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$\newcommand\la\lambda\newcommand\w{\mathfrak w}\newcommand\R{\mathbb R}$We have to show that $P(U<u)=u$ for $u\in(0,1)$, where $$U:=\min_{j\ge1} \frac{X_1+\cdots+X_j}j$$ and $X_1,X_2,\dots$ are iid exponential random variables with mean $1$. This minimum is attained almost surely (a.s.), because, by the strong law of large numbers, $\frac{X_1+\cdots+X_j}j\to1$ a.s. as $j\to\infty$, whereas $\inf_{j\ge1} \frac{X_1+\cdots+X_j}j<1$ a.s.

For each natural $j$ and each $u\in(0,1)$, $$\begin{aligned} U<u&\iff\exists j\ge1\ \;\sum_{i=1}^j X_i<ju \\ &\iff\exists j\ge1\ \;Y_{u,j}:=\sum_{i=1}^j(u-X_i)>0 \\ &\iff\bar Y_u>0, \end{aligned}\tag{1}$$ where $\bar Y_u:=\max_{j\ge0}Y_{u,j}$, with $Y_{u,0}=0$ (of course). By the formula $E e^{i\la\bar Y}=\w_+(\la)/\w_+(0)$ at the very end of Section 19 of Chapter 4 (p. 105) and Theorem 2 in this chapter (pp. 106--107) of Borovkov, $$g_u(\la):=E e^{i\la\bar Y_u}=\frac{(1-u)i\la}{1+i\la-e^{i\la u}}$$ for all real $\la$. Note also that $\bar Y_u\ge Y_{u,0}=0$. So, by Proposition 1 in this paper or its arXiv version , $$P(\bar Y_u>0)=E\,\text{sign}\,\bar Y_u =\frac1{\pi i}\,\int_\R \frac{g_u(\la)}\la\,d\la =\frac1{\pi i}\,\int_\R h_u(\la)\,d\la \tag{2} ,$$ where $$h_u(\la):=\frac{g_u(\la)-g_u(\infty-)}\la =(1-u)\frac{1-e^{i \la u}}{\la(e^{i \la u}-1-i\la)}$$ and the integrals are understood in the principal value sense.

In view of (1), it remains to show that the integrals in (2) equal $\pi i u$ for all $u\in(0,1)$. Numerical calculations suggest this is true. However, I have not been able to prove this, even with the help of Mathematica.

$\newcommand\la\lambda\newcommand\w{\mathfrak w}\newcommand\R{\mathbb R}$We have to show that $P(U<u)=u$ for $u\in(0,1)$, where $$U:=\min_{j\ge1} \frac{X_1+\cdots+X_j}j$$ and $X_1,X_2,\dots$ are iid exponential random variables with mean $1$. This minimum is attained almost surely (a.s.), because, by the strong law of large numbers, $\frac{X_1+\cdots+X_j}j\to1$ a.s. as $j\to\infty$, whereas $\inf_{j\ge1} \frac{X_1+\cdots+X_j}j<1$ a.s.

For each natural $j$ and each $u\in(0,1)$, $$\begin{aligned} U<u&\iff\exists j\ge1\ \;\sum_{i=1}^j X_i<ju \\ &\iff\exists j\ge1\ \;Y_{u,j}:=\sum_{i=1}^j(u-X_i)>0 \\ &\iff\bar Y_u>0, \end{aligned}\tag{1}$$ where $\bar Y_u:=\max_{j\ge0}Y_{u,j}$, with $Y_{u,0}=0$ (of course). By the formula $E e^{i\la\bar Y}=\w_+(\la)/\w_+(0)$ at the very end of Section 19 of Chapter 4 (p. 105) and Theorem 2 in this chapter (pp. 106--107) of Borovkov, $$g_u(\la):=E e^{i\la\bar Y_u}=\frac{(1-u)i\la}{1+i\la-e^{i\la u}}$$ for all real $\la$. Note also that $\bar Y_u\ge Y_{u,0}=0$. So, by Proposition 1 in this paper or its arXiv version , $$P(\bar Y_u>0)=E\,\text{sign}\,\bar Y_u =\frac1{\pi i}\,\int_\R \frac{g_u(\la)}\la\,d\la =\frac1{\pi i}\,\int_\R h_u(\la)\,d\la \tag{2} ,$$ where $$h_u(\la):=\frac{g_u(\la)-g_u(\infty-)}\la =(1-u)\frac{1-e^{i \la u}}{\la(e^{i \la u}-1-i\la)}$$ and the integrals are understood in the principal value sense.

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In view of (1), it remains to show that the integrals in (2) equal $\pi i u$ for all $u\in(0,1)$.

This is now proved at An integral identity

$\newcommand\la\lambda\newcommand\w{\mathfrak w}\newcommand\R{\mathbb R}$We have to show that $P(U<u)=u$ for $u\in(0,1)$, where $$U:=\min_{j\ge1} \frac{X_1+\cdots+X_j}j,$$ where $X_1,X_2,\dots$ are iid exponential random variables with mean $1$. This minimum is attained almost surely (a.s.), because, by the strong law of large numbers, $\frac{X_1+\cdots+X_j}j\to1$ a.s. as $j\to\infty$, whereas $\inf_{j\ge1} \frac{X_1+\cdots+X_j}j<1$.

For each natural $j$ and each $u\in(0,1)$, $$\begin{aligned} U<u&\iff\exists j\ge1\ \;\sum_{i=1}^j X_i<ju \\ &\iff\exists j\ge1\ \;Y_{u,j}:=\sum_{i=1}^j(u-X_i)>0 \\ &\iff\bar Y_u>0, \end{aligned}\tag{1}$$ where $\bar Y_u:=\max_{j\ge0}Y_{u,j}$, with $Y_{u,0}=0$ (of course). By the formula $E e^{i\la\bar Y}=\w_+(\la)/\w_+(0)$ at the very end of Section 19 of Chapter 4 (p. 105) and Theorem 2 in this chapter (pp. 106--107) of Borovkov, $$g_u(\la):=E e^{i\la\bar Y_u}=\frac{(1-u)i\la}{1+i\la-e^{i\la u}}$$ for all real $\la$. Note also that $\bar Y_u\ge Y_{u,0}=0$. So, by Proposition 1 in this paper or its arXiv version , $$P(\bar Y_u>0)=E\,\text{sign}\,\bar Y_u =\frac1{\pi i}\,\int_\R \frac{g_u(\la)}\la\,d\la =\frac1{\pi i}\,\int_\R h_u(\la)\,d\la \tag{2} ,$$ where $$h_u(\la):=\frac{g_u(\la)-g_u(\infty-)}\la =(1-u)\frac{1-e^{i \la u}}{\la(e^{i \la u}-1-i\la)}$$ and the integrals are understood in the principal value sense.

In view of (1), it remains to show that the integrals in (2) equal $\pi i u$ for all $u\in(0,1)$. Numerical calculations suggest this is true. However, I have not been able to prove this, even with the help of Mathematica.

$\newcommand\la\lambda\newcommand\w{\mathfrak w}\newcommand\R{\mathbb R}$We have to show that $P(U<u)=u$ for $u\in(0,1)$, where $$U:=\min_{j\ge1} \frac{X_1+\cdots+X_j}j$$ and $X_1,X_2,\dots$ are iid exponential random variables with mean $1$. This minimum is attained almost surely (a.s.), because, by the strong law of large numbers, $\frac{X_1+\cdots+X_j}j\to1$ a.s. as $j\to\infty$, whereas $\inf_{j\ge1} \frac{X_1+\cdots+X_j}j<1$ a.s.

For each natural $j$ and each $u\in(0,1)$, $$\begin{aligned} U<u&\iff\exists j\ge1\ \;\sum_{i=1}^j X_i<ju \\ &\iff\exists j\ge1\ \;Y_{u,j}:=\sum_{i=1}^j(u-X_i)>0 \\ &\iff\bar Y_u>0, \end{aligned}\tag{1}$$ where $\bar Y_u:=\max_{j\ge0}Y_{u,j}$, with $Y_{u,0}=0$ (of course). By the formula $E e^{i\la\bar Y}=\w_+(\la)/\w_+(0)$ at the very end of Section 19 of Chapter 4 (p. 105) and Theorem 2 in this chapter (pp. 106--107) of Borovkov, $$g_u(\la):=E e^{i\la\bar Y_u}=\frac{(1-u)i\la}{1+i\la-e^{i\la u}}$$ for all real $\la$. Note also that $\bar Y_u\ge Y_{u,0}=0$. So, by Proposition 1 in this paper or its arXiv version , $$P(\bar Y_u>0)=E\,\text{sign}\,\bar Y_u =\frac1{\pi i}\,\int_\R \frac{g_u(\la)}\la\,d\la =\frac1{\pi i}\,\int_\R h_u(\la)\,d\la \tag{2} ,$$ where $$h_u(\la):=\frac{g_u(\la)-g_u(\infty-)}\la =(1-u)\frac{1-e^{i \la u}}{\la(e^{i \la u}-1-i\la)}$$ and the integrals are understood in the principal value sense.

In view of (1), it remains to show that the integrals in (2) equal $\pi i u$ for all $u\in(0,1)$. Numerical calculations suggest this is true. However, I have not been able to prove this, even with the help of Mathematica.

$\newcommand\la\lambda\newcommand\w{\mathfrak w}\newcommand\R{\mathbb R}$We have to show that $P(U<u)=u$ for $u\in(0,1)$, where $$U:=\min_{j\ge1} \frac{X_1+\cdots+X_j}j,$$ where $X_1,X_2,\dots$ are iid exponential random variables with mean $1$. This minimum is attained almost surely (a.s.), because, by the strong law of large numbers, $\frac{X_1+\cdots+X_j}j\to1$ a.s. as $j\to\infty$, whereas $\inf_{j\ge1} \frac{X_1+\cdots+X_j}j<1$.

For each natural $j$ and each $u\in(0,1)$, $$\begin{aligned} U<u&\iff\exists j\ge1\ \;\sum_{i=1}^j X_i<ju \\ &\iff\exists j\ge1\ \;Y_{u,j}:=\sum_{i=1}^j(u-X_i)>0 \\ &\iff\bar Y_u>0, \end{aligned}\tag{1}$$ where $\bar Y_u:=\max_{j\ge0}Y_{u,j}$, with $Y_{u,0}=0$ (of course). By the formula $E e^{i\la\bar Y}=\w_+(\la)/\w_+(0)$ at the very end of Section 19 of Chapter 4 (p. 105) and Theorem 2 in this chapter (pp. 106--107) of Borovkov, $$g_u(\la):=E e^{i\la\bar Y}=\frac{(1-u)i\la}{1+i\la-e^{i\la u}}$$ for all real $\la$. Note also that $\bar Y_u\ge Y_{u,0}=0$. So, by Proposition 1 in this paper or its arXiv version , $$P(\bar Y_u>0)=E\,\text{sign}\,\bar Y_u =\frac1{\pi i}\,\int_\R \frac{g_u(\la)}\la\,d\la =\frac1{\pi i}\,\int_\R h_u(\la)\,d\la \tag{2} ,$$ where $$h_u(\la):=\frac{g_u(\la)-g_u(\infty-)}\la =(1-u)\frac{1-e^{i \la u}}{\la(e^{i \la u}-1-i\la)}$$ and the integrals are understood in the principal value sense.

In view of (1), it remains to show that the integrals in (2) equal $\pi i u$ for all $u\in(0,1)$. Numerical calculations suggest this is true. However, I have not been able to prove this, even with the help of Mathematica.

$\newcommand\la\lambda\newcommand\w{\mathfrak w}\newcommand\R{\mathbb R}$We have to show that $P(U<u)=u$ for $u\in(0,1)$, where $$U:=\min_{j\ge1} \frac{X_1+\cdots+X_j}j,$$ where $X_1,X_2,\dots$ are iid exponential random variables with mean $1$. This minimum is attained almost surely (a.s.), because, by the strong law of large numbers, $\frac{X_1+\cdots+X_j}j\to1$ a.s. as $j\to\infty$, whereas $\inf_{j\ge1} \frac{X_1+\cdots+X_j}j<1$.

For each natural $j$ and each $u\in(0,1)$, $$\begin{aligned} U<u&\iff\exists j\ge1\ \;\sum_{i=1}^j X_i<ju \\ &\iff\exists j\ge1\ \;Y_{u,j}:=\sum_{i=1}^j(u-X_i)>0 \\ &\iff\bar Y_u>0, \end{aligned}\tag{1}$$ where $\bar Y_u:=\max_{j\ge0}Y_{u,j}$, with $Y_{u,0}=0$ (of course). By the formula $E e^{i\la\bar Y}=\w_+(\la)/\w_+(0)$ at the very end of Section 19 of Chapter 4 (p. 105) and Theorem 2 in this chapter (pp. 106--107) of Borovkov, $$g_u(\la):=E e^{i\la\bar Y_u}=\frac{(1-u)i\la}{1+i\la-e^{i\la u}}$$ for all real $\la$. Note also that $\bar Y_u\ge Y_{u,0}=0$. So, by Proposition 1 in this paper or its arXiv version , $$P(\bar Y_u>0)=E\,\text{sign}\,\bar Y_u =\frac1{\pi i}\,\int_\R \frac{g_u(\la)}\la\,d\la =\frac1{\pi i}\,\int_\R h_u(\la)\,d\la \tag{2} ,$$ where $$h_u(\la):=\frac{g_u(\la)-g_u(\infty-)}\la =(1-u)\frac{1-e^{i \la u}}{\la(e^{i \la u}-1-i\la)}$$ and the integrals are understood in the principal value sense.

In view of (1), it remains to show that the integrals in (2) equal $\pi i u$ for all $u\in(0,1)$. Numerical calculations suggest this is true. However, I have not been able to prove this, even with the help of Mathematica.