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  1. It seems that linearly independent is totally different from independent of random variable concept. Non-zero vectors Orthogonality must imply linearly independence.

  1. In Statistics, the relation of Orthogonality and independence depends on how one defines the inner product. It's meaningless to say one implies another without defining what is the inner product(I've seen a saying inner product is zero implies it's orthogonal). If an inner product of random variables $X, Y$ defines as $E(XY)$, the orthogonality of two random variables is irrelevant to independence, whereas if the inner product is defined as $Cov(X, Y)$, then we could say orthogonality implies independence.

I am unsure if my above conclusion is correct, please just let me know if they are right or wrong.

Appreciate for any comments

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  • $\begingroup$ Which independence are you referring to in point 2: statistical or linear? What are X and Y? random variables or vectors? $\endgroup$
    – Dayne
    Mar 10, 2022 at 8:39
  • $\begingroup$ @Dayne Thank you for your comments. I meant statistical independence in #2, and they are random variables. Was I correct? $\endgroup$
    – LJNG
    Mar 10, 2022 at 10:50
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    $\begingroup$ Random variables are vectors. Thus, by default, whenever anyone refers to linear independence of random variables, they mean exactly the standard definition: namely, no nonzero linear combination of those variables is zero. Orthogonality is a concept related to additional mathematical structure: namely, an inner product. It is definitely not the case that orthogonality implies independence! $\endgroup$
    – whuber
    Mar 10, 2022 at 12:58
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    $\begingroup$ No, that's not possible. A scalar, by definition, is an element in a field--which means it's a kind of number subject to the usual laws of arithmetic. Random variables are more complicated than that. Random variables are functions from a probability space into some kind of space and that image space can consist of scalars, permitting us to refer to scalar-valued random variables. So beware of how terminology is used! $\endgroup$
    – whuber
    Mar 10, 2022 at 17:46
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    $\begingroup$ @LJNG Speaking to part 2 of your question, the inner product is not defined to be the covariance. Rather the covariance under suitable assumptions is an example of an inner product. Measure/mutual/statistical/stochastic independence is the equality of the joint CDF to the product of the marginal CDFs. A covariance of zero does not imply measure independence. $\endgroup$
    – Galen
    Mar 10, 2022 at 22:21

1 Answer 1

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First of all, $Cov(X,Y)=0$ does not implies independence unless $X$ and $Y$ are jointly Gaussian. So your orthogonality is way weaker than independence.

It looks disappointing but it is only because you are looking at it the wrong way. Independence does have an analogy to vector space orthogonality, but you have to treat $X$ and $Y$ as entire subspaces instead of individual vectors. Also, because the independence business happens in entirely in the $\sigma$-algebra world, which has less structures than an inner product space (e.g., there are no dimensions, basis nor spans in $\sigma$-algebras), it will be more fruitful to look for a weakened generalization of the vector space inner product idea rather than trying to find a vector space inner product that captures the entirely less structured $\sigma$-algebra situation.

Probability Theory View

Because independence is defined as $f(x,y)=f_x(x)f_y(y)$, you can also say that $X$ and $Y$ are independent if and only if for all functions $g(x)$ and $h(y)$, \begin{equation} \int\int g(x)h(y)f(x,y)dxdy = \left(\int g(x)f_x(x)dx\right)\left(\int h(x)f_y(y)dy\right). \end{equation} Without independence, the left hand side integral simply can't be factored out so nicely. In fact, it is easy to check that the left hand side is bilinear in both $g$ and $h$ and the right hand side is two "simple dot products," so we can write \begin{equation} \langle g, h\rangle_f = \langle g, f_x \rangle \langle h, f_y \rangle. \end{equation}

Remember this factorization and we will see how this has something to do with orthogonality in standard linear algebra.

(Note that the integral equation is same as $E(g(X)h(Y))=E(g(X))E(h(Y))$ but it's not really the point here.)

Linear Algebra Analogy

Now let's make an analogy with $\mathbb{R}^n$ vectors. Let $X$ be a subspace of $\mathbb{R}^n$ and $Y$ another subspace that is $X$'s orthogonal complement. Let $g$ and $h$ be any vectors from $\mathbb{R}^n$, and $F$ be an $n\times n$ rank-$r$ matrix which defines a bilinear form $\langle g, h\rangle_F = g^TFh$. Let $\mathcal{X}=\{x_i\}_i$ and $\mathcal{Y}=\{y_j\}_j$ be some orthogonal unit-length basis of $X$ and $Y$ respectively. Denote $\mathcal{E}=\mathcal{X}\cup\mathcal{Y}$ so that $\mathcal{E}$ is a basis that spans the entire space.

Obviously by basis expansion we can write $g$ and $h$ as so $g=\sum_{e_i\in\mathcal{E}}g_ie_i$ and $h=\sum_{e_j \in\mathcal{E}}h_je_j$.

Any rank-$r$ $F$ can be written as a sum \begin{equation} F=\sum_{k=1}^r u_k v_k^T \end{equation} by the SVD decomposition. Usually, the bilinear form \begin{equation} \langle g, h\rangle_F = \sum_{e_j\in\mathcal{E}}\sum_{e_i\in\mathcal{E}} (g_i e_i)^T F (h_j e_j) = \sum_{e_j\in\mathcal{E}}\sum_{e_i\in\mathcal{E}}\sum_{k=1}^r g_ih_j e_i^T u_kv_k^T e_j \end{equation} cannot be factorized into the product of two nice separate single sums. But if $F$ happens to be rank-$1$ with $F=uv^T$ such that $u\in X$ and $v\in Y$ then by basis expansion on $u$ and $v$, we can write $F$ as \begin{equation} F=\left(\sum_{x_\ell\in\mathcal{X}}{f_x}_{\ell}x_\ell\right)\left(\sum_{y_m\in\mathcal{Y}}{f_y}_{m}y_m^T\right), \end{equation} where ${f_x}_\ell$ and ${f_y}_m$ are scalar coordinates. So we can now factorize the double sum into a product of two single sums as follows: \begin{align} \langle g, h\rangle_F &= \sum_{e_j\in\mathcal{E}}\sum_{e_i\in\mathcal{E}} (g_i e_i)^T \left(\sum_{x_\ell\in\mathcal{X}}{f_x}_{\ell}x_\ell\right)\left(\sum_{y_m\in\mathcal{Y}}{f_y}_{m}y_m^T\right) (h_j e_j)\\ &= \sum_{y_j\in\mathcal{Y}}\sum_{x_i\in\mathcal{X}} (g_i {f_x}_i) (h_j {f_y}_j)\\ &= \left(\sum_{x_i\in\mathcal{X}} g_i {f_x}_i\right)\left(\sum_{y_j\in\mathcal{Y}} h_j {f_y}_j\right)\\ &= \langle g, f_x\rangle \langle h, f_y\rangle, \end{align} where the right hand side $\langle \cdot, \cdot\rangle$ are the "simple dot product." We have just proved that the bilinear form can be factored out for any arbitrary $g\in \mathbb{R}^n$ and $h\in \mathbb{R}^n$. The first line simplifies into the second line because $e_i^Tx_\ell$ is zero except when $e_i = x_\ell$; and in such case, $x_\ell^Tx_\ell = 1$ by definition.

Notice that in this argument:

  1. Orthogonality of $X$ and $Y$ together with $F$ being "compatible" with them is essential, because otherwise the second line of the simplification simply won't cancel out so nicely.
  2. Be cautious that $F$ is defined using $X$ and $Y$ just for illustration. In the probability version we define the joint density $f$ at the beginning when forming a probability space, and we call $X$ and $Y$ independent if the pre-defined $f$ can be factored out.
  3. Why can $g$ and $h$ be any vectors in the analogy, but in the probabilistic version $g$ is a function of only $X$ and $h$ only $Y$? Actually this is only because I have taken $\langle \cdot, \cdot \rangle_F$ to be $\mathbb{R}^n\times\mathbb{R}^n\to\mathbb{R}$ for illustration. The more general situation is one where $\langle \cdot, \cdot \rangle_F:V\times W\to\mathbb{R}$ where $V$ and $W$ are different spaces. Now $g$ must fit in the left slot and $h$ the right slot. But if $V$ and $W$ were any general vector spaces then we can't write down the decomposition of $F$ so easily any more, neither can we say a subspace of $V$ is "orthogonal" to a subspace of $W$ because they are from different spaces (so "inner product" equals to zero doesn't make sense any more). Nonetheless, because $F$ is pre-defined, we can still get around this problem by defining that some sub-spaces of $V$ and $W$ are "$F$-independent" if $\langle \cdot, \cdot \rangle_F$ can be factored out. This is somehow what you are seeing in the probabilistic definition.
  4. In fact, in advanced probability theory, statistical independence is defined on $\sigma$-algebras first, and then we say that two random variables are independent if the $\sigma$-algebras they generate are independent. So intuitively it also makes more sense to think of independence as a property about two sets rather than two vector elements.
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