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MInner
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For a univariate real-valued random variable I am pretty sure that the converse holds.

Consider a multivariate $\mathbf X$ with values in $\mathbb R^n$ with measure $\mu(X)$ and its multivariate CDF is:

$$F_X(a_0,\dots, a_n) = \int_{-\infty}^{a_0} \dots \int_{-\infty}^{a_n} \mu(d\mathbf x)$$

, then the univariate random variable $\tilde X = F_X(\mathbf X)$ is still distributed uniformly in $U_{[0;1]}$.

Is it true that if for some other $\mathbf Y$ holds $F_X(\mathbf Y) \sim U_{[0; 1]}$, then $F_X = F_Y$?

Or, in other words, can we use uniformity of $F_X(\mathbf Y)$ to test if $F_X = F_Y$ for multivariate $\mathbf X$ and $\mathbf Y$?

UPD

@whuber answered in the comment below that the "$\tilde X \sim U_{[0;1]}$" part is actually wrong, so the question does not make much sense and one interested in multivariate variables and uniformity should read about Copulas.

For a univariate real-valued random variable I am pretty sure that the converse holds.

Consider a multivariate $\mathbf X$ with values in $\mathbb R^n$ with measure $\mu(X)$ and its multivariate CDF is:

$$F_X(a_0,\dots, a_n) = \int_{-\infty}^{a_0} \dots \int_{-\infty}^{a_n} \mu(d\mathbf x)$$

, then the univariate random variable $\tilde X = F_X(\mathbf X)$ is still distributed uniformly in $U_{[0;1]}$.

Is it true that if for some other $\mathbf Y$ holds $F_X(\mathbf Y) \sim U_{[0; 1]}$, then $F_X = F_Y$?

Or, in other words, can we use uniformity of $F_X(\mathbf Y)$ to test if $F_X = F_Y$ for multivariate $\mathbf X$ and $\mathbf Y$?

For a univariate real-valued random variable I am pretty sure that the converse holds.

Consider a multivariate $\mathbf X$ with values in $\mathbb R^n$ with measure $\mu(X)$ and its multivariate CDF is:

$$F_X(a_0,\dots, a_n) = \int_{-\infty}^{a_0} \dots \int_{-\infty}^{a_n} \mu(d\mathbf x)$$

, then the univariate random variable $\tilde X = F_X(\mathbf X)$ is still distributed uniformly in $U_{[0;1]}$.

Is it true that if for some other $\mathbf Y$ holds $F_X(\mathbf Y) \sim U_{[0; 1]}$, then $F_X = F_Y$?

Or, in other words, can we use uniformity of $F_X(\mathbf Y)$ to test if $F_X = F_Y$ for multivariate $\mathbf X$ and $\mathbf Y$?

UPD

@whuber answered in the comment below that the "$\tilde X \sim U_{[0;1]}$" part is actually wrong, so the question does not make much sense and one interested in multivariate variables and uniformity should read about Copulas.

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MInner
MInner
  • 293
  • 2
  • 9

For a univariate real-valued random variable I am pretty sure that the converse holds.

Consider a multivariate $\mathbf X$ with values in $\mathbb R^n$ with measure $\mu(X)$ and its multivariate CDF is:

$$F_X(a_0,\dots, a_n) = \int_{-\infty}^{a_0} \dots \int_{-\infty}^{a_n} \mu(d\mathbf x)$$

, then the univariate random variable $\tilde X = F_X(\mathbf X)$ is still distributed uniformly in $U_{[0;1]}$.

Is it true that if for some other $\mathbf Y$ holds $F_X(\mathbf Y) \sim U_{[0; 1]}$, then $F_X = F_Y$?

Or, in other words, can we use uniformity of $F_X(\mathbf Y)$ as ato test statistic for hypothesisif $F_X = F_Y$ for multivariate $\mathbf X$ and $\mathbf Y$?

For a univariate real-valued random variable I am pretty sure that the converse holds.

Consider a multivariate $\mathbf X$ with values in $\mathbb R^n$ with measure $\mu(X)$ and its multivariate CDF is:

$$F_X(a_0,\dots, a_n) = \int_{-\infty}^{a_0} \dots \int_{-\infty}^{a_n} \mu(d\mathbf x)$$

, then the univariate random variable $\tilde X = F_X(\mathbf X)$ is still distributed uniformly in $U_{[0;1]}$.

Is it true that if for some other $\mathbf Y$ holds $F_X(\mathbf Y) \sim U_{[0; 1]}$, then $F_X = F_Y$?

Or, in other words, can we use uniformity of $F_X(\mathbf Y)$ as a test statistic for hypothesis $F_X = F_Y$ for multivariate $\mathbf X$ and $\mathbf Y$?

For a univariate real-valued random variable I am pretty sure that the converse holds.

Consider a multivariate $\mathbf X$ with values in $\mathbb R^n$ with measure $\mu(X)$ and its multivariate CDF is:

$$F_X(a_0,\dots, a_n) = \int_{-\infty}^{a_0} \dots \int_{-\infty}^{a_n} \mu(d\mathbf x)$$

, then the univariate random variable $\tilde X = F_X(\mathbf X)$ is still distributed uniformly in $U_{[0;1]}$.

Is it true that if for some other $\mathbf Y$ holds $F_X(\mathbf Y) \sim U_{[0; 1]}$, then $F_X = F_Y$?

Or, in other words, can we use uniformity of $F_X(\mathbf Y)$ to test if $F_X = F_Y$ for multivariate $\mathbf X$ and $\mathbf Y$?

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MInner
MInner
  • 293
  • 2
  • 9

It is true that $\mathbf X \sim F_X \Rightarrow F_X(\mathbf X) \sim U_{[0;1]}$; does the converse hold for multivariate $\mathbf X$?

For a univariate real-valued random variable I am pretty sure that the converse holds.

Consider a multivariate $\mathbf X$ with values in $\mathbb R^n$ with measure $\mu(X)$ and its multivariate CDF is:

$$F_X(a_0,\dots, a_n) = \int_{-\infty}^{a_0} \dots \int_{-\infty}^{a_n} \mu(d\mathbf x)$$

, then the univariate random variable $\tilde X = F_X(\mathbf X)$ is still distributed uniformly in $U_{[0;1]}$.

Is it true that if for some other $\mathbf Y$ holds $F_X(\mathbf Y) \sim U_{[0; 1]}$, then $F_X = F_Y$?

Or, in other words, can we use uniformity of $F_X(\mathbf Y)$ as a test statistic for hypothesis $F_X = F_Y$ for multivariate $\mathbf X$ and $\mathbf Y$?