attempts to use Higher-order statistics to force each component of $z$ to achieve Statistical Independence.

Summary in one sentence: Problem 2 tells us that if we only pursue $\text{Cov}=0$, we may still be unable to separate the mixed signals.

3. How does ICA separate signals? (Connect Problem 3: Kurtosis & CLT)

Since we can’t just look at the covariance, what objective function should we optimize to find $z$? This is where Problem 3 comes into play.

Here is a reverse thinking based on the Central Limit Theorem (CLT):

  1. CLT says: If you add (mix) several independent random variables together, their sum will tend to have a Gaussian distribution.
  • That is: $\text{Mixture} = z_1 + z_2 + \dots \to \text{Gaussian}$.
  1. Converse thinking: If mixing will make the distribution “Gaussian”, then Unmixing should make the distribution “Maximally Non-Gaussian”.

Problem 3 Derives the kurtosis of the sum:

$$\kappa(y_1 + y_2) = \frac{\sigma_1^4 \kappa(y_1) + \sigma_2^4 \kappa(y_2)}{(\sigma_1^2 + \sigma_2^2)^2}$$

This formula tells us that the kurtosis of a linear combination can be calculated. In ICA, we use Kurtosis as a measure of “non-Gaussianity”.

  • The Gaussian distribution has a kurtosis of 0.
  • Super-Gaussian distribution (peaks and fat tails, like human voices) kurtosis $>0$.
  • Sub-Gaussian distribution (flat-topped, such as a uniform distribution) kurtosis $<0$.

Algorithmic logic of ICA (e.g. FastICA): We need to find a projection vector $w$ that maximizes the kurtosis $|\kappa(y)|$ of $y = w^T X$. This is like rotating the coordinate axis in the data space. When the projection of the data on the axis looks “least like a normal distribution” (the sharpest or flattest), we think we have found an independent source signal $z_i$.

Summary in one sentence: Problem 3 provides the core “engine” of ICA - Maximizing Non-Gaussianity, which is usually achieved by maximizing the absolute value of kurtosis.

4. What are the costs of ICA? (Connect to Problem 1: Permutation Ambiguity)

When we worked hard to calculate the source signal, Problem 1 reminded us that this solution was “flawed”.

In Problem 1 we proved: If $X = Lz$, we introduce a permutation matrix $P$ (Permutation Matrix), and the model can be rewritten as:

$$X = (LP^{-1})(Pz) = \tilde{L}\tilde{z}$$

Mathematically, the observations $X$ produced by $Lz$ and $\tilde{L}\tilde{z}$ are exactly the same.

This means that ICA has two uncertainties (Ambiguities) that cannot be eliminated:

  1. Permutation Ambiguity: The first signal $\hat{z}_1$ you solve may be the original $z_3$ or $z_5$. You don’t know the original order. (Problem 1 proves this).
  2. Scaling Ambiguity: Because of $X = Lz = (L \cdot \alpha)(\frac{1}{\alpha} \cdot z)$. If the original sound is loud and $L$ is small, or if the original sound is small and $L$ is large, the observed $X$ is the same. ICA cannot restore the absolute volume (variance) of the original signal. Usually we will force the solved $z$ to be normalized to unit variance.

Summary in one sentence: Problem 1 tells us that ICA can only restore the waveform (Waveform), but it does not know where the waveform was originally ranked, nor how loud it was originally.

Summary (The Big Picture)

Based on these three questions, your ICA should look like this:

  1. Problem 2 said: Don’t just look at the covariance. Uncorrelatedness does not mean independence. We must pursue independence.
  2. Problem 3 said: How to find independence? Use the opposite of the central limit theorem. The less it follows the Gaussian distribution, the purer the signal. We want to maximize kurtosis.
  3. Problem 1 said: Don’t expect perfect reproduction. We can separate the waveforms, but the order and size are messed up.

That’s what ICA is: blindly separating non-Gaussian independent sources from a mixed signal in the presence of permutation and scale ambiguity using higher-order statistics such as kurtosis.

Permutation Matrix related

Here is the step-by-step solution in English, using explicit scalar matrix notation and formal mathematical terminology as requested.

1. Explicit Form of Matrix $P$

Concept: The matrix $P$ is an Elementary Matrix specifically representing a row switching operation (transposition). To obtain $P$, we perform the row swap operation on the Identity Matrix $I_p$.

The matrix $P$ has $1$s on the diagonal, except at positions $(i, i)$ and $(j, j)$ where it has $0$. Instead, the $1$s are placed at $(i, j)$ and $(j, i)$ to effectuate the swap.

$$ P = \begin{pmatrix} 1 & \cdots & 0 & \cdots & 0 & \cdots & 0 \\ \vdots & \ddots & \vdots & & \vdots & & \vdots \\ 0 & \cdots & 0 & \cdots & 1 & \cdots & 0 \\ \vdots & & \vdots & \ddots & \vdots & & \vdots \\ 0 & \cdots & 1 & \cdots & 0 & \cdots & 0 \\ \vdots & & \vdots & & \vdots & \ddots & \vdots \\ 0 & \cdots & 0 & \cdots & 0 & \cdots & 1 \end{pmatrix} \quad \begin{matrix} \\ \\ \leftarrow \text{row } i \\ \\ \leftarrow \text{row } j \\ \\ \\ \end{matrix} $$

Detailed Indices:

  • $P_{kk} = 1$ for all $k \neq i, j$.
  • $P_{ii} = 0$, $P_{jj} = 0$.
  • $P_{ij} = 1$, $P_{ji} = 1$.
  • All other entries are $0$.

2. Explicit Form of Inverse Matrix $P^{-1}$

Theorem: Properties of Elementary Permutation Matrices / Involutory Matrix.

Reasoning: Geometrically, if you swap the $i$-th and $j$-th items of a list, and then swap them again, you return to the original configuration. Therefore, the inverse operation of a swap is the swap itself. In linear algebra terms, $P$ is an Involutory Matrix, meaning $P^2 = I$. Thus, $P = P^{-1}$.

Additionally, since $P$ is symmetric ($P = P^T$) and orthogonal ($P^T = P^{-1}$), we also arrive at the same conclusion.

Explicitly:

$$ P^{-1} = P = \begin{pmatrix} 1 & \cdots & 0 & \cdots & 0 & \cdots & 0 \\ \vdots & \ddots & \vdots & & \vdots & & \vdots \\ 0 & \cdots & 0 & \cdots & 1 & \cdots & 0 \\ \vdots & & \vdots & \ddots & \vdots & & \vdots \\ 0 & \cdots & 1 & \cdots & 0 & \cdots & 0 \\ \vdots & & \vdots & & \vdots & \ddots & \vdots \\ 0 & \cdots & 0 & \cdots & 0 & \cdots & 1 \end{pmatrix} $$

3. Expressions for $\tilde{z}$ and $\tilde{L}$

a) For the column vector $\tilde{z} = Pz$: Left-multiplying a column vector by a permutation matrix permutes the rows (elements).

$$ \tilde{z} = \begin{pmatrix} z_1 \\ \vdots \\ z_j \\ \vdots \\ z_i \\ \vdots \\ z_p \end{pmatrix} \quad \begin{matrix} \\ \\ \leftarrow \text{position } i \text{ (now holds } z_j \text{)} \\ \\ \leftarrow \text{position } j \text{ (now holds } z_i \text{)} \\ \\ \end{matrix} $$

b) For the row vector $\tilde{L} = LP^{-1}$: Since $P^{-1} = P$, this is equivalent to $\tilde{L} = LP$. Right-multiplying a row vector by a permutation matrix permutes the columns (indices).

$$ \tilde{L} = (\ell_1, \dots, \ell_j, \dots, \ell_i, \dots, \ell_p) $$

(Note: The element $\ell_j$ is now at the $i$-th index, and $\ell_i$ is at the $j$-th index.)

4. Proof of $\tilde{L}\tilde{z} = Lz$

Theorem: Associativity of Matrix Multiplication.

Algebraic Proof: We substitute the definitions of $\tilde{L}$ and $\tilde{z}$ into the equation:

$$ \begin{aligned} \tilde{L}\tilde{z} &= (L P^{-1})(P z) \\ &= L (P^{-1} P) z \quad \text{(by Associativity)} \\ &= L (I) z \quad \text{(by Definition of Inverse Matrix)} \\ &= Lz \end{aligned} $$

Scalar Verification (Scalar Expansion): If we expand the scalar product (Inner Product), we can see that the summation terms are merely reordered:

$$ \begin{aligned} Lz &= \sum_{k=1}^p \ell_k z_k = \ell_1 z_1 + \dots + \mathbf{\ell_i z_i} + \dots + \mathbf{\ell_j z_j} + \dots + \ell_p z_p \\ \tilde{L}\tilde{z} &= \sum_{k=1}^p \tilde{\ell}_k \tilde{z}_k = \ell_1 z_1 + \dots + \underbrace{\mathbf{\ell_j}}_{\text{at pos } i} \underbrace{\mathbf{z_j}}_{\text{at pos } i} + \dots + \underbrace{\mathbf{\ell_i}}_{\text{at pos } j} \underbrace{\mathbf{z_i}}_{\text{at pos } j} + \dots + \ell_p z_p \end{aligned} $$

Since scalar addition is commutative, the total sum (the dot product) remains invariant under the permutation of indices.

3x3 Permutation Matrix multiplication = I operation example

Matrix Multiplication is the core cornerstone. The key formula here is: “left row” times “right column”.

Let’s construct a $3 \times 3$ permutation matrix $P$ that swaps rows 1 and 2 (leaving row 3 unchanged).

1. Our matrix $P$

According to the previous rules:

  • Row 1 wants the number in row 2 $\to$ Row 1, column 2 is 1 ($0, 1, 0$)
  • Row 2 wants the number in row 1 $\to$ Row 2, column 1 is 1 ($1, 0, 0$)
  • Row 3 remains unchanged $\to$ Row 3, column 3 is 1 ($0, 0, 1$)
$$ P = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$

2. Detailed process of calculating $P \times P$

We want to calculate:

$$\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} \times \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} c_{11} & c_{12} & c_{13} \\ c_{21} & c_{22} & c_{23} \\ c_{31} & c_{32} & c_{33} \end{bmatrix} $$

Calculation rules: The element $c_{ij}$ of the $i$ row and $j$ column of the result matrix is equal to the dot product of the th $i$ row of the left matrix and the th $j$ column of the right matrix (the corresponding positions are multiplied and then added).

Calculation of the first row (Row 1)

  • $c_{11}$ (left row 1 $\cdot$ right column 1): $(0, 1, 0) \cdot (0, 1, 0) = (0\times0) + (1\times1) + (0\times0) = 0 + 1 + 0 = \mathbf{1}$
  • $c_{12}$ (left row 1 $\cdot$ right column 2): $(0, 1, 0) \cdot (1, 0, 0) = (0\times1) + (1\times0) + (0\times0) = 0 + 0 + 0 = \mathbf{0}$
  • $c_{13}$ (left row 1 $\cdot$ right column 3): $(0, 1, 0) \cdot (0, 0, 1) = (0\times0) + (1\times0) + (0\times1) = 0 + 0 + 0 = \mathbf{0}$

Calculation of Row 2

  • $c_{21}$ (left row 2 $\cdot$ right column 1): $(1, 0, 0) \cdot (0, 1, 0) = (1\times0) + (0\times1) + (0\times0) = 0 + 0 + 0 = \mathbf{0}$
  • $c_{22}$ (left row 2 $\cdot$ right column 2): $(1, 0, 0) \cdot (1, 0, 0) = (1\times1) + (0\times0) + (0\times0) = 1 + 0 + 0 = \mathbf{1}$
  • $c_{23}$ (left row 2 $\cdot$ right column 3): $(1, 0, 0) \cdot (0, 0, 1) = (1\times0) + (0\times0) + (0\times1) = 0 + 0 + 0 = \mathbf{0}$

Calculation of Row 3

  • $c_{31}$ (left row 3 $\cdot$ right column 1): $(0, 0, 1) \cdot (0, 1, 0) = (0\times0) + (0\times1) + (1\times0) = 0 + 0 + 0 = \mathbf{0}$
  • $c_{32}$ (left row 3 $\cdot$ right column 2): $(0, 0, 1) \cdot (1, 0, 0) = (0\times1) + (0\times0) + (1\times0) = 0 + 0 + 0 = \mathbf{0}$
  • $c_{33}$ (left row 3 $\cdot$ right column 3): $(0, 0, 1) \cdot (0, 0, 1) = (0\times0) + (0\times0) + (1\times1) = 0 + 0 + 1 = \mathbf{1}$

3. Final result

Fill in the 9 numbers calculated above:

$$ P \times P = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I ext{ (identity matrix)} $$

4. Intuitive understanding (physical meaning)

If you don’t look at the calculation process, just look at the actions:

  1. First ride $P$: exchange the 1st and 2nd cards in your hand.
  2. Second Multiplication $P$: Exchange the 1st and 2nd cards in your hand again**.
  3. Result: The cards are returned to their original order (Identity).

That’s why $P^2 = I$, which is $P = P^{-1}$. **Only “symmetric” permutation matrix multiplications are $I$. **

Uncorrelated Does Not Implies Independence

This is a very classic question that touches the essence of probability theory. Your screenshot image_992c2b.png has already given the mathematical definition, let me help you fill in the logical gap.

To put it simply: ** “Uncorrelated” is a low-profile version of “Independent”. It is short-sighted and can only see straight lines. **

The following is a rigorous mathematical explanation and an intuitive physical counterexample:

1. Differences in mathematical definitions

We need to see clearly what these two concepts measure:

  • Uncorrelatedness: What is measured is Linear Relationship. Its definition is based on covariance:

    $$Cov(X, Y) = E[XY] - E[X]E[Y] = 0$$

    This means: there is no linear pull between $X$ and $Y$. On average, as $X$ gets larger, $Y$ does not get proportionally larger or smaller.

  • Independence: What is measured is Any Relationship, including linear, non-linear, and higher-order ones. Its definition is based on the probability density function:

    $$f_{X,Y}(x,y) = f_X(x)f_Y(y)$$

    This means: $P(Y|X) = P(Y)$. That is to say: Knowing the value of $X$ does not help in predicting $Y$ at all.

2. Fatal counterexample: $Y = X^2$

This is the most classic counterexample in the textbook that overturns “irrelevant $\Rightarrow$ independence”.

Assume that $X$ is a random variable uniformly distributed on $[-1, 1]$ (or standard normal distribution, as long as it is symmetric about 0). Let $Y = X^2$.

Obviously: $X$ and $Y$ are absolutely not independent. Because if you tell me $X=0.5$, I can be 100% sure about $Y=0.25$. This is a strongly deterministic dependency.

**But, are they related? Let’s calculate the covariance: **

  1. $E[X] = 0$ (because the distribution is symmetric about 0).
  2. $Cov(X, Y) = E[XY] - E[X]E[Y] = E[XY] - 0 \cdot E[Y] = E[XY]$。
  3. Substitute $Y=X^2$ into: $$E[XY] = E[X \cdot X^2] = E[X^3]$$
  4. Key point: $X^3$ is an odd function. Integrating an odd function over the symmetric interval $[-1, 1]$ results in 0. $$E[X^3] = \int_{-1}^{1} x^3 \cdot \frac{1}{2} dx = 0$$

in conclusion:

  • $Cov(X, Y) = 0$ $\rightarrow$ Not relevant.
  • $Y = X^2$ $\rightarrow$ Extremely dependent.

Physical intuition: If you draw a scatter plot, it’s a parabola.

  • Linear correlation coefficient (Pearson Correlation) is the slope of a fitted straight line.
  • On this parabola, the left slope is negative, the right slope is positive, and the average slope is 0.
  • So the “myopic” eye of covariance saw that the average slope was 0, and reported: “Sir, no (linear) relationship was found!”
  • But there is actually a huge non-linear relationship behind it.

3. Why is this critical to your ICA project?

This is the core difference between PCA (Principal Component Analysis) and ICA (Independent Component Analysis):

  • PCA (Whitening/Pre-processing)： Its goal is simply Decorrelation. It rotates the coordinate axis so that the covariance matrix of the data becomes a diagonal matrix (Covariance = 0).

  • Limitations: PCA can only handle second-order statistics (variance, covariance). If your sound signal mixing method is non-linear, or the signal itself contains high-order statistical information (audio is like this), PCA cannot separate it. It can only make the signal “orthogonal”, but it cannot separate the $Y=X^2$ relationship. There is a vertical relationship between the features extracted by PCA, but verticality does not mean independence. See the next chunk for specific derivation.

  • ICA： Its goal is Independence. Not only does it require $Cov(X,Y)=0$, it also requires higher-order statistics (such as fourth-order Kurtosis) to be decoupled as well. It tries to find that transformation so that $f(x,y)$ actually breaks down into $f(x)f(y)$.

4. The only exception: Gaussian distribution

There is a special case, and it would be disrespectful not to mention it:

**If $X$ and $Y$ are jointly Gaussian-distributed, then uncorrelated $\iff$ is independent. **

This is because the Gaussian distribution is so perfect that it only needs the mean (first order) and covariance (second order) to fully describe it. Without second-order correlations, the Gaussian distribution has nowhere else to hide its “dependencies.”

**but! ** Real-world sound signals (Voice, Music) are usually Super-Gaussian (with peaks and thick tails), definitely not Gaussian distribution. This is why for audio separation, PCA alone is useless, and independent ICA must be pursued.

This is the most common, but also the most dangerous, intuition trap in PCA understanding.

The first half of your sentence is correct, but the second half is wrong. It is this error that explains why audio separation must use ICA instead of just PCA.

We need to completely separate “Perpendicular/Orthogonal” and “Independent”.

Orthogonal implies Uncorrelated, but not independence.

In PCA, the extracted feature vectors (Principal Components) are indeed geometrically perpendicular to each other. Statistically, this is equivalent to them being Uncorrelated (i.e. having a covariance of 0).

**BUT: Vertical $\neq$ is independent. **

Let’s look at an extreme geometric example to give you a glimpse of the “vertical but dependent” phenomenon.

Example: The Cross Distribution

Imagine your data points are spread out on a two-dimensional plane, shaped like a perfect “cross” (or plus sign +), centered at the origin $(0,0)$. Data points fall only on the X and Y axes.

  • **What do you think about PCA? ** PCA will find two principal axes: one for the X-axis and one for the Y-axis.

  • Are they vertical? Yes (90 degree angle).

  • Are they related? Not relevant. $Cov(X,Y) = 0$ (because one of $x$ and $y$ is always 0, the product $xy$ is always 0, and the average is also 0).

  • **What do you think of statistics (independence)? ** Try to make a prediction:

  • If I tell you $x = 5$ (non-zero), can you predict $y$?

  • **able! ** You are 100% sure that $y$ must be 0 (since the points are only on the axis).

  • If they are independent, knowing $x=5$ should not help guessing $y$. But here, knowing $x$ completely locks $y$.

Conclusion: In this cross example, the features are strictly vertical, but they have an extremely strong dependency (mutually exclusive).

2. Why is the “verticality” of PCA a bad thing for audio separation?

Back to your ICA Blind Source Separation project.

Suppose you have two microphones and record the mixed sounds of two people (Voice A and Vocal B). In the signal space, the “direction” of the two vocals is determined by the physical environment (microphone placement).

  • Realistic situation (non-orthogonal mixing): Maybe microphone 1 is on the left and microphone 2 is on the right.

  • The direction vector of vocal A may be $\vec{v}_A = [1, 0.5]$.

  • The direction vector of vocal B may be $\vec{v}_B = [0.5, 1]$.

  • **Note: The angle between these two vectors is not 90 degrees! ** They are not vertical.

  • PCA’s violent approach: If you use PCA, it forces you to find two perpendicular axes to describe the data.

  • It will find a “strongest direction” (principal component 1) and a “vertical direction” (principal component 2).

  • Results: The axis found by PCA is neither Vocal A nor Vocal B, but some sort of “orthogonal mixture” of them.

  • ICA’s approach: ICA doesn’t care about verticality. The goal of ICA is to find directions that make the signal independent.

  • It will be found that only by projecting along the two non-vertical directions of $[1, 0.5]$ and $[0.5, 1]$, the two obtained signals are truly non-interfering (statistically independent).

3. Summary chart

ConceptGeometric significanceStatistical significanceAlgorithm representationAbility
Orthogonal (Orthogonal)Vector dot product is 0 ($\vec{u} \cdot \vec{v} = 0$)Uncorrelated (Uncorrelated, Cov=0)PCACan only remove second-order linear correlations
Independent (Independent)Joint probabilistic decoupling ($P(xy) = P(x)P(y)$)Mutual information is 0 (Mutual Info=0)ICACan remove all high-order dependencies

One sentence summary for engineers (yourself):

PCA is an “obsessive-compulsive” algorithm that must break the coordinate axis of the data into a 90-degree right angle. But in the real world, sound sources (Sources) are often mixed together at any angle (obliquely). So you need ICA, which allows the axes to be non-orthogonal as long as this decouples the signal.

Clarification of misunderstandings between PDF and CDF

Here you come across the most confusing concept in probability theory, but the most fatal conceptual misunderstanding for continuous signal processing (such as ICA).

Your understanding is half right and half wrong. The wrong half is precisely the fundamental difference between “discrete” and “continuous” in mathematical analysis.

Let’s cut these two concepts apart like calculus surgery:

1. For PDF ($f$): your equation is wrong in the continuous case

You wrote: $P(X = \text{constant}) = f_X(x = \text{constant})$.

  • In the discrete case (PMF): This is correct. For example, throwing dice, $P(X=1) = 1/6$.
  • In the Continuous Case (PDF): This is so wrong.

**The fact is: for a continuous random variable (such as your audio signal), the probability of any specific point is 0. **

$$P(X = \text{constant}) = 0$$

**Why? ** Because $f_X(x)$ represents Density, not Probability. *Imagine an iron rod this meter long (total mass 1).

  • $f(x)$ is its density at $x$. Density can be 10, or 100 (as long as it’s very narrow).
  • But, what is the mass of this single point $x$? is 0. Because a single point has no width.

Correct physical meaning: $f_X(x)$ is the “rate of change” of the probability. Only when multiplied by a microelement $dx$ does it become a probability:

$$P(x < X < x + dx) \approx f_X(x) \cdot dx$$

2. For CDF ($F$): Your understanding is correct

You wrote: cdfyesintegral(value to value)fx(x).

**Exactly correct. ** The CDF (cumulative distribution function) $F_X(x)$ is essentially the variable upper limit integral of PDF:

$$F_X(x) = P(X \le x) = \int_{-\infty}^{x} f_X(t) \, dt$$

This is exactly the application of the Fundamental Theorem of Calculus:

  • CDF is the original function (integral) of PDF: $F(x) = \int f(x)$
  • PDF is the derivative of CDF: $f(x) = F'(x)$

If you ask for the probability of a certain interval (for example, the volume is between 0.5 and 0.8), that is the integral of what you call “value to value”:

$$P(a < X < b) = \int_{a}^{b} f_X(x) \, dx = F_X(b) - F_X(a)$$

3. Why is this distinction important in ICA?

When doing derivation of ICA (such as calculating negentropy), you will frequently encounter the expected value $E[G(y)]$.

According to the definition of expectation (continuous):

$$E[G(y)] = \int_{-\infty}^{+\infty} G(y) f_Y(y) \, dy$$

Note that $f_Y(y)$ (density) is used as the weight for integration. If you mistake $f_Y(y)$ for $P(Y=y)$, when doing variable substitution or derivation (Score Function), the dimensions will not match, and the entire gradient descent formula will be derived incorrectly.

To summarize the revised understanding:

  1. PDF ($f$) = Height (Height/Density). Values can be greater than 1, or even infinity (such as Dirac delta). The probability of a single point is 0.
  2. CDF ($F$) = Area (Area/Probability). The value is always between 0 and 1. Monotonically increasing.
  3. Probability = Volume/Mass = $f(x)dx$ (density $\times$ width).

Covariance = 0 <-> Correlation = 0

This is a very keen intuition, but from a strictly mathematical perspective, their meanings are Mathematically Equivalent if and only if they are both equal to 0. **

Their meanings (mainly in terms of magnitude) differ only if they are not equal to 0**.

We can directly pierce this layer of window paper through the formula:

1. “Parent-child relationship” in mathematics

$$ \text{Correlation} (\rho_{X,Y}) = \frac{\text{Covariance} (Cov(X,Y))}{\sigma_X \sigma_Y} $$
  • Numerator is the covariance.
  • The denominator is the product of the standard deviations of the two variables, which is just a scaling factor.

**As long as the standard deviations $\sigma_X$ and $\sigma_Y$ are not 0 (i.e. the variables are not constants), then it is obvious: **

$$ \text{Covariance} = 0 \iff \text{Correlation} = 0 $$

Conclusion: On the qualitative question of judging “whether there is a linear relationship” (that is, when judging whether it is 0), they are exactly the same. If the covariance is 0, the correlation coefficient must be 0, and vice versa.

2. Since they are the same, why have two names?

The difference is how you “read” the number when they are not 0.

Covariance: “Absolute value” with units

  • Physical meaning: It is “the degree of common variation of the original data”.
  • Units: Its units are the units of $X$ multiplied by the units of $Y$.
  • Example: If $X$ is height (meters), $Y$ is weight (kg).
  • The unit of $Cov(X, Y)$ is “meter·kilogram”.
  • Question: If it is not 0, such as $Cov = 500$. Is 500 considered large or small?
  • **Unable to answer. ** Because if you change the height unit to “mm”, the covariance will instantly become 1000 times larger, becoming 500,000. But the correlation between the two heights has not actually changed.

Correlation: Normalized “relative value”

  • Physical meaning: It is “the linear correlation strength after removing the unit influence”.
  • Unit: Unitless.
  • In the formula, the unit of the numerator (meter·kilogram) is perfectly canceled by the unit of the denominator (meter $\times$ kilogram).
  • Advantage: It is enforced between $[-1, 1]$.
  • $0.9$ must be strong, $0.1$ must be weak. Whether you use meters or millimeters, it never changes.

3. The only mathematical special case (Edge Case)

There is only one case, Covariance = 0 but Correlation is undefined (or not equal to 0):

**When the variable is a constant (Variance = 0). **

  • If $X$ is always equal to 5 (constant).
  • Covariance: $E[(5-5)(Y-\mu_y)] = 0$. The covariance is 0.
  • Correlation: Denominator $\sigma_X = 0$. Division by Zero, Undefined/NaN.

But in your ICA audio processing scenario, the signal obviously fluctuates (variance is not 0), so this special case can be ignored.

Summarize

For the Preprocessing (Whitening) steps in your ICA project:

  • The state you want to achieve is Uncorrelated.
  • At this time, you can say “I want to make the covariance matrix diagonal (off-diagonal lines are 0)”, or you can say “I want to make the correlation coefficient matrix diagonal”.
  • **In this context, they refer to the same geometric state: vectors are orthogonal. **

What do we get under the assumption that $y_1$ and $y_2$ are Independent?

Simply put: **When two variables are independent of each other, their “sum of the variances” is equal to the “sum of the variances”. **

Let us break down why $\text{Var}(y_1 + y_2) = \sigma_1^2 + \sigma_2^2$ in detail through mathematical derivation and intuitive understanding.

1. Mathematical derivation (using the definition of expectation)

We have assumed in the previous step that without loss of generality (WLOG) the mean is 0 ($E[y_1]=0, E[y_2]=0$). In this case, the variance is defined as the second moment of origin: $\text{Var}(y) = E[y^2]$.

Let’s look at the variance of $y_1 + y_2$:

$$\text{Var}(y_1 + y_2) = E[(y_1 + y_2)^2]$$

Step 1: Expand the square term According to the algebraic formula $(a+b)^2 = a^2 + b^2 + 2ab$, we expand the brackets:

$$E[(y_1 + y_2)^2] = E[y_1^2 + y_2^2 + 2y_1y_2]$$

Step 2: Take advantage of the linear properties of expectations Expect $E[\cdot]$ to be linear and unpackable:

$$= E[y_1^2] + E[y_2^2] + 2E[y_1y_2]$$

Step 3: Key point - processing cross-term (Cross-term) $2E[y_1y_2]$ appears here. This term actually corresponds to Covariance.

  • Because $y_1$ and $y_2$ are independent and have a mean of 0.
  • According to the nature of independence: $E[y_1y_2] = E[y_1] \cdot E[y_2]$.
  • Because $E[y_1]=0$ and $E[y_2]=0$, therefore: $$E[y_1y_2] = 0 \cdot 0 = 0$$

Step 4: Draw Conclusion Because the cross terms disappear, only:

$$= E[y_1^2] + E[y_2^2]$$

$$= \text{Var}(y_1) + \text{Var}(y_2)$$

$$= \sigma_1^2 + \sigma_2^2$$

2. What will happen if we are not independent?

If $y_1$ and $y_2$ are not independent (that is, they are related), then the middle term $2E[y_1y_2]$ will not be 0, and the formula will become:

$$\text{Var}(y_1 + y_2) = \sigma_1^2 + \sigma_2^2 + 2\text{Cov}(y_1, y_2)$$

It is precisely because the question emphasizes “independence” that the covariance $\text{Cov}(y_1, y_2)$ is 0, and the formula can be simplified to what it looks like in your picture.

3. Intuitive understanding (analogy to Pythagoras’ theorem)

You can think of independent random variables as mutually perpendicular (orthogonal) vectors in a geometric space.

  • $\sigma_1$ is the length of vector A.
  • $\sigma_2$ is the length of vector B.
  • Because they are “independent”, their angle is 90 degrees.
  • Their sum ($y_1+y_2$) is the hypotenuse. *According to the Pythagorean theorem: The square of the hypotenuse = the square of the right-angled side A + the square of the right-angled side B.

That is: $\text{total variance} = \sigma_1^2 + \sigma_2^2$.

Summarize

The reason for this step is that the expectation of the expanded cross term $2y_1y_2$ becomes 0 due to the “independence” and “zero mean” of the variables. **

Start with joint = PDF1 * PDF2:

This is exactly the point! The Joint PDF = PDF1 * PDF2 you mentioned is actually the source of this expected property.

To put it simply: **$E[y_1 y_2] = E[y_1] \cdot E[y_2]$ This conclusion is obtained by integrating the PDF product formula you are familiar with. **

Let me walk you through it using the most intuitive mathematical derivation (integration), and you will understand it right away.

1. Start from the definition

You already know that the definition of independence is:

$$f(y_1, y_2) = f(y_1) \cdot f(y_2)$$

The essence of the expectation $E[\cdot]$ is the “weighted average”, which is the integral in continuous variables. We want to calculate the expectation of the product of $y_1$ and $y_2$, which by definition means multiplying $y_1 \cdot y_2$ by the joint probability density and then integrating over the entire space:

$$E[y_1 y_2] = \iint (y_1 \cdot y_2) \cdot f(y_1, y_2) \, dy_1 \, dy_2$$

2. Substitute the formula you know well

Because they are independent, we replace $f(y_1, y_2)$ with the product form you mentioned:

$$E[y_1 y_2] = \iint (y_1 \cdot y_2) \cdot [f(y_1) \cdot f(y_2)] \, dy_1 \, dy_2$$

3. Witness the miraculous moment (split points)

Now we want to exploit the properties of integrals. Note that some items in the formula are only related to $y_1$, and some are only related to $y_2$. We can separate them completely:

$$E[y_1 y_2] = \left( \int y_1 \cdot f(y_1) \, dy_1 \right) \cdot \left( \int y_2 \cdot f(y_2) \, dy_2 \right)$$

The reason why this step can be split is entirely because the $f(y_1, y_2)$ in the middle can be split into products (that is, the independence definition you mentioned). ** If it cannot be disassembled, this step will not be possible.

4. Recognize who they are

Look at what’s left in parentheses:

  • The left bracket $\int y_1 f(y_1) \, dy_1$ is exactly the definition of $E[y_1]$.
  • The right bracket $\int y_2 f(y_2) \, dy_2$ is exactly the definition of $E[y_2]$.

so:

$$E[y_1 y_2] = E[y_1] \cdot E[y_2]$$

Summary and Supplements

  1. Intuitive understanding:
  • The properties of your PDFproduct indicate that the two variables do not interfere with each other in the probability distribution.
  • This property of Expected product indicates that the two variables do not interfere with each other in the average trend of values.
  • If they are not independent (for example, if $y_1$ becomes larger, $y_2$ will also become larger), then $E[y_1 y_2]$ will be larger than simple $E[y_1]E[y_2]$ (because large numbers multiplied by large numbers will pull up the average), and the difference between them is Covariance.
  1. Role in this question:
  • The title not only says that they are independent ($E[y_1 y_2] = E[y_1]E[y_2]$), but also says that they are zero-mean, that is, $E[y_1]=0, E[y_2]=0$.
  • So the result directly becomes: $0 \times 0 = 0$.
  • This is why when calculating the variance $(y_1+y_2)^2$ expansion, the middle cross term $2y_1y_2$ disappears directly.