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Let $X_1,X_2,...,X_n$ be a fixed number of Bernoulli random variables. My problem is to find a distribution for $Y$ such that for some function $f$, we have $Y=f(X_1,X_2,...,X_n)$. There are two candidate functions to use, $max$ or $avg$. I have no idea if an average function would work here or not but I think it'd give me a meaningful result.

I have looked into similar problems and mostly found the cases where $X_i$ are continuous random variables. Any hint on this problem is highly appreciated.

Thanks!

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Are the variables $X_1,\dotsc, X_n$ independent? –  Liviu Nicolaescu Apr 4 '12 at 19:40
    
I don't think this Q is right for this site - try math.stackexchange.org –  Anthony Quas Apr 4 '12 at 20:50
    
Or possibly stats.stackexchange.com :-) –  David Roberts Apr 4 '12 at 21:41
    
Yes, the variables are independent. Why this is not a right question for this site? I have seen a similar question but with differences on this site! –  isec Apr 5 '12 at 0:16
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1 Answer

I agree that this question is rather elementary for this site but since Easter is around the corner...

Assume the $X_i$'s are independent. Suppose that $X_i$ takes the values $a_i$ and $ b_i $, with probabilities $p_i(a_i)$ and respectively $p_i(b_i)$, where $p_i(a_i)+p_1(b_i)=1$. Without loss of generality we can assume $a_i< b_i$.

The vector valued r.v. $\vec{X}=(X_1,\dotsc, X_n)$ is distributed on the set $V$ of vertices of the parallelepiped

$$ P=\prod_{i=1}^n [a_i,b_i]. $$

A vertex $\vec{v}$ of this parallelepiped has coordinates

$$ \vec{v}=(v_1,\dotsc, v_n),\;\;v_i\in\lbrace a_i,b_i\rbrace. $$

The probability that $\vec{X}=\vec{v}$ is

$$p(\vec{v})=\prod_{i=1}^n p_i(v_i). $$

In other words, the probability distribution of $\vec{X}$ the measure

$$\vec{\mu}=\sum_{\vec{v}\in V} p(\vec{v})\delta_{\vec{v}}, $$

where $\delta_{\vec{v}}$ denotes the Dirac measure on $\mathbb{R}^n$ concentrated at $\vec{v}$. The distribution $\mu$ of $f(\vec{X})$ is a sum of Dirac measures

$$ \mu=f_*(\vec{\mu})=\sum_{t\in \mathbb{R}} w_t \delta_t, $$

where

$$w_t =\sum_{f(\vec{v})=t} p(\vec{v}). $$

In the end the problems reduces to identifying which of the vertices of $V$ lies on a given level set of $f$ which may not be easy for a complicated $f$. If $a_1=\cdots =a_n=a$, $b_1=\cdots =b_n=b$, $p_1(a)=\cdots =p_n(a)=p$ and $p_1(b)=\cdots = p_n(b)=q=1-p$ the above formula simplifies somewhat.

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Well, thanks for the answer but I do not see why this should be such an elementary question for the site. Anyhow, I'd say the problem is already simplified since a and b of each X_i is known. We only have $X_i: D -> {0,1}$ and I can also fix $f$ to be a $max$ function. –  isec Apr 5 '12 at 21:00
    
This question is an exercise typically assigned as homework in an introductory graduate course on probability. –  Liviu Nicolaescu Apr 6 '12 at 0:23
    
For me this was a question to answer as part of a research project. So, with limited knowledge of probability, I needed a hint not even a complete answer. If it was a homework, then I'd have known the answer by reading my book and listening to the lectures. –  isec Apr 9 '12 at 13:58
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