Can multivariate randomness be reduced to one dimensional randomness?

Suppose to have two random vectors $X,Y \in \mathbf{R}^n$ and a function $t: X,Y \rightarrow R^n$. Let $f$ be the joint p.d.f. of the resulting vector $W \in \mathbf{R}^n$, the following occurs:

$$ f(w_1,\ldots,w_n) = f(w_1) = \ldots = f(w_n) $$

Thus joint probability is equal to that of one of its marginals, so that components uniquely determine each other.

So if I know $f(w_1)$ then I know also that $f(w_2)$ has exactly the same distribution and the same probability to present a given value. They are totally dependent.

I can imagine for example that each marginal computes some invariant operation on vectors, depending on a parameter that doesn't affect probability.

For example, if vectors are taken independently from the standard normal $N(0,\mathbb{I}_n)$, they are rotation invariant. Suppose that each $i$-th marginal is computed rotating $X$ and $Y$ respectively by angles $\theta_i$, and their norms are summed:

$$ W=\begin{cases} w_1 = \left\|Q_1 X\right\| + \left\|Q_1Y\right\| \\ w_2 = \left\|Q_2 X\right\| + \left\|Q_2Y\right\| \end{cases} $$

$Q_1$ and $Q_2$ are distinct rotation matrices.

Even without trying to know the exact distribution of $w_1$ and $w_2$ may we argue that both are exactly the same? And how joint distribution can be properly described in such cases?

  • $\begingroup$ I'm a bit confused about your statement that $f(w_1,\cdots,w_n)=f(w_1)=\cdots=f(w_n)$. This implies that $f(w_i)$ doesn't depend on $w_i$, so that the only possible distribution is uniform on an interval of length 1. $\endgroup$
    – Alex R.
    May 19 '17 at 17:18
  • $\begingroup$ I agree with @Alex: Your overloading of the symbol "$f$", other abuses of mathematical notation, and inconsistencies (e.g., originally $W$ is an $n$-vector but at the end it appears to be a $2$-vector), make it difficult to determine what you're trying to ask. Is there any way you could edit this post to clarify the situation and your question? $\endgroup$
    – whuber
    May 19 '17 at 17:43

I'm not sure what you're asking with the first part of the question, but at the end it seems that you really just want to know how to describe the joint distribution of a random vector made up of perfectly dependent elements, so that's what I'm going to do.

Let $(\Omega, \mathscr F, P)$ be a probability space, and let $X: \Omega \to \mathbb R$ be a random variable. Define a random vector $Y : \Omega \to \mathbb R^n$ by $Y(\omega) = (X(\omega), \dots, X(\omega))$, i.e. we just copy $X(\omega)$ $n$ times. In your example since the $Q_i$ are orthogonal we have $w_1 = w_2$ so you have exactly the situation that I'm describing.

You want the joint distribution of $Y$. First let's think about what $P(Y \in A)$ is for some set $A = A_1 \times \dots \times A_n \subseteq \mathbb R^n$.

$$ P_Y(A) := P(Y \in A) = P(X \in A_1, X \in A_2, \dots, X \in A_n) = P\left(X \in \bigcap_{i=1}^n A_i\right) = P_X(\cap_i A_i) $$ so we have just shown that $P_Y(\prod_i A_i) = P_X(\cap_i A_i)$. This exactly specifies the distribution of $Y$ and lets us calculate probabilities for $Y$ (indeed, the official name of the measure $P_Y$ is the distribution of $Y$).

But let's say you're not happy with this and you really want a density for $Y$. Suppose that $X$ has a Lebesgue pdf $f$, which means that $P_X \ll \lambda$ where $\lambda$ denotes the Lebesgue measure. Consider the set $A = [0,1] \times [2,3] \times \dots \times [2(n-1), 2n-1]$. Note how $\lambda(A) > 0$ but $P_Y(A) = 0$. This means that $Y$ does not have a Lebesgue pdf so even though $X$ has a perfectly sensible Lebesgue pdf we do not inherit one for $Y$. This is not to say that $Y$ has no density of any sort, but rather in general I wouldn't expect it to have an interpretable, recognizable density since it may be with respect to a strange dominating measure.

Nevertheless, we can still use $X$'s Lebesgue pdf to compute $P_Y(A)$ via $$ P_Y(A) = P_X(\cap_i A_i) = \int_{\cap_i A_i} f d\lambda $$

so to me it seems that the only density we need is the one we already have, namely the univariate density $f$ for $X$.

A concrete example of this is if $Z \sim \mathcal N(0,1)$ and $W = (Z, Z)$. We know that the covariance matrix of $W$ is singular so bivariate pdf is undefined, but if we want to find $P(W \in [1,2] \times [1,3])$, say, we just need to compute $$ P(W \in [1,2] \times [1,3]) = P(Z \in [1,2] \cap Z \in [1,3]) = P(Z \in [1,2]) = \int_1^2 \frac{1}{\sqrt{2\pi}} e^{-\frac 12 z^2} dz. $$

  • $\begingroup$ thanks for your accurate answer @Chaconne. My example was too simplistic. The point, in facts, is not having exactly the same value shared among marginals, but the fact that a given value in any marginal implies that values in other marginals would have the same probabilty to occur. Then probability for a given value in $w_1$ determines exactly the probability of having any corresponding value in $w_2$, $w_3$ and so on. $\endgroup$
    – user136737
    May 20 '17 at 7:20
  • $\begingroup$ @user136737 well if we have functions $h_i : \mathbb R \to \mathbb R$ for $i = 1, \dots, n$ and define $Y(\omega)_i = h_i(X(\omega))$ then we have $P(Y \in \prod_i A_i) = P(h_1 \circ X \in A_1 \cap \dots \cap h_n \circ X \in A_n) = P\left( X \in \bigcap_i h_i^{-1}(A_i) \right)$ so nothing fundamental has changed. Does that address your question better? $\endgroup$
    – jld
    May 21 '17 at 17:21

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