Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I came across a problem involving the following limit:

$\lim_{n \to \infty} (\frac{1}{n} \sum\limits_{i=1}^n \mathbf{1}_{x_i>0}), \mbox{ where } X \sim N(\mu, \sigma)$

How would you approach evaluating the limit of this sum? I thought about applying some form of Riemann integral, but got stuck with the indicator function... Also, is it possible to say something about the distribution of the sum?

Thanks a lot!

share|improve this question
3  
Can you interpret this limit as the probability of some event? (Hint: yes.) –  Michael Lugo Aug 25 '12 at 15:01
    
This is actually part of the background to my question - the above is part of the broader problem. I've got an event which is determined by the sum $y_i \equiv k - c_i - \phi l$ being positive, where $C \sim N(\mu, \sigma)$ and k and l being constant. I want to express the probability of this event occuring depending on $\phi$, which shifts the mean of the distribution of $Y$. From this, can I simply claim that $Y \sim N(\mu + k - \phi l, \sigma)$? As a more general question, would it make sense at all to look for a derivative of something like $Prob(Y)*m$, m being some constant? Thanks a lot! –  jush Aug 25 '12 at 15:55
    
if the $X_i$ are i.i.d, as I suppose the are, the summands are i.i.d random variables. use the SLLN –  mike Aug 25 '12 at 19:32
add comment

1 Answer

up vote 0 down vote accepted

One is considering $T_n=\frac{1}{n} \sum\limits_{i=1}^nY_i$, with $Y_i= \mathbf{1}_{X_i>0}$, for some random variables $(X_i)_i$. If the random variables $(X_i)_i$ are i.i.d., the random variables $(Y_i)_i$ are, and the strong law of large numbers shows that $T_n\to\mathrm E(Y)=\mathrm P(X\gt0)$, almost surely and in every $L^p$ for $p$ finite.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.