# Sum of independent random variables almost surely constant

I am trying to solve the following problem:

Let $(\Omega, \mathbb{A}, \mathbb{P})$ be a probability space and $X_1, X_2, \ldots, X_n$ independent real random variables. Prove that the sum $X_1 + X_2 + ... + \ldots X_n$ is $\mathbb{P}$-almost surely constant iff each $X_i$ is $\mathbb{P}$-almost surely constant.

Do you have any ideas or hints how to tackle with this problem? Thanks.

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It suffices to show: $X+Y=c$ almost surely $\Rightarrow X$ constant almost surely where $X,Y$ are independent random variables.

Let $\xi,\eta \in \mathbb{R}$. By the independence we have

$$\mathbb{E}e^{\imath \, (X,X+Y) \cdot (\xi,\eta)} = \mathbb{E}e^{\imath \, \xi \cdot X} \cdot \mathbb{E}e^{\imath \, (X+Y) \cdot \eta} = \mathbb{E}e^{\imath \, \xi \cdot X} \cdot \mathbb{E}e^{\imath \, \eta \cdot X} \cdot \mathbb{E}e^{\imath \, \eta \cdot Y}$$

On the other hand

$$\mathbb{E}e^{\imath \, (X,X+Y) \cdot (\xi,\eta)} = \mathbb{E}e^{\imath \, (\xi+\eta) \cdot X} \cdot \mathbb{E}e^{\imath \, \eta \cdot Y}$$

Since $$\mathbb{E}e^{\imath \, \eta \cdot X} \cdot \mathbb{E}e^{\imath \, \eta \cdot Y} = \mathbb{E}e^{\imath \, (X+Y) \cdot \eta} = e^{\imath \, \eta \cdot c} (\not=0)$$ we have $\mathbb{E}e^{\imath \, \eta \cdot Y} \not= 0$ for all $\eta$. Hence we obtain from the first two equations

$$\mathbb{E}e^{\imath \, (X,X) \cdot (\eta,\xi)} = \mathbb{E}e^{\imath \, (\xi+\eta) \cdot X} = \mathbb{E}e^{\imath \, X \cdot \eta} \cdot \mathbb{E}e^{\imath \, X \cdot \xi}$$

This means that $X$ is independent of $X$ and therefore almost surely constant.

Remark In the last step we used the following theorem: Two random variables $U,V$ are independent $$\Leftrightarrow \forall \xi,\eta: \mathbb{E}e^{\imath (U,V) \cdot (\xi,\eta)} = \mathbb{E}e^{\imath \, U \cdot \xi} \cdot \mathbb{E}e^{\imath \, V \cdot \eta}$$

Another approach: Let $S_n := \sum_{j=1}^n X_j$. We have $0=\mathbb{V}S_n= \sum_{j=1}^n \mathbb{V}X_j$. This implies $\mathbb{V}X_j=0$ and therefore $X_j = \mathbb{E}X_j$ a.s.. The problem is that one has to show $X_j \in L^2$ (to do these calculations).

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I suppose $\mathbb{V}$ indicates variance? –  Nate Eldredge Nov 25 '12 at 13:53
Exactly... totally forgot to mention it. –  saz Nov 25 '12 at 15:39

I will prove the $\implies$ part. Assume that $X+Y$ is a constant a.s now since $\mathbb{R}=\cup_n[n,n+1]$ so we can find an interval $I_1$ such that $P(X\in I_1)>0$ by the same idea we can divide $I_1$ into two parts so we can find an interval $I_2\subset I_1$ such that $P(X\in I_2)>0$. By induction we can find a nested sequence of intervals $I_n$ such that $P(X\in I_n)>0$. Since $\cap_nI_n$ is a single point by our construction call it $a$ so $P(X=a)=\lim_nP(X\in I_n)$ so if $X$ is not a constant a.s then we can find $n_0$ such that $0<P(X\in I_{n_0})<1$. now since $X+Y$ is equal to a constant $c$ a.s so $P(X\in I_{n_0}\cap Y\notin c-I_{n_0})=0$ and since $X,Y$ are independent so $P(Y\notin c-I_{n_0})=0$ also using that $P(X\notin I_{n_0}\cap Y\in c-I_{n_0})=0$ we get $P(Y\in c-I_{n_0} )=0$ contradiction.

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Thanks a lot for your comments. Got it. –  eugen1806 Nov 25 '12 at 19:50