Do $X,Z$ and $Y,Z$ have the same density if $X, Y \sim \text{Unif}(0,1)$, $X \perp \!\!\! \perp Y$, and $Z = X + Y$

Question

Let

• $X, Y \sim \text{Unif}(0,1)$
• $X \perp \!\!\! \perp Y$, and
• $Z = X + Y$

Is it true that $f_{X\mid Z=z}(s)$ and $f_{Y\mid Z=z}(s)$ for all $s$ (i.e. they have the same density)?

Disclosure: this is part of a homework question, but the TA said we can just claim this is true "by symmetry" without any additional work. There is more to the problem. I wanted to know if there is a more rigorous proof that this is true.

Intuitively, this seems true since $Z = X + Y$. (It doesn't seem true in general, for example of $Z = 2X + Y$). My approach to show this is to show

$$ \mathbb{P}\{X \leq s, Z \leq t\} = \mathbb{P}\{Y \leq s, Z \leq t\} \quad \forall s, t $$

and so

\begin{align} \mathbb{P}\{X \leq s, Z \leq t\} &= \mathbb{P}\{Y \leq s, Z \leq t\} \\ \mathbb{P}\{X \leq s, X+Y \leq t\} &= \mathbb{P}\{Y \leq s, X+Y \leq t\} \\ \mathbb{P}\{X \leq s, Y \leq t-X\} &= \mathbb{P}\{Y \leq s, X \leq t-Y\} \end{align}

Now we've put the joint CDF of $X$ and $Y$. But at this point I am stuck.

Answer 1

I would agree with your TA on this one. I'm not saying it's a bad idea to consider other arguments. My point is that using symmetry is rigorous, even though it might not feel like it.

If you take all the assumptions and swap $X$ and $Y$, we get the exact same assumptions. (This is why $Z=2X+Y$ doesn't work, because $2X+Y \ne 2Y+X$). We say that the problem is "symmetric in $X$ and $Y$". Now, if the assumptions doesn't change when swapping $X$ and $Y$, neither can any conclusion we draw from them. Therefore we can safely say that $(X,Z)$ has the same joint distribution as $(Y,Z)$.

EDIT: If you're not convinced, think of it like this: We first work out the distribution of $(X,Z)$. Then we start over in order to calculate the distribution of $(Y,Z)$. But before we begin, we can rewrite all the assumptions so $X$ and $Y$ are swapped, i.e. $Y,X\sim {\rm Unif}(0,1)$, $Y \perp \!\!\! \perp X$ and $Z = Y + X$. But now everything is like before, except we call the variables different names. Therefore all calculations must be the same as before, so we will get the same answer.

Answer 2

For $0\leq t-s \leq 1$, we have \begin{align} \mathbb{P}\left\{X\leq s,Y\leq t-X\right\}&=\mathbb{P}\left\{X\leq s,X\leq t-Y\right\} \\ &= \int_{0}^{t-s} \mathbb{P}\left\{X\leq s\right\}dy + \int_{t-s}^{1} \mathbb{P}\left\{X\leq t-y\right\}dy \\ &= \int_{0}^{t-s} s dy + \int_{t-s}^{1} (t-y)dy \\ &= s(t-s) + t(1-t+s)-(1-(t-s)^2)/2 = st-s^2/2-(1-t)^2/2. \end{align} So, we have, \begin{equation} \mathbb{P}\left\{X\leq s,Y\leq t-X\right\}=\begin{cases} st-s^2/2-(1-t)^2/2,& \text{ if } 0\leq t-s\leq 1\\ 0, & \text{ otherwise}. \end{cases} \end{equation}

Answer 3

Would a geometric proof be rigorous enough for you? $X, Y \sim U(0,1)$ and $X \perp Y$ means you're picking a uniform random dot inside the square $[0,1]^2$. Conditioning on $Z\le z$ means you are restricted to a "feasible" subset of the square on the "lower left" side of the line $X+Y=z$ which is a $-45^\circ$ line.

When $z \le 1$ the feasible region is a lower-left triangle of the square, while if $z > 1$ the feasible region is the square with an upper-right triangle cut off. In both cases you can evaluate $P(X\le x|Z\le z)$ by geometry (it's probably a standard homework problem). But the point is of course that the feasible is region is symmetric w.r.t. exchanging $X$ and $Y$ (i.e. reflection through the $X=Y$ line, i.e. the $+45^\circ$ line through the origin).

Answer 4

$\newcommand{\P}{\mathbf P}\newcommand{\PM}{\mathbb P}$This holds more generally we don't need that the random variables are uniformly distributed for that. Let $X,Y$ i.i.d. random variables, say with marginal density $f$. Then $(X,X+Y)$ and $(Y,X+Y)$ have the same distribution. Indeed, let $g(x,y):=(x,x+y)$ and notice that for a (measurable) set $A\subset\mathbb R^2$ \begin{align} \mathbb P(g(X,Y)\in A)=\PM((X,Y)\in g^{-1}(A))=\iint_{g^{-1}(A)}f(x)f(y)\,d(x,y) \end{align} But now we call $x=y'$ and $y=x'$ (this is the symmetry that is referred to) to see that \begin{align} \mathbb P(g(X,Y)\in A)=\iint_{g^{-1}(A)}f(y')f(x')\,d(y',x')=\PM((Y,X)\in g^{-1}(A))=\PM(g(Y,X)\in A) \end{align} This shows $g(X,Y)$ has the same distribution as $g(Y,X)$ which is what we were aiming for.

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Measure Theory approach. This holds even more generally, say even if there is no density. In that case we write \begin{align} \mathbb P(g(X,Y)\in A)=\PM((X,Y)\in g^{-1}(A))=\P_{X,Y}(g^{-1}(A)) \end{align} where $\P_{X,Y}$ is the law of $(X,Y)$. Due to i.i.d. we know that the law of $X$ and $Y$ are equal, say, equal to $P$. Then for all sets of the form $U=(-\infty,u]$ and $V=(-\infty,v]$ \begin{align} \P_{X,Y}(U\times V)\stackrel{\text{i.i.d.}}{=}P(U)P(V)=P(V)P(U)=\P_{Y,X}(V\times U) \end{align} But since sets of the form $U\times V$ form a $\pi$-system generating the Borel $\sigma$-algebra, we conclude $\P_{X,Y}=\P_{Y,X}$ due to uniqueness of measure. In particular \begin{align} \PM(g(X,Y)\in A)=\P_{X,Y}(g^{-1}(A))=\P_{Y,X}(g^{-1}(A))=\PM(g(Y,X)\in A) \end{align}