Let $A,B,C$ be iid Unif(0,1). Let $X,Y$ be random variables:

  • $X=(A-B)1_{A-B>0}+(1+(A-B))1_{A-B<0}$

  • $Y=(C-B)1_{C-B>0}+(1+(C-B))1_{C-B<0}$

I am able to show that $X,Y$ are id Unif(0,1). My problem is showing they are iid (i.e. I'm missing the independent).

(I'm not allowed to use measure theory here, but I actually don't see how I would anyway since both $X$ and $Y$ have a '$B$' in the formula.)

Okay so elementary probability stuff only. Let's compute joint cdf and hope it's uniform on unit square. This is

$$P(X \le x, Y \le y) = 1_{x,y > 1} + x1_{0 < x < 1, y > 1} + y1_{0 < y < 1, x > 1} + xy1_{0 < x,y < 1}$$

I believe I get everything except the $xy1_{0 < x,y < 1}$ part.

It seems we have to take cases

  • Case 1:$ X=A-B, Y=C-B$
  • Case 2: $X=(A-B)+1, Y=C-B$
  • Case 3: $X=A-B, Y=(C-B)+1$
  • Case 4: $X=(A-B)+1, Y=(C-B)+1$

Okay let's try Case 1. (Update: The bounds are wrong, but the questions in re the tags are still valid, I believe.)

$$P(0 < X = A - B \le x, 0 < Y = C - B \le y)$$

What I think is conditional probability but of 2 random variables conditioned on 1. Instead of the usual $$P(z_1 < Z < z_2 | B=b) := \int_{z_1}^{z_2} f_{Z|B=b}(z) dz,$$ with $f_{Z|B=b}(z)=\frac{f_{Z,B}(z,b)}{f_B(b)},$ it looks like we'll have like $$P(z_1 < Z < z_2, u_1 < U < u_2 | B=b) := \int_{u_1}^{u_2} \int_{z_1}^{z_2} f_{(Z,U)|B=b}(z,u) dz du,$$ with $f_{(Z,U)|B=b}(z,u) = $, I think, $\frac{f_{Z,U,B}(z,b,u)}{f_B(b)}$

  • Note: if any of the definitions ':=' are in fact not definitions, then you'll have to explain conditioning on an event of probability zero to me please.

So here's what I think is next:

$$P(0 < X = A - B \le x, 0 < Y = C - B \le y) = P(B < A \le x+B, B < C \le y+B)$$

$$ = \int_{b=0}^{b=1} P(B < A \le x+B, B < C \le y+B | B=b) f_B(b) db \tag{1?}$$

$$ = \int_{b=0}^{b=1} P(b < A \le x+b, b < C \le y+b | B=b) f_B(b) db \tag{2????}$$

$$ = \int_{b=0}^{b=1} \int_{b}^{x+b} \int_{b}^{y+b} f_{(A,C)|B=b}(a,c) dc da f_B(b) db \tag{3 part 1?}$$

$$ = \int_{b=0}^{b=1} \int_{b}^{x+b} \int_{b}^{y+b} \frac{f_{(A,C,B)}(a,c,b)}{f_B(b)} dc da f_B(b) db \tag{3 part 2?}$$

$$ \text{[details omitted because actually the bounds are wrong]}$$

$$ \text{[details omitted because actually the bounds are wrong]}$$

$$ \text{[details omitted because actually the bounds are wrong]}$$

$$ = xy $$

And then assuming all of the above is correct and all the question mark parts are justified, repeat for the other 3 cases and it looks like we have $xy$ in each. Are these cases supposed to be added up and so I'm missing $\frac14$? Or what?

  • Case 2: $b-1<A<x+b-1, b<C<y+b$, so again just $xy$

  • Case 3: $b<A<x+b, b-1<C<y+b-1$, so again just $xy$

  • Case 4: $b-1<A<x+b-1, b-1<C<y+b-1$, so again just $xy$

About the question marks:

  • For $(1?)$, I think the rule is like for an event $E$ and continuous random variable $B$, we have $P(E)=\int_{\mathbb R} P(E|B=b) f_B(b) db$. Is this correct?

    • Oh wait a minute wiki says we can't quite do this. What I understand is that we can't do it for arbitrary $E$, but we can do it when (but not only when I guess) $E=\{Y \in \ \text{some interval or Borel set I guess}\}$, for some continuous random variable $Y$ s.t. the joint pdf $f_{X,Y}$ is well-defined? (I forgot if any 2 continuous random variables necessarily have a well-defined joint pdf.)
  • For $(2????)$, I think we're doing something like for events $E$, $H$ and $G$ and continuous random variable $B$: we have $P(E|H)=P(E \cap H|H)$, but $P(G|H)$ is defined only for $P(H)>0$. What is being done here when technically $P(H)=0$? I mean of course in the 1st place when we say like '$P(E|B=b)$', this is notational, we're not really conditioning on the $P$-null event $\{B=b\}$. But I still don't get exactly what's being done here.

  • For $(3 \ \text{parts 1 and 2})$, I'm actually just guessing here, what's the definition of conditional joint cdf of 2 random variables given a 3rd? And please provide a reference.

    • Wiki just says $F_{(X,Y)|Z=z}(x,y):=P(X \le x, Y \le y|Z=z)$, but it doesn't quite define $P(X \le x, Y \le y|Z=z)$.

    • For just 1 continuous random variable conditioned on 1 continuous random variable, it's $P(X \le x, |Z=z) := \int_{-\infty}^{x} f_{X|Z=z}(t) dt$, where $f_{X|Z=z}(t) := \frac{f_{(X,Z)}(x,z)}{f_{Z}(z)}$.

    • For 2 continuous random variables conditioned on 1 continuous random variable, I think it's $P(X \le x, Y \le y|Z=z) := \int_{-\infty}^{x} \int_{-\infty}^{y} f_{(X,Y)|Z=z}(t,u) du dt$, but then...

    • What's $f_{(X,Y)|Z=z}(t,u)$? (I guess we do the elementary probability way of thinking: define the pdf before the cdf...) According to this site (see problems 1 and 16), it's $f_{(X,Y)|Z=z}(t,u): = \frac{f_{(X,Y,Z)}(t,u,z)}{f_Z(z)}$. So, I guess I'm right about joint cdf/pdf stuff. I'm just hoping for a reference please.

  • 2
    $\begingroup$ As I wrote in an answer to a previous question like this one: draw a picture. $\endgroup$
    – whuber
    Mar 18, 2021 at 13:33
  • $\begingroup$ @Xi'an Although certainly your diagrams qualify, I don't see any 3D versions there. $\endgroup$
    – whuber
    Mar 18, 2021 at 18:55
  • $\begingroup$ @whuber wait I made a huge error over something apparently so simple (maybe). Please see edit. Is it correct? I have $xy$ now instead of $xy \pm$ (other stuff) $\endgroup$
    – BCLC
    Mar 22, 2021 at 5:57
  • $\begingroup$ @whuber updated question. it's not just about the bounds now. i get the bounds are wrong. i figured out the bounds on my own. now i'm wondering about the joint pdf/cdf stuff mainly. $\endgroup$
    – BCLC
    Mar 24, 2021 at 5:37
  • 1
    $\begingroup$ Cross-post: math.stackexchange.com/q/4066689/321264. $\endgroup$ Mar 25, 2021 at 14:55

3 Answers 3


The joint distribution of $(A^-,C^ -)=(A-B,C-B)$ is given by its density \begin{align} f(a^-,c^-)&=\int f_A(a^-+b)f_C(c^-+b)f_B(b)\,\text d b\\ &=\int_0^1 \mathbb I_{(0,1)}(a^-+b)\mathbb I_{(0,1)}(c^-+b)\,\text d b\\ &=[\min(1,1-a^-,1-c^-)-\max(0,-a^-,-c^-)]^+\\ &=\mathbb I_{(0,1)}(a^-)\mathbb I_{(0,1)}(c^-)[\min(1,1-a^-,1-c^-)-\max(0,-a^-,-c^-)]^+\\ &\ +\mathbb I_{(-1,0)}(a^-)\mathbb I_{(0,1)}(c^-)[\min(1,1-a^-,1-c^-)-\max(0,-a^-,-c^-)]^+\\ &\ +\mathbb I_{(0,1)}(a^-)\mathbb I_{(-1,0)}(c^-)[\min(1,1-a^-,1-c^-)-\max(0,-a^-,-c^-)]^+\\ &\ +\mathbb I_{(-1,0)}(a^-)\mathbb I_{(-1,0)}(c^-)[\min(1,1-a^-,1-c^-)-\max(0,-a^-,-c^-)]^+\\ &=\mathbb I_{(0,1)}(a^-)\mathbb I_{(0,1)}(c^-)[\min(1-a^-,1-c^-)-0]\\ &\ +\mathbb I_{(-1,0)}(a^-)\mathbb I_{(0,1)}(c^-)[\min(1,1-c^-)+a^-]^+\\ &\ +\mathbb I_{(0,1)}(a^-)\mathbb I_{(-1,0)}(c^-)[\min(1,1-a^-)+c^-]^+\\ &\ +\mathbb I_{(-1,0)}(a^-)\mathbb I_{(-1,0)}(c^-)[1+\min(a^-,c^-)]\\ &=\mathbb I_{(0,1)}(a^-)\mathbb I_{(0,1)}(c^-)[1-\max(a^-,c^-)]\\ &\ +\mathbb I_{(-1,0)}(a^-)\mathbb I_{(0,1)}(c^-)[1-c^-+a^-]^+\\ &\ +\mathbb I_{(0,1)}(a^-)\mathbb I_{(-1,0)}(c^-)[1-a^-+c^-]^+\\ &\ +\mathbb I_{(-1,0)}(a^-)\mathbb I_{(-1,0)}(c^-)[1+\min(a^-,c^-)]\\ &=\mathbb I_{(0,1)}(a^-)\mathbb I_{(0,1)}(c^-)\mathbb I_{a^->c^-}[1-a^-]\\ &\ +\mathbb I_{(0,1)}(a^-)\mathbb I_{(0,1)}(c^-)\mathbb I_{a^-<c^-}[1-c^-]\\ &\ +\mathbb I_{(-1,0)}(a^-)\mathbb I_{(0,1)}(c^-)\mathbb I_{c⁻-a^-<1}[1-c^-+a^-]\\ &\ +\mathbb I_{(0,1)}(a^-)\mathbb I_{(-1,0)}(c^-)\mathbb I_{a^--c⁻<1}[1-a^-+c^-]\\ &\ +\mathbb I_{(-1,0)}(a^-)\mathbb I_{(-1,0)}(c^-)\mathbb I_{a^-<c^-}[1+a^-]\\ &\ +\mathbb I_{(-1,0)}(a^-)\mathbb I_{(-1,0)}(c^-)\mathbb I_{a^->c^-}[1+c^-]\\ \end{align}

The joint density thus writes differently on the 8 rectangular triangles of the $(-1,1)^2$ square as drawn below (with two triangles where it is null):

                                        enter image description here

If one now considers $(X,Y)$ i.e. $$X=A^-+\mathbb I_{A^-<0}\qquad Y=C^-+\mathbb I_{C^-<0}$$ its distribution is defined by the probabilities of arbitrary squares $$\mathbb P((X,Y) \in [x_1,x_2]\times[y_1,y_2]) \qquad 0< x_1,x_2,y_1,y_2<1$$ which are equal to \begin{align}&\mathbb P((A^-,C^-) \in [x_1,x_2]\times[y_1,y_2])\\ &\quad+\mathbb P((A^-+1,C^-) \in [x_1,x_2]\times[y_1,y_2])\\ &\quad+\mathbb P((A^-,C^-+1) \in [x_1,x_2]\times[y_1,y_2])\\ &\quad+\mathbb P((A^-+1,C^-+1) \in [x_1,x_2]\times[y_1,y_2])\\ &=\mathbb P((A^-,C^-) \in [x_1,x_2]\times[y_1,y_2])\\ &\quad+\mathbb P((A^-,C^-) \in [x_1-1,x_2-1]\times[y_1,y_2])\\ &\quad+\mathbb P((A^-,C^-) \in [x_1,x_2]\times[y_1-1,y_2-1])\\ &\quad+\mathbb P((A^-,C^-) \in [x_1-1,x_2-1]\times[y_1-1,y_2-1]) \end{align}

Using the symmetries between the four translated squares, as in the figure below (feel free to move the squares!), the non-constant terms just cancel each other and one ends up with the uniform distribution on $(0,1)^2$.

                                        enter image description here

A much faster resolution goes as follows: since $A^-$ and $C^-$ are independent given $B=b$ \begin{align} &P((X,Y)\in [x_1,x_2]\times[y_1,y_2]|B=b)\\ &=\mathbb P((A^-+\mathbb I_{A^-<0},C^-+\mathbb I_{C^-<0}) \in [x_1,x_2]\times[y_1,y_2]|B=b)\\ &=\mathbb P(A^-+\mathbb I_{A^-<0}\in [x_1,x_2]|B=b)\times \mathbb P(C^-+\mathbb I_{C^-<0}\in [y_1,y_2]|B=b)\\ &=\mathbb P(A-b+\mathbb I_{A<b} \in [x_1,x_2]|B=b)\times \mathbb P(C-b+\mathbb I_{C<b}\in [y_1,y_2]|B=b)\\ &=\left\{\mathbb P(A-b \in [x_1,x_2]\cap[0,1-b]|B=b)\right.\\ &\qquad\qquad\left.+ \mathbb P(A-b+1 \in [x_1,x_2]\cap[1-b,1]|B=b)\right\}\\ &\quad\times \left\{\mathbb P(C-b \in [y_1,y_2]\cap[0,1-b]|B=b)\right.\\ &\qquad\qquad\left.+ \mathbb P(C-b+1 \in [y_1,y_2]\cap[1-b,1]|B=b)\right\}\\ &=(x_2-x_2)(y_2-y_1) \end{align} the conditional joint distribution is Uniform, hence the marginal is also Uniform.

  • 1
    $\begingroup$ oh wow. thanks Xi'an. had no idea this would be so complicated. 1 - am i going somewhere with my thought 1? 2 - do you definitely disagree that $A-B$ and $C-B$ are independent for any independent $A,B,C$? 3 - Are $A-B$ and $C-B$ something 'conditionally' independent on $B$ for any independent $A,B,C$? (I didn't learn that much conditional independence in grad or undergrad.) $\endgroup$
    – BCLC
    Mar 18, 2021 at 10:19
  • $\begingroup$ Thanks xi'an 1 - am i going somewhere with my thought 1? $\endgroup$
    – BCLC
    Mar 18, 2021 at 11:28
  • 1
    $\begingroup$ Xi'an, 4 - wait so i'm going somewhere in my thought 3 re 'promote' (and thought 1 too perhaps)? i'd think to say joint cdf conditional on $B$ is uniform and then say the unconditional joint cdf is uniform $\endgroup$
    – BCLC
    Mar 18, 2021 at 11:36
  • $\begingroup$ (damn it xi'an i accidentally deleted a comment aside from just -5 and -6...) Edit: (Oh good thing i had another tab open. the other comment i deleted was the update about the $xy$ thingy) $\endgroup$
    – BCLC
    Mar 24, 2021 at 5:38
  • $\begingroup$ 7 - thanks xi'an! updated question. it's not just about the bounds now. i get the bounds are wrong. i figured out the bounds on my own. now i'm wondering about the joint pdf/cdf stuff mainly. may you please help with the joint pdf/cdf stuff and other stuff in the tag question marks parts? $\endgroup$
    – BCLC
    Mar 24, 2021 at 5:38

I would like the share the results of staring at a diagram. I promise to do almost no calculation (and the calculations that are performed involve only multiplications by $0,$ $1,$ and $-1$ along with additions).

Start by re-expressing $X$ as

$$X = A-B \mod 1,$$

which is just the fractional part of $A-B.$ This function is defined on the entire $(A,B)$ plane, where its contours look like those in the first figure.

Figure 1

On this figure I have outlined the unit square. This is a fundamental domain for the group of unit translations of the square in the plane. The group has other fundamental domains, though, such as this one:

Figure 2

These domains differ by two congruent regions in which the variables $A\mod 1,$ $B \mod 1,$ and $A-B \mod 1$ have matching values:

Figure 3

Therefore the distributions of $A-B \mod 1$ are the same on both domains.

However, there's an area (=probability)-preserving map between the domains. The linear transformation defined by the matrix

$$\pmatrix{1 & -1\\0 & 1}$$

is a skew transformation that tilts the second domain (the parallelogram) sideways back into the first domain (the unit square). Because it is a one-to-one area-preserving transformation and does not change the values of $A \mod 1$ or $B \mod 1,$ it does not change their distribution. But the image of $A-B \mod 1$ under this transformation is just $A:$

Figure 4

We may perform a similar set of operations in the $(C,B)$ plane where $Y=C-B\mod 1,$ and achieve a comparable result. The conclusion? The combined transformation

$$\pmatrix{1 & -1 & 0 \\ 0 & 1 & 0 \\0 & 0 & 1} \pmatrix{1 & 0 & -1\\0 & 1 & 0 \\ 0 & 0 & 1} = \pmatrix{1 & -1 & -1\\0 & 1 & 0\\0 & 0 & 1}$$

is a one-to-one distribution-preserving transformation that maps $(X,B,Y)$ into $(A,B,C).$ Since the latter is a set of iid uniform$(0,1)$ variables, so is the former. In particular,

$X$ and $Y$ have uniform distributions on $[0,1)$ and are independent (as well as being independent of $B$).

As a check, here's a scatterplot matrix based on 1,000 realizations of $(A,B,C).$

A <- seq(0,1, length.out=11)[-1]
X <- expand.grid(A=A, B=A, C=A) + runif(length(A)^3*3, A[1]-A[2], 0)
X$X <- with(X, (A-B) %% 1)
X$Y <- with(X, (C-B) %% 1)
pairs(subset(X, select=c(X,B,Y)), pch=19, col="#00000010")

Figure 5

All bivariate marginal distributions, at least, appear to be uniform on the square, supporting these results.


I am able to show that $X,Y$ are id Unif(0,1). My problem is showing they are iid (i.e. I'm missing the independent).


Besides $X \sim U(0,1)$ and $Y \sim U(0,1)$ you can also show the independence from $B$ like $X|B \sim U(0,1)$ and $Y|B \sim U(0,1)$.

  • $B$ is the only common variable in the equations for $X$ and $Y$
  • The distributions of $X$ and $Y$ are independent from this $B$

This leads to

  • the $X$ and $Y$ are also independent from each other.

I believe that this intuitive view is sufficient. The rest, computing a joint distribution, is more like an awkward exercise that obscures the insight.

You could compute the joint distribution conditional on $B$ as a 2D uniform variable $X,Y|B \sim U(0,1) \times U(0,1)$ to show independence of the joint distribution from $B$ to conclude that also $X,Y \sim U(0,1) \times U(0,1)$

(I am using the symbol $\times$ to indicate the product of the pdf's, I do not know if there's an official notation or standard)

  • 1
    $\begingroup$ +1 This is by far the best answer. $\endgroup$ Mar 24, 2021 at 9:49
  • 1
    $\begingroup$ @JarleTufto Xian also sort of gives the same answer at the end of his post (but it is obscured by lots of lines of formulae and that's why I added my short post). $\endgroup$ Mar 24, 2021 at 9:57
  • $\begingroup$ Thanks Sextus Empiricus! 2 follow-up questions: 1 - is the the $X,Y|B \sim$ stuff below the line supposed to be the precise version of the intuitive stuff above the line? or is it an alternative prooff? 2 - in re the $X,Y|B \sim$ stuff below the line, my next question then is what exactly is conditional joint distribution of 2 (or more) random variables on a single random variable (at least assuming they all have continuous pdfs). I ask about this here on maths se and sort of here $\endgroup$
    – BCLC
    Mar 25, 2021 at 5:20
  • $\begingroup$ @JohnSmithKyon 1 The stuff below the line is an alternative. It is how you can make the approach if you insist on using a joint distribution. 2 To make computations about this joint distribution easier, you could first compute the joint distribution of $X,Y$ with $B$ fixed to some particular value to gain intuition about the problem (particularly $B=0$ is easy and you could start with that, but for the intuition try others too), second do the same but consider it as a parameter. Third, you will realize that the value of the parameter $B$ does not influence the distribution of $X,Y$. $\endgroup$ Mar 25, 2021 at 6:53
  • 2
    $\begingroup$ @Xi'an the second part of my answer is the same as your second part, but in the first part I give an argument that is different and simplifies the problem: there is no need to think about a joint distribution. I disagree with you that mathematical formulas can not be verbose and obscuring intuition, that doesn't mean I am against the use of formulas on this forum (But there are differences in styles). Personally I found the question post full of too much spaghetti, and that's what made me create this simple post. It was not criticism of your post. $\endgroup$ Mar 25, 2021 at 22:39

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