Let A be a point that moves in the plane $[0, \infty) \times[0, \infty)$ starting from $(\mathbf{0}, \mathbf{0})$. The $\mathrm{x}$ - and $\mathrm{y}$ -coordinates of the point follow two independent Poisson processes with parameters $\lambda_{1}$ and $\lambda_{2}$ respectively.
I want to determine the probability that the point will ever visit the point $(\boldsymbol{i}, \boldsymbol{j}),$ where $\boldsymbol{i}$ and $\boldsymbol{j}$ are some given positive integers? I tried to go from the independence of the two random processes and express the event but I didn't know how to proceed.
Any suggestions??
I am assuming $i,j\geq1$. The case in which one of them is $0$ is easier and produces the same formula (where $\sum_0^{-1} = 0$ by definition).
Let $(S_n^x)_n$ and $(S_n^y)_n$ be two independent sequences of exponential random variables with parameters $\lambda_x$ and $\lambda_y$ respectively.
Let $T_n^x = \sum_{m=1}^n S_m^x$ and $T_n^y = \sum_{m=1}^nS_m^y$.
Finally, if we pose $$X_t = \sum_{n=1}^\infty 1_{T_n^x\leq t}$$ and $$Y_t = \sum_{n=1}^\infty 1_{T_n^y\leq t}$$ we obtain two independent Poisson processes with the desires parameters.
Moreover, the desired probability is just $$\mathbb P(T_i^x < T_{j+1}^y, T_j^y < T_{i+1}^x) = \mathbb P(-S_{i+1}^x < R_{i,j} < S_{j+1}^y),$$ where $R_{i,j} = T_i^x-T_j^y$.
Now, since everything is independent, we know the distribution of $R_{i,j}$, $S_{j+1}^y$ and $S_{i+1}^x$. I believe that with some calculations we can find the desired probability.
For example: $T_i^x \sim \Gamma(i,\lambda_x)$ and $T_j^y \sim \Gamma(j,\lambda_y)$. Therefore the answers is equal to $$ANS = \frac1{\Gamma(i)\Gamma(j)}\iint_{-w < x-y < z}\lambda_x^{i+1}\lambda_y^{j+1}x^{i-1}y^{j-1}e^{-\lambda_x(x+w)}e^{-\lambda_y(y+z)}dxdydzdw,$$ where $x,y,z,w$ are assumed to be positive.
We remember that $$\int_u^\infty \lambda e^{-\lambda \zeta}d\zeta = e^{-\lambda u}.$$ Therefore, with $\zeta = z$ (since $-w<0<z$ is necessarily true), $$ ANS = \frac1{\Gamma(i)\Gamma(j)}\iint_{-w < x-y \& x>y}\lambda_x^{i+1}\lambda_y^{j}x^{i-1}y^{j-1}e^{-\lambda_x(x+w)}e^{-\lambda_yy}e^{-\lambda_y(x-y)}dxdydw + \frac1{\Gamma(i)\Gamma(j)}\iint_{-w < x-y \& x<y}\lambda_x^{i+1}\lambda_y^{j}x^{i-1}y^{j-1}e^{-\lambda_x(x+w)}e^{-\lambda_yy}dxdydw.$$ Doing the same for $w$, we obtain that $$ ANS = \frac1{\Gamma(i)\Gamma(j)}\iint_{x>y}\lambda_x^{i}\lambda_y^{j}x^{i-1}y^{j-1}e^{-\lambda_xx}e^{-\lambda_yy}e^{-\lambda_y(x-y)}dxdy + \frac1{\Gamma(i)\Gamma(j)}\iint_{x<y}\lambda_x^{i}\lambda_y^{j}x^{i-1}y^{j-1}e^{-\lambda_xx}e^{-\lambda_x(y-x)}e^{-\lambda_yy}dxdy.$$ Simplifying, $$ ANS = \frac1{\Gamma(i)\Gamma(j)}\iint_{x>y}\lambda_x^{i}\lambda_y^{j}x^{i-1}y^{j-1}e^{-(\lambda_x+\lambda_y)x}dxdy + \frac1{\Gamma(i)\Gamma(j)}\iint_{x<y}\lambda_x^{i}\lambda_y^{j}x^{i-1}y^{j-1}e^{-(\lambda_x+\lambda_y)y}dxdy.$$ Applying the well-knonw property $$\int_u^\infty \zeta^{l-1}\lambda^le^{-\lambda\zeta}d\zeta = e^{-\lambda u}\Big((\lambda u)^{l-1} + (l-1)(\lambda u)^{l-2} + \ldots + (l-1)!\Big) = e^{-\lambda u}\sum_{a = 1}^{l} \frac{\Gamma(l)}{\Gamma(a)}(\lambda u)^{a-1}$$with $\zeta = x$ for the first integral and $\zeta = y$ for the second one, we obtain $$ ANS = \frac{\lambda_x^i\lambda_y^j}{(\lambda_x+\lambda_y)^{i+j}}\cdot \Bigg(\sum_{a=0}^{i-1}\binom{a+j-1}{a} \frac1{(\lambda_x+\lambda_y)^a} + \sum_{a=0}^{j-1}\binom{a+i-1}{a} \frac1{(\lambda_x+\lambda_y)^a}\Bigg).$$
I hope I didn't make any stupid mistake!