Q1: Regarding the derivation of $E[zz^T] = I$

This is a great mathematical detail question! This step is often confusing because it compresses together concepts from linear algebra (matrix operations) and statistics (expectation and variance).

To understand why $E[zz^T] = I$, we need to break it down into three steps:

  1. What exactly is **$zz^T$? **(It is not a number, it is a matrix)
  2. What did **$E[\dots]$ do when entering this matrix? **
  3. Why did ** finally become $I$? **

Step 1: Disassemble $zz^T$ (outer product vs inner product)

This is where things get most confusing.

  • $z^T z$ is the inner product and the result is a number (scalar).
  • $zz^T$ is the outer product, and the result is a matrix** of **$k \times k$.

Let’s expand the vector $z$ and assume that $z$ has only 2 elements (for ease of understanding):

$$z = \begin{bmatrix} z_1 \\ z_2 \end{bmatrix}$$

Then the operation process of $zz^T$ is as follows:

$$ zz^T = \begin{bmatrix} z_1 \\ z_2 \end{bmatrix} \begin{bmatrix} z_1 & z_2 \end{bmatrix} = \begin{bmatrix} z_1 \cdot z_1 & z_1 \cdot z_2 \\ z_2 \cdot z_1 & z_2 \cdot z_2 \end{bmatrix} = \begin{bmatrix} z_1^2 & z_1 z_2 \\ z_2 z_1 & z_2^2 \end{bmatrix} $$

You see, in the resulting matrix:

  • On the diagonal is the square ($z_1^2, z_2^2$) of each element.
  • Off-diagonal is the product of different elements ($z_1 z_2$).

Step 2: Add expectations $E[\dots]$

When we say $E[zz^T]$, we are actually putting the desired symbol $E$ into each cell of the matrix:

$$ E[zz^T] = \begin{bmatrix} E[z_1^2] & E[z_1 z_2] \\ E[z_2 z_1] & E[z_2^2] \end{bmatrix} $$

Now, each element in this matrix requires us to calculate a specific value based on known conditions.

Step 3: Utilize the assumptions of $z \sim \mathcal{N}(0, I)$

The core here lies in the conditions given in the question: $z$ obeys the standard normal distribution, and each dimension is independent.

what does that mean?

  1. Mean is 0: $E[z_1] = 0, E[z_2] = 0$.

  2. The variance is 1 (diagonal of the identity matrix $I$): By the variance formula $Var(X) = E[X^2] - (E[X])^2$, because $E[X]=0$, so $Var(X) = E[X^2]$. Because the variance of the standard normal distribution is 1, then:

    $$E[z_1^2] = 1$$

    $$E[z_2^2] = 1$$

    (This means that the diagonal of the matrix is ​​all 1)

  3. Independent and uncorrelated (the off-diagonal of the identity matrix $I$ is 0): Because $z_1$ and $z_2$ are independent, their expectations can be unpacked:

    $$E[z_1 z_2] = E[z_1] \cdot E[z_2] = 0 \cdot 0 = 0$$

    (This means that the off-diagonals of the matrix are all 0)

in conclusion

Fill the numbers calculated above back into the matrix just now:

$$ E[zz^T] = \begin{bmatrix} \mathbf{1} & \mathbf{0} \\ \mathbf{0} & \mathbf{1} \end{bmatrix} = I $$

That’s why $E[zz^T] = I$.

To summarize the derivation chain:

  1. Expansion: $zz^T$ produces a matrix containing square terms and cross product terms.
  2. Zero mean: Because the mean is 0, the expectation of the square term is equal to the variance, and the expectation of the cross term is equal to the covariance.
  3. Standard distribution: Because it is a standard normal distribution, the variance is 1 (diagonal) and the covariance is 0 (off-diagonal).
  4. Result: Form the identity matrix $I$.

Q3: The Inner/Outer Products of the orthogonal matrix $Q$ are all $I$

As long as $Q$ is an Orthogonal Matrix, then $Q^T Q = I$ and $Q Q^T = I$ are true at the same time.

Here is a detailed explanation to help you clarify the logic:

1. Why do you think $Q^T Q = I$ is right?

Your intuition is very accurate. This is the original definition of an orthogonal matrix: if the column vectors (Columns) of a matrix are orthogonal and have length 1 (orthonormal basis), then $Q^T Q = I$.

2. Why is $Q Q^T = I$ in the picture also correct?

In mathematical definitions, when we refer to an “orthogonal matrix”, by default it is a Square Matrix.

For square matrices, there is a very key property: **If $Q^T$ is the left inverse of $Q$, then it must also be the right inverse of $Q$. **

The derivation logic is as follows:

  1. It can be seen from the definition: $Q^T Q = I$.
  2. This means that $Q$ is invertible, and its inverse matrix is ​​the transposed matrix: $Q^{-1} = Q^T$.
  3. According to the definition of inverse matrix, $A A^{-1} = A^{-1} A = I$.
  4. Since $Q^{-1} = Q^T$, then substitute $Q^T$ into the position of $Q Q^{-1}$ and you will get: $$Q Q^T = I$$

3. Intuitive understanding

  • $Q^T Q = I$ represents: the column vectors of $Q$ are perpendicular to each other and have a length of 1.
  • $Q Q^T = I$ represents: the row vectors of $Q$ are perpendicular to each other and have a length of 1.

For orthogonal square matrices, if the column vectors are orthogonal, it follows that the row vectors are also orthogonal.

Easily confusing special cases

The reason why you have questions may be because you have seen non-square matrix situations (that is, rectangular matrices).

If $Q$ is a matrix of $m \times n$ (e.g. $m > n$, a tall and thin matrix), and the column vectors are orthogonal:

  • $Q^T Q = I$ still holds (columns are orthogonal).
  • But $Q Q^T \neq I$ (it becomes a projection matrix).

Summary: Since the picture clearly states that the properties of “orthogonal matrix” are used, it refers to the square matrix by default, so $Q Q^T = I$ is completely correct.

Clarify expression: Is $L$ a Scaler in the context of $Lq_i$?

No, $L$ is not a scalar, and the calculation result $Lq_j$ is not a scalar either.

This is actually a way of writing Block Matrix Multiplication.

Let’s break down the dimensions and you’ll get it very clear:

1. Disassemble the dimensions

Suppose we are doing factor analysis:

  • $L$ (Loadings Matrix): It is a matrix** of **$p \times r$ ($p$ is the number of variables, $r$ is the number of factors).
  • $Q$ (Rotation Matrix): is a matrix** of **$r \times r$.
  • $q_j$: It is the th $j$ column of matrix $Q$. Because it comes from the matrix of $r \times r$, $q_j$ is a column vector of $r \times 1$.

2. Operation process $Lq_j$

When you calculate $L$ times $q_j$, the dimensions change as follows:

$$ \underbrace{L}_{p \times r} \times \underbrace{q_j}_{r \times 1} = \underbrace{\text{result}}_{p \times 1} $$

Therefore, $Lq_j$ is a column vector of $p \times 1$ (that is, a new load column), not a scalar value.

3. Understand the meaning of that formula

The formula is written as:

$$ \tilde{L} = (\tilde{\ell}_1, ..., \tilde{\ell}_r) = (Lq_1, ..., Lq_r) $$

This is not listing a bunch of numbers, but assembling a matrix:

  1. Calculate the first column vector $Lq_1$ (this is a vertical bar).
  2. Find the second column vector $Lq_2$.
  3. …
  4. Finally calculate the $r$th column vector $Lq_r$.
  5. Put these “vertical bars” (column vectors) side by side.

Summarize

  • $L$ is still the large load matrix.
  • $Lq_j$ is the rotated loading vector** for the $j$th factor (an entire column).
  • $\tilde{L}$ is a new $p \times r$ matrix (that is, the rotated loading matrix) formed by combining these newly calculated columns.

Proof that the load matrix is ​​redistributed after Varimax rotation, and each component is not necessarily consistent.

Your understanding is very accurate. This is the core geometric property of “quadratic form” in linear algebra.

You understand the problem very thoroughly. Let me dismantle your sentence and confirm the details (slightly correcting the symbolic expression):

1. Your conclusion is completely correct

You say: “The result is unique only if $L^T L$ is $I$.” **Yes. **

More precisely:

  • If $L^T L = I$ (or $kI$, which is a scalar matrix): No matter how you rotate $q_j$, the calculated result (variance/modulus squared) is constant.
  • If $L^T L \neq I$ (and the eigenvalues ​​are not all equal): The calculated result must fluctuate depending on the angle of $q_j$.

2. “Isotropy” in mathematics

Let us write your logic out as a formula, and you will find that it is as beautiful as a circle.

Assumption A: $L^T L = I$ (perfect sphere) We want to calculate $\text{Value} = q_j^T (L^T L) q_j$. Substitute $I$ in:

$$ \text{Value} = q_j^T (I) q_j = q_j^T q_j $$

Because $q_j$ is a column of an orthogonal matrix, which is a unit vector, so $q_j^T q_j \equiv 1$. Conclusion: No matter how you turn it, the result will always be 1. It’s like you’re touching a perfect ball. No matter which angle you touch it from, the radius is the same.

What-if B: $L^T L \neq I$ (football/ellipsoid) For example $L^T L = \text{diag}(10, 2)$.

$$ \text{Value} = q_j^T \begin{bmatrix} 10 & 0 \\ 0 & 2 \end{bmatrix} q_j $$

At this time, the matrix in the middle plays a “weighting” role. It amplifies some components of $q_j$ ($\times 10$) and reduces some components ($\times 2$). Conclusion: The result is no longer simply $q_j^T q_j$ (i.e. 1), but depends on whether the component of $q_j$ hits “10” or “2”.

3. What does this mean for factor analysis?

This actually reveals a very profound truth:

Why can we do rotation? (Why Rotation works?) It is precisely because the $L^T L$ ** calculated from the actual data is usually not $I$**.

  • Among the initially extracted factors, the first factor often has a particularly large variance (large eigenvalue), and the second factor has a small variance.
  • This means that the shape of the data is an “ellipsoid”.
  • Precisely because it is an ellipsoid, we have the opportunity to rotate the coordinate axis so that $q_j$ points in different directions, thus changing the projection length in this direction (that is, changing the variance contribution of the factor), achieving the effect of “maximizing the gap between the rich and the poor” that Varimax wants.

on the contrary: If the original data has a $L^T L = I$ (completely spherical distribution), then no matter how you do the Varimax rotation, it will be in vain. Because the ball looks round no matter how it spins, the load can never be optimized.

Identifiability related: Is the corresponding Q unique after determining the diagonal matrix?

The short answer is: **In the vast majority of cases, yes, it is the only one. **

But if you want to dig deeper from a rigorous mathematical perspective, the answer is: **It is unique under the premise of “regardless of the sign (positive and negative sign)” and “the order of the columns”. **

Let me explain why $Q$ (and subsequently $L$) is unique based on the Diagonalization constraint:

1. The essence is “Eigendecomposition”

That formula you see in the picture:

$$\tilde{L}^T \Psi^{-1} \tilde{L} = \Delta \quad (\text{Diagonal Matrix})$$

This step is mathematically equivalent to doing eigenvalue decomposition.

  • The $Q$ (or the initial $L$) we are looking for is actually a matrix composed of Eigenvectors.
  • The diagonal elements in $\Delta$ (diagonal matrix) are Eigenvalues.

2. Three prerequisites for uniqueness

As long as the data is normal (not artificial data where everyone scores exactly the same), the direction of the feature vector is fixed. However, there are three little “cheap” places in mathematics:

A. Sign Indeterminacy - “It doesn’t matter whether the sign is positive or negative”

Mathematical formulas cannot distinguish between $v$ and $-v$.

  • For example: one factor represents “intelligence”.
  • $Q$ may work out to be positive: the loading is 0.8 (meaning the higher the intelligence, the higher the score).
  • $Q$ may also work out to be negative: the loading is -0.8 (meaning the higher the “non-intelligence”, the lower the score).
  • These two cases are mathematically $LL^T$ identical. Therefore, the positive and negative signs of the results calculated by the computer may be random, but this does not affect the uniqueness (it is essentially a line).

B. Order Indeterminacy - “It doesn’t matter who is the boss”

The element $d_1, d_2, ...$ in the diagonal matrix can be placed in any position.

  • In order to solve this problem, all statistical software will force sorting: put the one with the largest feature value in the first column (Factor 1), the second largest one in the second column, and so on.
  • **With this artificial regulation, the solution is the only one. **

C. Repeated Eigenvalues ​​- “Extremely rare circles”

If there are two numbers in $\Delta$ that are exactly the same (such as $d_1 = d_2 = 5$), then $Q$ is not unique at this time.

  • Just like a perfect circle, no matter how you turn it, it is always diagonal.
  • But in real world real data, this situation is almost impossible to happen.

Summarize

The purpose of the “diagonalization” step you mentioned is to give the computer a deadline command: “Among countless possible solutions, please give me the specific solution that is sorted by variance contribution and is mathematically perpendicular to each other.”

**So, as long as you ignore sign flipping and accept the rules of sorting by size, this solution is unique. ** This is why we can always get a fixed “Initial Solution” before doing Varimax rotation.

What does $r=1$ mean in the context of FA?

What does r represent in the context of this question? Why does r=1 get the expression of that covariance matrix? If r is not equal to 1, what will I see?

answer:

Alex, this is a very pertinent question. Understanding the meaning of $r$ is key to understanding factor analysis matrix operations.

Let’s look at these three questions in turn:

1. What does $r$ mean in this context?

$r$ represents the number of common factors, which is the dimension of the latent variable $z$.

  • $p = 3$: We have 3 observation variables ($X_1, X_2, X_3$), which is the number of rows/columns of the matrix.
  • $r = 1$: We assume that the changes behind these 3 variables are driven only by 1 hidden root cause (factor $z_1$).

In a practical context, assume that these 3 variables are students’ three test scores (mathematics, physics, chemistry).

  • If $r=1$: We assume that only one “science talent” factor determines the results of these three courses.
  • If $r=2$: We may assume that there is a “logical ability” factor and a “memory ability” factor that jointly determine the score.

2. Why does $r=1$ get the expression of that covariance matrix?

This is because when $r=1$, the factor loading matrix $L$ degenerates into a column vector.

Let’s look at the details of matrix multiplication:

  • Dimensions: The dimensions of $L$ are $p \times r$.
  • When $r=1$: $L$ is the matrix of $3 \times 1$, which is $L = \begin{pmatrix} \ell_{11} \\ \ell_{21} \\ \ell_{31} \end{pmatrix}$.

Now we calculate $LL^T$ (which is an Outer Product):

$$ LL^T = \begin{pmatrix} \ell_{11} \\ \ell_{21} \\ \ell_{31} \end{pmatrix} \begin{pmatrix} \ell_{11} & \ell_{21} & \ell_{31} \end{pmatrix} = \begin{pmatrix} \ell_{11}\ell_{11} & \ell_{11}\ell_{21} & \ell_{11}\ell_{31} \\ \ell_{21}\ell_{11} & \ell_{21}\ell_{21} & \ell_{21}\ell_{31} \\ \ell_{31}\ell_{11} & \ell_{31}\ell_{21} & \ell_{31}\ell_{31} \end{pmatrix} $$

This explains why the off-diagonal elements of the matrix you see are very simple, just the simple product of two numbers (e.g. $\Sigma_{12} = \ell_{11}\ell_{21}$).

Intuitive meaning: Because there is only one factor, $z_1$, all correlations between variable 1 and variable 2 ($\Sigma_{12}$) must be completely transmitted through this single factor.

3. If $r$ is not equal to 1 (such as $r=2$), what will I see?

Things get more complicated (and more flexible) if $r=2$.

  • The shape of $L$: becomes the matrix of $3 \times 2$.

    $$ L = \begin{pmatrix} \ell_{11} & \ell_{12} \\ \ell_{21} & \ell_{22} \\ \ell_{31} & \ell_{32} \end{pmatrix} $$

    (Each row has two numbers, representing the influence of factor 1 and factor 2 on the variable)

  • Calculation of $LL^T$: Matrix multiplication now becomes “row times column” (dot product). For example, let’s look at the element $\Sigma_{12}$ in the first row and second column:

    $$ \Sigma_{12} = (\text{Row } 1 \text{ of } L) \cdot (\text{Row } 2 \text{ of } L)^T $$

    $$ \Sigma_{12} = \ell_{11}\ell_{21} + \ell_{12}\ell_{22} $$

Changes You Will See: The covariance is no longer a simple multiplication of two numbers, but the sum of the products of two sets of numbers.

  • Physical meaning: Why are variable 1 and variable 2 related?
  • Partly because they are both affected by factor 1 ($\ell_{11}\ell_{21}$).
  • The other part is because they are both affected by factor 2 ($\ell_{12}\ell_{22}$). *The total correlation is the sum of these two components.

The impact of this on problem solving: Back to the unsolvable question just now. If $r=2$, we have a lot more free parameters ($\ell_{12}, \ell_{22}, \ell_{32}$). With these extra “tuning knobs” we can usually easily find solutions without the awkward situation of $\ell_{11} > 1$ or negative variance.

Simple summary: The smaller **$r$ is, the simpler the model is and the more restrictive it is (it is easy to have no solution); the larger $r$ is, the more flexible the model is and the easier it is to fit the data. **

Q: Are variable 1 and variable 2 data points or features?

answer:

This is a very important conceptual distinction, Alex.

**Not a data point. **

Here, “Variable 1” and “Variable 2” refer to the Features or Dimensions of the data, that is, the Columns in the data table.

To give you a thorough understanding, we map this concept to a specific Excel table:

1. Intuitive example: student performance table

Suppose you are analyzing the test scores of a class.

Variable 1 ($X_1$)Variable 2 ($X_2$)Variable 3 ($X_3$)
MeaningMath ScorePhysics ScoreChinese Score
Data Point 1 (Student A)859070
Data Point 2 (Student B)606580
Data Point 3 (Student C)………
  • Variables ($X_1, X_2$): Refers to “Math” and “Physics” at the top of the table.
  • Data Points: refers to each row in the table, that is, the specific “Student A” and “Student B”.

2. Return to the $\Sigma$ matrix in the question

The covariance matrix $\Sigma$ in the question is a matrix of $3 \times 3$:

$$ \Sigma = \begin{pmatrix} 1 & 0.9 & 0.7 \\ 0.9 & 1 & 0.4 \\ 0.7 & 0.4 & 1 \end{pmatrix} $$
  • This matrix describes the relationship between “columns and columns”.
  • That 0.9 ($\Sigma_{12}$) means: ** There is a strong positive correlation between “math scores” ($X_1$) and “physics scores” ($X_2$).
  • It does not mean that “Student A” is similar to “Student B”.

3. What is $\ell_{11}, \ell_{21}$ in the $L$ matrix?

When we say $\ell_{11}\ell_{21} \approx \Sigma_{12}$, we are explaining why the grades in these two courses are related:

  • $\ell_{11}$: How much influence factor $z$ (e.g. “logical thinking ability”) has on math ($X_1$).
  • $\ell_{21}$: How much influence does factor $z$ (again the “logical thinking ability”) have on physics ($X_2$).

Because logical thinking ability strongly affects both mathematics ($\ell_{11}$ is very large) and physics ($\ell_{21}$ is very large), when we observe the data, we will find that people who are good at mathematics are usually good at physics (that is, $X_1$ and $X_2$ are highly correlated at 0.9).

Summarize:

  • Variables = Test Subject ($X_1, X_2, X_3$) $\leftarrow$ The object we are discussing here
  • Data Points = The specific student $\leftarrow$ is the raw material used to calculate the covariance, but is hidden behind the scenes after the $\Sigma$ matrix is ​​formed.

Is this distinction clear?

Question: So this covariance matrix actually has nothing to do with students. It just means that when we are in fa, the task r=1, which means that we think that the three features are actually explained by a single z. Is this understanding okay?

Answer: Absolutely correct! Your understanding is spot on.

We can summarize what you just learned like this:

  1. About Covariance Matrix: It’s really an abstraction and aggregation of “students” (data points). Once we calculate $\Sigma$, the specific students and their scores no longer appear in the formula. We only care about the “property” (that is: when the math scores are high, are the physics scores usually also high?).

  2. Assumptions about $r=1$: This is the core philosophy of Factor Analysis. When we set $r=1$, we are forcing the model to accept this worldview:

“Although these 3 courses (3 Features) look different, there is only one driving force behind them.”

Why does this assumption lead to “no solution” to the question just now?

Combining your understanding just now, we can take a deeper look at the “unsolvable” question:

  • Real world (covariance matrix $\Sigma$) says: *The correlation between mathematics and physics is 0.9 (very strong).

  • The correlation between mathematics and Chinese is 0.7 (strong).

  • The correlation between Physics and Chinese is 0.4 (moderate).

  • Your hypothesis ($r=1$) says:

  • Because there is only one $z$ (such as “IQ”), the high correlation between mathematics and physics must mean that they both rely heavily on this $z$ ($\ell$ is large).

  • Since they are very dependent on $z$, their correlation with language should also be very regular (the product relationship of $\ell_i \ell_j$ must be satisfied).

  • Conflict Point: When you try to use a $z$ to satisfy the three sets of relationships of 0.9, 0.7, 0.4 at the same time, the mathematics “collapses”. Just like you are trying to tell everyone: “The reason why Xiao Ming is good at math, good at physics, and average at Chinese is all because he is ‘smart’ (single factor).” *Explains good math and physics $\to$ deduces that he is “extremely smart”.

  • But his Chinese skills are only average $\to$ Since he is extremely smart, his Chinese skills should not be just average.

  • Conclusion: A single “smart” factor cannot be explained. $r=2$ may be needed (e.g. introducing a “Language Talent” factor and a “Logic Factor”).

Therefore, doing Factor Analysis is actually asking: “Can I use fewer stories ($z$) to round out these complex phenomena ($X$)?”

This chapter does a great job establishing your current intuition, especially the mental leap that separates the Matrix from specific Data Points. If there are no other questions, can we move on to the next question?

$E(Z|X)$ is a good estimate mentioned in Q5, what is its mathematical proof?

The question you asked is one of the most famous theorems in statistical decision theory: the minimum mean square error (MMSE) estimator is the conditional expectation. **

I will explain it to you in three levels: Intuitive Example $\to$ Mathematical Proof (Core) $\to$ Geometric Explanation.

1. Intuitive understanding: Why choose “mean”?

Suppose we don’t talk about formulas at all.

Scenario A: Blind Guessing I want you to guess what the average $z$ score of the class in this course is. There is a penalty (squared error) for guessing wrong. You don’t know any information. What would you guess? You would guess the average score for the class (say 75). Because if you guess 0 or 100, the risk is too great. **The mean is the center point that balances all possibilities. **

Scenario B: Guessing with clues ($x$) Now let me tell you a piece of information $x$: “This classmate never attends class and never turns in homework.” What is your current guessing strategy? You throw away the “class average” and think instead: “What’s the average grade among students who never attend class?” Maybe the average score for this particular group is only 40 points.

This is the essence of $E[z|x]$:

  • It’s not just “expected”, it’s “that is, the local average under this $x$ condition”.
  • Picking any value that deviates from this “local center” will cause your predictions to skew to one side, thus increasing the average squared error.

2. Mathematical derivation: Why does it have to be it? (core part)

We need to prove it mathematically.

Goal: We want to find a function $f(x)$ that minimizes the mean square error (MSE).

$$J(f) = E \left[ (z - f(x))^2 \right]$$

Proof technique: This is a classic “Add and Subtract” (Add and Subtract). We force insert a term $E[z|x]$ into the square term, splitting it into two parts.

Let $g(x) = E[z|x]$ be the best answer we want to verify. We rewrite $z - f(x)$ as:

$$z - f(x) = \underbrace{(z - g(x))}_{\text{Part A}} + \underbrace{(g(x) - f(x))}_{\text{Part B}}$$

Substitute the above equation into the MSE formula to expand:

$$ \begin{aligned} MSE &= E \left[ \left( (z - g(x)) + (g(x) - f(x)) \right)^2 \right] \\ &= \underbrace{E[(z - g(x))^2]}_{\text{First term}} + \underbrace{E[(g(x) - f(x))^2]}_{\text{Second term}} + \underbrace{2E \left[ (z - g(x))(g(x) - f(x)) \right]}_{\text{Cross term}} \end{aligned} $$

Key step: Prove that the “cross term” is 0

Let’s look at the third term (the cross term):

$$2E \left[ (z - E[z|x]) \cdot (E[z|x] - f(x)) \right]$$

Utilize the Law of Iterated Expectations: $E[Y] = E[E[Y|x]]$. We first fix $x$ for internal expectations:

$$ E[\text{cross term}] = E_x \left[ \quad E_{z|x} \left[ (z - E[z|x]) \cdot (E[z|x] - f(x)) \;\bigg|\; x \right] \quad \right] $$

Conditioned on $x$:

  1. $E[z|x]$ is a constant.
  2. $f(x)$ is also a constant (because $x$ is known).
  3. Therefore, this part of $(E[z|x] - f(x))$ is just like the constant $k$ for internal expectations and can be mentioned directly to the outside.

So the internal expectation becomes:

$$ (E[z|x] - f(x)) \cdot E_{z|x} [ (z - E[z|x]) \mid x ] $$

Look at this little $E_{z|x} [ z - E[z|x] \mid x ]$ at the back:

$$ = E[z|x] - E[E[z|x]|x] = E[z|x] - E[z|x] = 0 $$

(because the expectation of $z$ in $x$ condition, minus its own expectation, of course the average deviation is 0)

**So, the cross term completely disappears! **

Final Result Analysis

The MSE now becomes the sum of two parts:

$$ MSE = \underbrace{E[(z - E[z|x])^2]}_{\text{Irreducible error}} + \underbrace{E[(E[z|x] - f(x))^2]}_{\text{Reducible error}} $$
  1. First term (irreducible error): This is the conditional variance of $z$ itself $\text{Var}(z|x)$. No matter how you choose $f(x)$, this item has nothing to do with you and you cannot change it (this is the noise of the data itself).
  2. Second term (reducible error): This is a square term, so it is constant $\ge 0$.

in conclusion: To minimize the MSE, the only thing we can do is make the second term equal to 0. To get $E[(E[z|x] - f(x))^2] = 0$, this requires:

$$ f(x) = E[z|x] $$

This is why conditional expectation is best estimate in the MSE sense.

3. Geometric interpretation (orthogonal projection)

If you are used to thinking in linear algebra, this is actually the Projection Principle (Orthogonality Principle).

*Imagine all random variables forming a space.

  • $z$ is a point in space.
  • Observation data $x$ constitute an “information plane”.
  • We need to find a point on this plane that is closest to $f(x)$ and $z$. *The shortest distance is the perpendicular.
  • $E[z|x]$ is the orthogonal projection of $z$ on $x$ space.

Summarize

Back to your notes:

  • Question in the note: Why is it a good estimate? *Answer: Because if you want to minimize the sum of squares of prediction errors (which is the most commonly used loss function), mathematical proof tells you that any other function except $E[z|x]$ will bring additional errors (that is, the second term in the above formula).

This is why in the handwritten notes in the fifth picture, we spent so much effort to use the joint distribution to derive the distribution of $z|x$ - ** As long as the posterior mean is calculated, we will get the most accurate estimate in theory. **

Is this logical chain (from “why to calculate” to “how to calculate”) smooth now?