Let $U_n=\sum_{i=1}^n X_i,V_n=\sum_{i=1}^n Y_i$, $n\geq 1$, be a two-dimensional random walk with i.i.d. increments $(X_n, Y_n)$, where $X_n, Y_n$ are discrete random variables with joint pmf $P_{X,Y}$. $X_n,Y_n$ have the following properties: \begin{align} 0 < \mathbb{E}[X_n]=\mu_X, \quad 0 < \mathbb{E}[Y_n]=\mu_Y,\quad |X_n/\mu_X| \leq K_X \text{ and } |Y_n/\mu_Y| \leq K_Y \end{align} for finite $K_X,K_Y$. The stopping time $\tau(t)$ is given by \begin{align} \tau(t) = \min(n \geq 0: U_n/\mu_X \geq t, V_n/\mu_Y \geq t) \end{align}

I am looking for an upper bound for $E[\tau(t)]$ that captures the asymptotic behavior as $t\rightarrow \infty$. I hope for an upper bound similar to \begin{align} E[\tau(t)] \leq t + \frac{1}{\sqrt{2\pi}} \sqrt{\text{Var}\left(\frac{X_1}{\mu_X} - \frac{Y_1}{\mu_Y}\right)}\sqrt{t} + \mathcal{O}(1), \end{align} as simulations suggest. However, a higher constant in front of $\sqrt{\text{Var}\left(\frac{X_1}{\mu_X} - \frac{Y_1}{\mu_Y}\right)}\sqrt{t}$ or faster growing remainder terms are also sufficient, i.e. $\mathcal{O}(t^{1/4})$ instead of $\mathcal{O}(1)$ is fine.

In the one-dimensional cases, with stopping times $ \tau_1(t)=\min(n\geq 0: U_n/\mu_X \geq t)$, $\tau_2(t)=\min(n\geq 0: V_n/\mu_Y \geq t)$ and $\tau_{12}(t)=\min(n \geq0 : \frac{1}{2}(U_n/\mu_X + V_n/\mu_Y) \geq t)$, the following bounds hold \begin{align} \mathbb{E}[\tau_1(t)] &= \mathbb{E}[U_{\tau_1(t)}/\mu_X] \leq \mu_X t + K_X\\ \mathbb{E}[\tau_2(t)] &= \mathbb{E}[V_{\tau_2(t)}/\mu_Y] \leq \mu_Y t + K_Y,\\ \mathbb{E}[\tau_{12}(t)] &= \frac{1}{2}\mathbb{E}[U_{\tau_{12}}/\mu_X+V_{\tau_{12}(t)}/\mu_Y] \leq t + \max(K_X,K_Y), \end{align} for $t > 0$, where the equalities follow from Wald's equality and the inequalities follow since $X_n$ and $Y_n$ are bounded.

# Partial Solution

My main idea is to write the stopping time $\tau(t)$ in two terms; the time until $\frac{1}{2}(U_n/\mu_X + V_n/\mu_Y)$ hits the boundary $t$ plus the time until $U_n$ hits the boundary $\mu_X t$ starting from $U_{\tau_{12}(t)}$ or $V_n$ hits the boundary $\mu_Y t$ starting from $V_{\tau_{12}(t)}$: \begin{align} \mathbb{E}[\tau(t)] &\stackrel{(a)}{\leq} \mathbb{E}[\tau_{12}(t) + \tau_1(t-U_{\tau_{12}(t)}/\mu_X)+\tau_2(t-V_{\tau_{12}(t)}/\mu_Y)]+\mathcal{O}(1)\\ &\leq t +\mathbb{E}\left[1\{U_{\tau_{12}(t)}\leq \mu_X t\}\left(t- U_{\tau_{12}(t)}/\mu_X\right)\right]\nonumber\\ &\quad+\mathbb{E}\left[1\{V_{\tau_{12}(t)}\leq \mu_Y t\}\left(t- V_{\tau_{12}(t)}/\mu_Y\right)\right]+\mathcal{O}(1)\\ \end{align} However, I have not been able to come up with an argument for whether/under what conditions (a) is true, since the random variables $X_n$ and $Y_n$ are allowed to take negative values. My main concern is that $U_n$ may have decreased below $\mu_X t$ when $V_n$ hits the boundary $\mu_Y t$ or visa versa.

Assuming that (a) is correct, I was able to obtain the bound \begin{align} \mathbb{E}[\tau(t)] \leq t+ \frac{1}{2}\sqrt{\text{Var}\left(\frac{X_1}{\mu_X} - \frac{Y_1}{\mu_Y}\right)}\sqrt{t}+\mathcal{O}(t^{1/4}), \end{align} which is sufficient for my application.

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Any suggestions or ideas are appreciated.