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Let $X_1,X_2,X_3.\dots$ be iid Uniform $(0,1)$ and let $N_k$ be the minimum $n$ such that $$X_1^k+X_2^k+\cdots +X_n^k\le 1<X_1^k+X_2^k+\cdots +X_{n+1}^k, \quad k\in\mathbb{N}$$ How to find $\displaystyle \lim_{k\to\infty}\frac{E(N_k)}{k}$?

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If $X_i$ is uniformly distributed over $(0,1)$, then $$ \mathbb{P}[X_i^k\leq t] = t^{\frac{1}{k}}$$ hence the pdf of $X_i^k$ is supported on $(0,1)$ and given by $\frac{1}{k}\,t^{\frac{1}{k}-1}$. Moreover, $$\begin{eqnarray*}\mathbb{P}[N_k\geq n]=\mathbb{P}[X_1^k+\ldots+X_n^k<1]&=&\int_{(0,1)^n}\mathbb{1}_{x_1^k+\ldots+x_n^k<1}\,d\mu\\&=&\frac{1}{k^n}\int_{(0,1)^n}\left(z_1\cdot\ldots\cdot z_n\right)^{\frac{1}{k}-1}\cdot\mathbb{1}_{z_1+\ldots+z_n<1}\,d\mu\\&=&\frac{\Gamma\left(\frac{1}{k}\right)^{n}}{k^n\cdot \Gamma\left(1+\frac{n}{k}\right)}\end{eqnarray*}$$ where the last step follows from the properties of the Dirichlet distribution.

It gives: $$\mathbb{E}[N_k]=\sum_{n\geq 1}\frac{\Gamma\left(1+\frac{1}{k}\right)^n}{\Gamma\left(1+\frac{n}{k}\right)}$$ but since the limit for $k\to +\infty$ of $\Gamma\left(1+\frac{1}{k}\right)^k$ equals $e^{-\gamma}$, by a Riemann sum argument:

$$ \lim_{k\to +\infty}\frac{\mathbb{E}[N_k]}{k}=\int_{0}^{+\infty}\frac{e^{-\gamma x}}{\Gamma(1+x)}\,dx = \color{red}{1.24941661444\ldots}.$$

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  • $\begingroup$ You're quite right, my answer was too handwaving. However, numerical simulations seem to indicate a limit near $1.25$ -- if your result is correct, the convergence is extremely slow -- the numerical result is still above $1.25$ for $k=200$. $\endgroup$ – joriki Oct 17 '15 at 15:17
  • $\begingroup$ Why can you replace $\Gamma\left(1+\frac1k\right)^n$ by the limit of $\Gamma\left(1+\frac1k\right)^k$? $\endgroup$ – joriki Oct 17 '15 at 15:22
  • $\begingroup$ @joriki: there was an error, now the computation is fixed and the limit is close to $\frac{5}{4}$ as numerical computations suggest. I write $\Gamma\left(1+\frac{1}{k}\right)^n$ as $\left[\Gamma\left(1+\frac{1}{k}\right)^k\right]^{\frac{n}{k}}$, then exploit the dominated convergence theorem and a Riemann sum argument. $\endgroup$ – Jack D'Aurizio Oct 17 '15 at 15:30
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    $\begingroup$ I see -- very nice! $\endgroup$ – joriki Oct 17 '15 at 15:33

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