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Let $x_n$, with $n=1,2,\ldots$, be uniformly distributed random variables in $(0,1)$

What is the expected value and probability distribution of the sum

$$\sum_{n=1}^\infty x_n^{n^n}$$

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Can you provide some motivation for this question? – Shai Covo Mar 30 '11 at 0:54
up vote 5 down vote accepted

Elaborating on Steven's answer.

Since $\sum\nolimits_{n = 1}^\infty {{\rm E}[X_n^{n^n } ]} < \infty $ and $\sum\nolimits_{n = 1}^\infty {{\rm Var}[X_n^{n^n } ]} < \infty $, the series $\sum\nolimits_{n = 1}^\infty {X_n^{n^n } } $, assuming that the $X_n$ are independent, converges with probability $1$ (see Theorem 1 on page 130 of the book Probability theory: collection of problems). Call the limit $X$. Then, by the monotone convergence theorem, $$ {\rm E}[X] = \mathop {\lim }\limits_{n \to \infty } {\rm E}\bigg[\sum\limits_{k = 1}^n {X_k^{k^k } } \bigg] = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {{\rm E}[X_k^{k^k } ]} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{1}{{k^k + 1}}} = \sum\limits_{n = 1}^\infty {\frac{1}{{n^n + 1}}} . $$

EDIT: The fact that the series $\sum\nolimits_{n = 1}^\infty {X_n^{n^n } } $ converges with probability $1$ follows from the monotone convergence theorem (MCT), and we don't have to assume that the $X_n$ are independent (it suffices that they are positive). Indeed, by MCT, the limit $X:=\sum\nolimits_{n = 1}^\infty {X_n^{n^n } }$ must be almost surely finite, since the integral $\int_\Omega {XdP} \,(= \sum\nolimits_{n = 1}^\infty {\frac{1}{{n^n + 1}}})$ is finite.

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Alpha [^n%29%29+from+1+to+infi‌​nity] shows this as 0.739948. It gives more digits, but I couldn't copy them. – Ross Millikan Mar 30 '11 at 3:11
@Ross: Thanks for this information. – Shai Covo Mar 30 '11 at 3:15

Your answer almost certainly has no closed form, but it's easy enough to write the EV as a sum; just use the linearity of expectation, and use the defining integral to calculate the expectation of each of the individual terms.

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