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

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.

• 1. It's better to write text without special formatting. 2. My prof told me not to write quantifiers in words in a sentence. Apr 26, 2018 at 1:02
• @GNUSupporter Can u teach me how to highlight? Apr 26, 2018 at 1:04
• $\rm\LaTeX$ 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** Apr 26, 2018 at 1:07
• 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. Apr 26, 2018 at 1:16
• @GNUSupporter I edited the question Apr 26, 2018 at 1:52

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.

$$\to c=X+Y$$

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

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

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

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

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

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

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

Thanks to @Daniel Schepler, here's my answer:

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

• If X and Y are independent, then X and Y are square integrable?
– BCLC
May 1, 2018 at 11:38
• @BCLC You are right... May 1, 2018 at 11:40
• izimath, why? 
– BCLC
May 1, 2018 at 11:40
• @BCLC It is possible to show that they are square integrable May 1, 2018 at 11:56

$$P(Y=c-X)= 1$$

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