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Let $x_i$ be uniformly distributed random variables in the interval $[0,a]$, with $a>0$.

Let f(x) be equal to 1 if x=0, and 0 otherwise.

Let $$S(a)=\sum_{n=1}^\infty f(x_n)$$

What is S(1)?

What is $\lim_{a->0} S(a)$ from positive side?

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Can you clarify what exactly you are intending with your statement regarding the "limit"? –  cardinal Mar 30 '11 at 13:20

3 Answers 3

up vote 3 down vote accepted

The series $S(a)=\sum_{n=1}^\infty f(x_n)$ has countably many terms, and any countable set has measure $0$ while any interval has positive measure, so $S(1) = 0$ and I believe the limit $\lim_{a->0} S(a) = 0$ as well.

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For any fixed $a > 0$, $$ \renewcommand{\Pr}{\mathbb{P}} \Pr(S(a) = 0) = \Pr\Big(\sum_{n=1}^\infty f(X_n) = 0\Big) = \Pr(\cap_n \{X_n > 0\}) = 1 $$ where the last line follows since $\Pr(X_n > 0) = 1$ for all $n$. So $S(a) = 0$ almost surely for each $a > 0$.

But, it's not immediately clear to me how you intend to define your limit.

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First note the following equality of events: $$ \big \{ \sum\nolimits_{n = 1}^\infty {f(X_n )} > 0 \big \} = \cup _{n = 1}^\infty \{ f(X_n ) = 1\}. $$ Now, for each $n$, $$ {\rm P}[f(X_n) = 1] = {\rm P}[X_n = 0] = 0. $$ Hence, $$ {\rm P}\big[\sum\nolimits_{n = 1}^\infty {f(X_n )} > 0\big] = {\rm P}[ \cup _{n = 1}^\infty \{ f(X_n ) = 1\} ] \le \sum\nolimits_{n = 1}^\infty {{\rm P}[f(X_n ) = 1]} = 0. $$ So, for any $a>0$, ${\rm P}[S(a)>0] = 0$, hence $S(a)=0$ with probability $1$.

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Oops. Looks like we almost simultaneously posted the same answer. –  cardinal Mar 30 '11 at 2:42
    
I can take mine down if you'd like since I'm not sure it adds much (if anything) over yours. –  cardinal Mar 30 '11 at 2:49
    
@cardinal: No need for that. –  Shai Covo Mar 30 '11 at 2:54

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