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This is an exercise in Jacod's Probability Essentials:

Let $X$ and $Y$ be independent random variables and $P(X+Y=\alpha)=1$ where $\alpha\in{\Bbb R}$ is some constant. Show that both $X$ and $Y$ are constant random variables.

What I think is that one might use Borel-Cantelli theorem here. Since $$ \bigcup_{i=0}^{\infty}\{X=i,Y=\alpha-i\}\subset\{X+Y=\alpha\}=\bigcup_{\beta\in{\Bbb R}}\{X=\beta,Y=\alpha-\beta\}, $$ we have $$ \begin{align} P\bigg(\bigcup_{i=0}^{\infty}\{X=i,Y=\alpha-i\}\bigg)=\sum_{i=0}^{\infty}P(X=i,Y=\alpha-i)<\infty \end{align} $$ But this seems to give nothing. Also, I'm surprised about the result is that $X=\gamma$ for some constant $\gamma$ instead of $P(X=\gamma)=1$. Any idea about how I can go on?

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I think that what is meant by $X$ being a constant variable is exactly that $P(X=\gamma)=1$ for some $\gamma$ and not that $X=\gamma$ pointwise. – Stefan Hansen Oct 17 '12 at 19:01

2 Answers 2

If $X+Y=\alpha$ then $X=\alpha-Y$. The random variable $Y$ cannot be independent of $\alpha-Y$, since as $Y$ gets bigger, $\alpha-Y$ gets smaller and as $Y$ gets smaller, $\alpha-Y$ gets bigger. Unless, of course, $Y$ never gets bigger or smaller.

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up vote 0 down vote accepted

By assumption, $X=\alpha-Y $ almost surely. Since $X$ and $Y$ are independent, so are $X$ and $\alpha-Y$ by properties of independent random variables. It follows that $X$ is independent with itself, which implies that $$ P(X\leq x)=P(X\leq x,\ X\leq x)=[P(X\leq x)]^2,\quad x\in{\Bbb R}. $$ Therefore, $P(X\leq x)=0$ or $1$. Using the properties of probability distribution function one can show (exercise) that $X$ be a constant (almost surely).

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