# Show: $X_1, \dots, X_n$ are almost surely constant $\Leftrightarrow\sum_{i=1}^n X_i$ is almost surely constant.

Let $X, X_1, \dots, X_n$ be independent, real random variables. I need to show:

1. The pair $(X,X)$ is independent $\Leftrightarrow$ $X$ is almost surely constant
2. $X_1, \dots, X_n$ are almost surely constant $\Leftrightarrow\sum_{i=1}^n X_i$ is almost surely constant.

I know that if a random variable $X$ is almost surely constant it means that there exist a real number $a$ with $P(X=a)=1$. I've proven the $\Rightarrow$ direction of 2. but I don't get anywhere with the rest. Also I'm not quite sure what it means that the "pair $(X,X)$ is independent"
Does anyone have some tipps or ideas? Thanks in advance!

• can I ask where you find this exercise? – Masacroso Jul 8 '17 at 12:38
• It's not from a book, but from an exercise sheet of my university – DeltaChief Jul 8 '17 at 12:42
• math.stackexchange.com/q/464264/9464 – Jack Jul 8 '17 at 12:52
• You might want to be aware that the solution to 2. you accepted is not the one you are supposed to come with. As usual, quickly accepting answers effectively shuts the question and prevents others to post alternative, possibly more pertinent, answers. – Did Jul 9 '17 at 22:14
• should i unmark it then and wait for a more complete anser? – DeltaChief Jul 9 '17 at 22:23

To deal with first point, it's nesessary to know what definition of independence of two random values you can use.

Say, for the following definition: $X$ is idependent on $X$, iff for any $x,y\in\mathbb R$ $$\mathbb P(X\leq x, X\leq y)=\mathbb P(X\leq x)\mathbb P(X\leq y).$$ Таke $x=y$ be any real number. Then $$\mathbb P(X\leq x)=\left(\mathbb P(X\leq x)\right)^2,$$ that is CDF $F_X(x)=\mathbb P(X\leq x)$ is either $0$ or $1$. Since it is non-decreasing and have zero limit at $-\infty$, unit limit at $+\infty$ and right-continuous at any point, there exists some $a$ s.t. $F_X(x)=0$ for $x<a$ and $F_X(x)=1$ for $x\geq a$. This is the CDF of degenerate distribution $\mathbb P(X=a)=1$.

For second point. If this is allowed, you can use the properties of variance. $$0=\text{Var}(X_1+\ldots+X_n)=\text{Var}(X_1)+\ldots+\text{Var}(X_n)\geq 0$$ The r.h.s. is zero only if all the summands are zero. And $\text{Var}(X_i)=0$ implies that $X_i$ is a.s. constant.

EDIT: As Did pointed out, the equality between variance of the sum and te sum of variances requires square integrability of summands.

Prove the second point without additional requrements. Consider the case of two summands only.

Let $X$ and $Y$ are independent and $X+Y$ is a constant a.s. Prove that both $X,Y$ are a.s. constant.

By contradiction, let the distribution of $X$ is non-degenerate. Then there exists two disjoint intervals $[a,b]$ and $[c,d]$, where $b<c$ s.t. $$\mathbb P(X\in[a,b])>0\quad \text{ and } \quad \mathbb P(X\in[c,d])>0.$$

Fix $\varepsilon>0$ s.t. $b+\varepsilon<c$. For $Y$, there exists some $z$ s.t. $$\mathbb P(Y\in[z,z+\varepsilon])>0.$$ Then the intersection of independent events $\{a\leq X \leq b\}$ and $\{z\leq Y\leq z+\varepsilon\}$ implies that $$a+z \leq X+Y \leq b+z+\varepsilon < c+z,$$ and therefore $$\mathbb P(a+z \leq X+Y < c+z)\geq \mathbb P(a+z \leq X+Y\leq b+z+\varepsilon )\geq \mathbb P(a\leq X \leq b)\mathbb P(z\leq Y\leq z+\varepsilon)>0.$$ Also $$\mathbb P(c+z \leq X+Y\leq d+z+\varepsilon )\geq \mathbb P(c\leq X \leq d)\mathbb P(z\leq Y\leq z+\varepsilon)>0.$$ We prove that there exists two disjoint intervals $[a+z, c+z)$ and $[c+z,d+z+\varepsilon]$ s.t. probabilities of $X+Y$ belong to any of these intervals are positive, and then the distribution of $X+Y$ is non-degenerate too. Contradiction.

• The solution suggested for 2. is certainly not the one those who asked the exercise have in mind since it assumes square integrability and 2. is valid for every random variables. – Did Jul 9 '17 at 22:10
• @Did If we have that sum is a constant a.s., it is square integrable. – NCh Jul 10 '17 at 3:22
• @Did But I'm really stuck proving that the finiteness of the variance of the sum of independent r.v.'s implies the existence of variances of each. – NCh Jul 10 '17 at 4:21
• "f we have that sum is a constant a.s., it is square integrable" Wrong argument. The sum is square integrable, yes, but you need its parts to be square integrable. – Did Jul 10 '17 at 5:27
• @Did I added the proof that does not required square integrability. Thanks! – NCh Jul 10 '17 at 7:44