Dgdrxrt

Let

• $X, Y sim textUnif(0,1)$


• 
  $X perp !!! perp Y$, and


• $Z = X + Y$

Is it true that $f_Xmid Z=z(s)$ and $f_Ymid Z=z(s)$ 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

$$
mathbbPX leq s, Z leq t = mathbbPY leq s, Z leq t quad forall s, t
$$

and so

beginalign
mathbbPX leq s, Z leq t &= mathbbPY leq s, Z leq t
\
mathbbPX leq s, X+Y leq t &= mathbbPY leq s, X+Y leq t
\
mathbbPX leq s, Y leq t-X &= mathbbPY leq s, X leq t-Y
endalign

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

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. $Y,Xsim rm Unif(0,1)$, $Y perp !!! perp X$ 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.

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 $Zle z$ 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 $P(Xle x|Zle z)$ 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).

$newcommandPmathbf PnewcommandPMmathbb P$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 $Asubsetmathbb R^2$
beginalign
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)
endalign
But now we call $x=y'$ and $y=x'$ (this is the symmetry that is referred to) to see that
beginalign
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)
endalign
This shows $g(X,Y)$ has the same distribution as $g(Y,X)$ which is what we were aiming for.

Measure Theory approach. This holds even more generally, say even if there is no density. In that case we write
beginalign
mathbb P(g(X,Y)in A)=PM((X,Y)in g^-1(A))=P_X,Y(g^-1(A))
endalign
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 $V=(-infty,v]$
beginalign
P_X,Y(Utimes V)stackreltexti.i.d.=P(U)P(V)=P(V)P(U)=P_Y,X(Vtimes U)
endalign
But since sets of the form $Utimes V$ form a $pi$-system generating the Borel $sigma$-algebra, we conclude $P_X,Y=P_Y,X$ due to uniqueness of measure. In particular
beginalign
PM(g(X,Y)in A)=P_X,Y(g^-1(A))=P_Y,X(g^-1(A))=PM(g(Y,X)in A)
endalign

For $0leq t-s leq 1$, we have
beginalign
mathbbPleftXleq s,Yleq t-Xright&=mathbbPleftXleq s,Xleq t-Yright \
&= int_0^t-s mathbbPleftXleq srightdy + int_t-s^1 mathbbPleftXleq t-yrightdy \
&= 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.
endalign
So, we have,
beginequation
mathbbPleftXleq s,Yleq t-Xright=begincases
st-s^2/2-(1-t)^2/2,& text if 0leq t-sleq 1\
0, & text otherwise.
endcases
endequation