5
$\begingroup$

Good day,

In class we said that if a random variable $X-Y$ is independent of random variables $X$ and $Y$ then $X-Y$ is almost sure constant, i.e. there exists a $c \in \mathbb{R}$ such that $P(X-Y=c)=1$.

First, I don't exactly how to prove this. I know that $X$ is constant if it is independent of itself. Therefore I could prove that $X-Y$ is independent of itself (But the other directions doesn't hold I suppose). Do I know that $X-Y$ is independent of itself?

Is it correct to say: If $Z$ is independent of $X$ and $Y$ then it is independent of $g(X,Y)$ where $g$ is a measurable function.

I don't think so. The definition of independence doesn't give this property.

Then how do I prove that $X-Y$ is almost sure constant? Another approach through expectations:

$$E(X-Y|X)=E(X-Y|Y)=E(X-Y)=EX-EY $$

But is seems not leading me to a goal.

So: Why is the random variable $X-Y$ almost sure constant if it is independent of $X$ and $Y$?

Is this valid for a general random variable $f(X,Y)$ (where $f$ is measurable for example)? i.e. $f(X,Y)$ is almost sure constant it it is independent of $X$ and $Y$?

If not I would ask for a counterexample.

Thanks a lot for your help, Marvin

$\endgroup$
5
$\begingroup$

Let $\phi_A(t)$ be the characteristic function of random variable $A$.Then you know that if $A,B$ are independent, then $\phi_{A+B}(t)=\phi_A(t)\phi_B(t)$.

You have that $X-Y$ is independent of $X$ and $Y$. Noting that $X=(X-Y)+Y$, you have that $\phi_X(t)=\phi_{(X-Y)+Y}(t)=\phi_{X-Y}(t)\phi_Y(t)$ [using the fact that $X-Y$ and $Y$ are independent] for every $t\in\mathbb R$. Thus $\phi_{X-Y}(t)=\dfrac{\phi_X(t)}{\phi_Y(t)}$.

Also you have, similarly, that $\phi_Y(t)=\phi_{Y-X}(t)\phi_X(t)$ implying $\phi_{Y-X}(t)=\dfrac{\phi_Y(t)}{\phi_X(t)}$.

Let us call $Z:=X-Y$ for brevity of notation. Then the above two give that $\phi_Z(t)=\dfrac{\phi_X(t)}{\phi_Y(t)}$ and $\phi_{-Z}(t)=\dfrac{\phi_Y(t)}{\phi_X(t)}$.

Hence $\phi_Z(t)\phi_{-Z}(t)=1$, for every $t\in\mathbb R$.

Using the fact that $\phi_{-Z}(t)=\overline{\phi_{Z}(t)}$, we have that $|\phi_Z(t)|=1$ for all $t\in\mathbb R$.

Thus, $\phi_Z(t)=e^{ig(t)}$ for some function $g$, for all $t\in\mathbb R$.

Now observe that for all $t$, $e^{ig(t)}=E(e^{itZ})$ implies $E(e^{i(tZ-g(t))})=1$ for all $t$. Noting that $|e^{i(tZ-g(t))}|=1$ we must have that $tZ-g(t)=2k(t)\pi$ for an integer valued function $k$, almost surely.

Thus almost surely, $Z=\dfrac{g(t)+2k(t)\pi}{t}$ for all $t\in\mathbb R$.

Since the LHS is independent of $t$, we may choose any $t$, say $t=1$. Then $Z=g(1)+2k(1)\pi$ which is afterall, a constant, almost surely.

Hence $X-Y$ is constant almost surely.

Now let us see your other questions. You want to know if $Z$ is independent of $X$ and $Y$ then is it true that $Z$ is independent of $g(X,Y)$ for any measurable function $g$?

So consider for Borel sets $A,B$ the following: $P(Z\in A, g(X,Y)\in B)=P[Z\in A, (X,Y)\in g^{-1}(B)]$. This can be written as the product $P[Z\in A]P[(X,Y)\in g^{-1}(B)]$ if and only if $Z$ is JOINTLY INDEPENDENT with $X,Y$. There exist examples where $Z$ is independent of each $X,Y$ but maybe not jointly. Here's one:

Throw two dice independently. Define $A$ to be the event that $7$ is obtained as the sum of the two throws, $B$ be the event that $3$ is obtained on first throw and $C$ be the event that $4$ is obtained on second throw. Then you can check that $A,B,C$ are pairwise independent but not jointly independent ($A$ is NOT jointly independent with $B$ and $C$.) For example with random variables, take $X=1_B,Y=1_C,Z=1_A$.

As you see, we crucially used the structure of $f(X,Y)=X-Y$. I cannot say right now if it can always be said that if $f(X,Y)$ is independent of both $X$ and $Y$ then it is a.s. constant.

$\endgroup$
  • $\begingroup$ Great answer, I like the approach via the characteristic function. Thank you for the explanation to my 2nd question und the counterexample, pretty understandable. $\endgroup$ – Fritz Sep 27 '16 at 14:41
1
$\begingroup$

Extending Landon Carter's answer,

i.e. $f(X,Y)$ is almost sure constant it it is independent of $X$ and $Y$?

No. Flip a coin twice, let $X_i = \mathbf{1}_{\text{H on $i$th flip}}$, $i=1,2$, $Z = \mathbf{1}_{X_1=X_2}$. Then $Z$ is independent of both $X_1$ and $X_2$, but is not constant. (Note that here $Z$ is moreover a function of the difference $X_1-X_2$.)

As Landon Carter wrote, joint independence is enough. A weaker sufficient condition is that $f(X,Y)$ is independent of the vector $(X,Y)$. Indeed, in this case $f(X,Y)$ is independent of any transformation of the vector $(X,Y)$. In particular, it is independent of itself, which means that it is constant a.s.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.