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Let $0 < x < 1$ and $1 < y < \infty$. Let $X = (x_1,...)$ an equally distributed random variable from the set of all infinite tuples over $[x,y]$ (that is, $[x,y]^\infty$). Let $X_n = x_1\cdot\ldots\cdot x_n$ the product of the first $n$ entries to $X$.

Now if $0 < x < 1$ was given, but $1 < y < \infty$ was variable, I want to find to make $y$ so that $\lim\limits_{n\to \infty} E\left[{X_n}\right] = 1$.

How would I do that?

EDIT: the $\lim$ must be outside to make sense

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What's $Y_n{}$? –  joriki Oct 13 '12 at 15:24
    
Was edited away... fixed –  Jo So Oct 13 '12 at 15:29
    
I think you want the expectation and the limit the other way around? Or in what sense are you taking the limit of $X_n$ such that it would be well-defined? –  joriki Oct 13 '12 at 15:42
    
Just what I edited this moment. embarassed –  Jo So Oct 13 '12 at 15:44
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You can use $\TeX$ on this site by enclosing formulas in dollar signs; single dollar signs for inline formulas and double dollar signs for displayed equations. You can see the source code for any math formatting you see on this site by right-clicking on it and selecting "Show Math As:TeX Commands". Here's a basic tutorial and quick reference. –  joriki Oct 13 '12 at 15:54

1 Answer 1

up vote 1 down vote accepted

Nice question (after it was rendered legible). Perhaps somewhat surprisingly, you cannot choose $y$ in this manner.

Multiplying the $x_i$ corresponds to adding their logarithms. By the central limit theorem, the distribution of $\log X_n$ will tend to a Gaussian with mean $n\mu$ and variance $n\sigma^2$, where $\mu$ and $\sigma^2$ are the mean and variance, respectively, of $\log x_i$. There are three possibilities:

If you choose $y$ such that $\mu\lt0$, eventually almost the entire distribution of $\log X_n$ will be concentrated at negative values, so $E[X_n]\lt1$.

If you choose $y$ such that $\mu\gt0$, eventually almost the entire distribution of $\log X_n$ will be concentrated at positive values, so $E[X_n]\gt1$.

If you choose $y$ such that $\mu=0$, the distribution of $\log X_n$ will remain concentrated symmetrically around $0$, but it will spread out and the upper part will blow up $E[X_n]$ while the lower part eventually contributes $0$. Thus $E[X_n]$ will increase without bound.

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Thank you! Was able to understand it, also to get an intuitive understanding. –  Jo So Oct 23 '12 at 13:27

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