If $X+Y$ is independent from $X$ and independent from $Y$, then $X+Y$ is deterministic

Question

Let $X,Y$ be two real random variables such that $X+Y$ is independent from $X$ and also independent from $Y$. Show that $X+Y$ is deterministic

If $X$ and $Y$ have a variance, then we can compute : $$\mathbb E[(X+Y)^2] = \mathbb E[ X (X+Y)] + \mathbb E[Y(X+Y)] = \mathbb E[ X ]\mathbb E[X+Y] + \mathbb E[Y]\mathbb E[X+Y] = (\mathbb E[X+Y])^2$$ and therefore $\operatorname{Var}(X+Y) = 0$ and $X+Y$ is almost surely constant.

In the general case however, I don't see how to prove this. Using the characteristic function $f_X(t) = \mathbb E[e^{itX}]$, I can show that : $$\forall t,s\in\mathbb R,f_X(t) f_{X+Y}(s) = f_{X+Y}(s+t) f_Y(-t)$$ I am not sure if and how this helps, though.

Answer 1

Notation: I use $Z\perp Y$ to denote that $Z$ and $Y$ are independent random variables, and $\phi_Z(t)=E[\exp(itZ)]$ for the characteristic function of the random variable $Z$.

Assume $X+Y\perp X$ and $X+Y\perp Y$. Then, for all $s,t\in\mathbb{R}$

\begin{align} \phi_{X+Y}(s+t)\phi_Y(-t)&=E[\exp(i(X+Y)(s+t))\exp(-itY)]\\ &=E[\exp(is(X+Y))\exp(itX)]\\ &=\phi_{X+Y}(s)\phi_X(t) \end{align}

Similarly,

\begin{align} \phi_{X+Y}(s+t)\phi_X(-t)&=E[\exp(i(X+Y)(s+t))\exp(-itX)]\\ &=E[\exp(is(X+Y))\exp(itY)]\\ &=\phi_{X+Y}(s)\phi_Y(t) \end{align}

Hence, for all $s,t$ in a neighborhood of $0$

\begin{align} \phi_{X+Y}(s+t)&=\phi_{X+Y}(s)\frac{\phi_X(t)}{\phi_Y(-t)}=\phi_{X+Y}(t+s)=\phi_{X+Y}(t)\frac{\phi_X(s)}{\phi_Y(-s)}\\ \phi_{X+Y}(s+t)&=\phi_{X+Y}(s)\frac{\phi_Y(t)}{\phi_X(-t)}=\phi_{X+Y}(t+s)=\phi_{X+Y}(t)\frac{\phi_Y(s)}{\phi_X(-s)} \end{align}

Thus $\phi_{X+Y}(s)=\frac{\phi_X(s)}{\phi_Y(-s)}=\frac{\phi_Y(s)}{\phi_X(-s)}$ for all $s$ in a neighborhood of $0$. Notice that $|\phi_{X+Y}(s)|^2=\frac{\phi_X(s)\phi_X(-s)}{\phi_Y(s)\phi_Y(-s)}=1$. Then $|\phi_{X+Y}(s)|=1$ in a neighborhood of $0$; hence $X+Y$ is constant.

The conclusion is based on the following results:

Lemma: Suppose $\mu$ is a Borel probability measure on $\mathbb{R}$. If $|\widehat{\mu}(t)|=1$ for some $t\neq0$, then there are $b\in\mathbb{R}$ and $h>0$ such that $\operatorname{supp}\mu\subset b+h\mathbb{Z}$.

Proof: Without loss of generality assume $t>0$. Then form some $\theta\in(-\pi,\pi]$, \begin{align} 1=e^{-i\theta}\widehat{\mu}(t)=\int \cos(xt-\theta)\,\mu(dx) \end{align} Then $x\mapsto\cos(xt-\theta)=1$ $\mu$--a.s. Hence, $\operatorname{supp}\mu\subset \frac{\theta}{t}+\frac{2\pi}{t}\mathbb{Z}$.

Theorem Suppose $\mu$ is a probability measure on $\mathbb{R}^n$. If $|\widehat{\mu}({\bf t})|=1$ in a small neighborhood of ${\bf 0}$, then $\mu=\delta_{\boldsymbol{b}}$ for some $\boldsymbol{b}\in\mathbb{R}^n$.

Proof: By considering the law of each component of $\mathbb{R}^n$, it is enough to consider the case $n=1$. For every $t$ in a neighboorhood of $0$, there is a number $b_t$ such that $\operatorname{supp}(\mu)\subset b_t+\frac{2\pi}{|t|}\mathbb{Z}$. If $\mu$ is not a trivial distribution, then there are distinct points $x_1$ and $x_2$ of positive measure $\mu$. It follows that $|x_1-x_2|\geq\frac{2\pi}{|t|}$. This is not possible as $|t|$ can be taken to be very small.

Answer 2

If $S= X+Y$ is independent of $X$ and $Y$, any measurable function of $S$ is independent of any measurable function of $X$ and $Y$. So, $S$ is independent of $f(X,Y) = X+Y$, and so $S$ is independent of itself, hence almost-surely constant.

EDIT: The original poster has clarified their problem, which is that $X+Y$ is independent of $\sigma(X)$ and $\sigma(Y)$ separately, instead of $\sigma(X,Y)$. In the former case, this explanation does not hold.

Answer 3

A small addition @Julius' answer and a slight caution to Solublefish is that the fact that "if $X$ is an r.v. such that $P(X\in A)=0$ or $1$ for each Borel set $A$ implies $X$ is a constant" is not directly apparent. One has to go by Cantor's Intersection Theorem to prove this fact and is infact a nice problem in measure theory. Well, just keep shrinking compact intervals of probability $1$ and you'll get a unique point $x_{0}$ by Cantor's Intersection Theorem to which the distribution of $X$ will give mass $1$. I cannot seem to find a link to this in this site so I'll just write it up myself.

So let $\mu$ be the distribution of $X+Y$ in this case, i.e. $\mu(A)=P((X+Y)\in A)$ for all $A\in\mathcal{B}(\Bbb{R})$. As $X+Y$ is independent of itself (By Julius') answer.

So now consider $\mu([n,n+1))$ where $n\in\Bbb{Z}$. Then atleast one of these $[n,n+1)$ has to have non-zero measure, i.e. $\mu([n,n+1))\neq 0$ as otherwise $\mu(\bigcup_{n\in\Bbb{Z}}[n,n+1))=\mu(\Bbb{R})=0$.

So assume $\mu([n,n+1))=1$ . Now if $\mu(\{n\})=1$, then you are done and $X+Y=n$

Otherwise $\mu((n,n+1))=1$. Now divide $[n,n+1]$ into $[n,n+\frac{1}{2}]$ and $[n+\frac{1}{2},n+1]$ . Again, by the above argument, we will asume that $\mu(\{n+\frac{1}{2}\})=0$ as otherwise the proof is completed automatically.

So atleast one and exactly one of $[n,n+\frac{1}{2}]$ or $[n+\frac{1}{2},n+1]$ must be given mass $1$. Suppose it is $[n,n+\frac{1}{2}]$.

Then again you divide this interval into two parts and carry on the same procedure inductively. If you stop at a finite time, then you have found your constant $c$ which is an end point of one of such interval.

Otherwise, if the process goes on and on, then you have a decreasing sequence of compact intervals $I_{k}$ such that $I_{k}\subseteq I_{k-1}$ and hence by Cantor's Intersection Theorem, you'll find an unique $x_{0}$ such that $\{x_{0}\}=\bigcap_{k}I_{k}$ and

$\mu(\{x_{0}\})=\mu(\bigcap_{k}I_{k})=\lim_{m\to\infty}\mu(\bigcap_{k=1}^{m}I_{k})=\lim_{m\to\infty}\mu(I_{m})=1$

NOTE Using (hyper)cubes instead of intervals, you can see that this result will also be true for $\Bbb{R}^{n}$ valued random variables.