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I would like to show:

Let $X,Y$ be independent RVs. If there exists $c \in \mathbb R $ s.t. $P(X+Y=c)=1$, then $X,Y$ are constants a.s..

What I tried:

Since X,Y are indep,$$1=P(X+Y=c)=\int 1_{x+y=c} \, d \mu _x \, d \mu_y = \int P(X=c-y) \, d \mu_y$$ but it gets nowhere from here.

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  • $\begingroup$ 1. It's better to write text without special formatting. 2. My prof told me not to write quantifiers in words in a sentence. $\endgroup$ Apr 26, 2018 at 1:02
  • $\begingroup$ @GNUSupporter Can u teach me how to highlight? $\endgroup$
    – izimath
    Apr 26, 2018 at 1:04
  • $\begingroup$ $\rm\LaTeX$ provides \emph{...} to emphasize text (default in italics). To render italics in Markdown, you surround the text *like this*; to render boldface in Markdown, you surround the text **like this** $\endgroup$ Apr 26, 2018 at 1:07
  • $\begingroup$ This is false. $X,Y$ constant a.s. is the best you can get. Think about $X$, $Y$ taking other values on a measure-zero set. $\endgroup$ Apr 26, 2018 at 1:16
  • $\begingroup$ @GNUSupporter I edited the question $\endgroup$
    – izimath
    Apr 26, 2018 at 1:52

3 Answers 3

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If $X$ and $Y$ are square integrable (*) then we may consider $Var(X+Y)$ to say:

$$0 = Var(X+Y) =Var(X)+Var(Y) \to 0 = Var(X)=Var(Y)$$

Otherwise:

Two integrals for two random variables!

$$1=P(X+Y=c)=\int\int 1_{x+y=c} \, d \mu _x \, d \mu_y $$

$$=\int\int 1_{d+y=c,x=d} \, d \mu _x \, d \mu_y (d \in \operatorname{Range}(X))$$

$$=\int\int 1_{d+y=c}1_{x=d} \, d \mu _x \, d \mu_y $$

$$=\int 1_{d+y=c}\int1_{x=d} \, d \mu _x \, d \mu_y $$

$$=\int 1_{d+y=c}\mu_x(x=d) \, d \mu_y $$

$$=\mu_x(x=d) \int 1_{d+y=c} \, d \mu_y $$

$$=\mu_x(x=d) \mu_y(d+y=c)$$

$$=\mu_x(x=d) \mu_y(y=c-d)$$

$$\to 1 =\mu_x(x=d) = \mu_y(y=c-d)$$


(*) Hmmm...I guess if $Z=c$ a.s. then $E[Z], E[|Z|], E[Z^2], Var(Z) < \infty$.

But if $\exists$ independent $X, Y$ s.t. $Z=X+Y$, then does that mean that $X$ and $Y$ are square integrable? I was thinking $\infty - \infty$, but I guess that's undefined. Thus, $X,Y < \infty$ a.s.

$$\to c=X+Y$$

$$\to c^2=(X+Y)^2$$

$$\to E[c^2]=E[(X+Y)^2]$$

$$\to c^2=E[X^2+2XY+Y^2]$$

$$\to c^2=E[X^2]+2E[XY]+E[Y^2]$$

$$\to c^2=E[X^2]+2E[X]E[Y]+E[Y^2]$$

$$\to E[X^2], E[XY], E[X]E[Y], E[X], E[Y], E[Y^2] < \infty$$

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    $\begingroup$ Is $d$ arbitrary? $\endgroup$
    – izimath
    May 1, 2018 at 12:08
  • $\begingroup$ @izimath Good question. Edited $\endgroup$
    – BCLC
    May 1, 2018 at 12:43
  • $\begingroup$ Fubini implies that $E(X+y)^2 <\infty$ for some real $y$, which implies $X$ is square integrable. $\endgroup$
    – izimath
    Sep 3, 2020 at 5:56
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Thanks to @Daniel Schepler, here's my answer:

$$Var(X) + Var(Y) = Var(X+Y)=E[(X+Y)^2]-{E[(X+Y)]}^2 = c^2 -c^2 =0 \\ \Rightarrow Var(X)=0=Var(Y)$$

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  • $\begingroup$ If X and Y are independent, then X and Y are square integrable? $\endgroup$
    – BCLC
    May 1, 2018 at 11:38
  • $\begingroup$ @BCLC You are right... $\endgroup$
    – izimath
    May 1, 2018 at 11:40
  • $\begingroup$ izimath, why? $ $ $\endgroup$
    – BCLC
    May 1, 2018 at 11:40
  • $\begingroup$ @BCLC It is possible to show that they are square integrable $\endgroup$
    – izimath
    May 1, 2018 at 11:56
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$$P(Y=c-X)= 1$$

Thus $$\sigma(Y) \subseteq \sigma(X)$$

But $\sigma(Y)$ and $\sigma(X)$ are independent.

Thus $\sigma(Y)$ is independent of itself!

$\to P(A) \in \{0,1\} \ \forall A \in \sigma(Y)$. Choose $$A=\{Y=\inf\{y \mid F_Y(y)=1\}\}.$$

Convince yourself $P(A) > 0$.

Thus $P(A) = 1 \to P(\{X=\inf\{x \mid F_X(x)=1\}\})=1=P(X=c-\inf\{y \mid F_Y(y)=1\})$

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