Probability Calculation over Disjoint Partition of Sample Space

Question

Let $X \sim \mathcal{N}(0,\sigma^2_x)$. Let $N_1,N_2 \sim \mathcal{N}(0,\sigma^2_N)$ be two independent random variables.

Define two other random variables, $Y_1=X +N_1$ and $Y_2=X+N_2$ respectively. Note that independence of $N_1,N_2$ makes $Y_1,Y_2$ conditionally independent given $X$. Let us consider a disjoint partition $\{Z_1,Z_2,Z_3,\dots,Z_M\}$ of $\mathbb{R}$. I am interested in calculating the probability that $Y_1$ and $Y_2$ are in two different partitions. Precisely, consider the event,

Define the event $E=1 \left (\displaystyle \bigcup_{l \neq m}\{Y_1 \in Z_l, Y_2 \in Z_m \} \right )$.

I am interested in calculating the probability,$P(E)$.

My try:

$P(E) = \displaystyle \sum\limits_{k=1}^MP(Y_1 \in Z_k)P(Y_2 \notin Z_k|Y_1 \in Z_k)$ $=\displaystyle \sum\limits_{k=1}^M\pi^{Y_1}_k P(Y_2 \notin Z_k|Y_1 \in Z_k)$ $=\displaystyle \sum\limits_{k=1}^M\pi^{Y_1}_k \displaystyle \sum \limits_{j\neq k}P(Y_2\in Z_j|Y_1 \in Z_k) $

where $\pi^{Y_1}_k= \int_{Z_k}P_{Y_1}dY_1$ is the marginal probability of $Y_1$ over the partition $Z_k$, $1 \le k \le M$.

I wish to rewrite the last expression in terms of $N_1,N_2$. I find it difficult to calculate as the calculations turns messy.

Any help would be appreciated.

Thanks in advance.

Answer 1

Under the usual convention that $Z_k-x =\{z-x\mid z\in Z_k\} $

$E_{lm}$ appears to be the event of the variables being in two specific partitions, $Z_l ,Z_m$, rather than any two different partitions

$$\begin{align}\mathsf P(E_{lm})~=~&\mathsf P(Y_1\in Z_l, Y_2\in Z_m) \\[1ex] ~=~& \int_\Bbb R f_X(x)\mathsf P(N_1\in Z_l-x)\mathsf P(N_2\in Z_m-x)\operatorname d x \\[1ex] ~=~& \int_\Bbb R f_X(x) \Big(\int_{Z_l-x}f_N(s)\operatorname d s\Big)\Big(\int_{Z_m-x}f_N(t)\operatorname d t\Big)\operatorname d x\end{align}$$

It's "any two different partitions" and NOT a "specific partition". I will edit it.

$$\begin{align} \underset{l\neq m}{\sum_{l=1}^M\sum_{m=1}^M}\mathsf P(E_{lm})~=~&\int_\Bbb R f_X(x) \sum_{k=1}^M \Big(\int_{Z_k-x}f_N(s)\operatorname d s\Big)\Big(1-\int_{Z_k-x}f_N(s)\operatorname d s\Big)\operatorname d x \\[1ex] ~=~&\int_\Bbb R f_X(x)\Big(1-\sum_{k=1}^M\big(\int_{Z_k-x}f_N(s)\operatorname d s\big)^2\Big)\operatorname d x \\[1ex] ~=~&1-\int_\Bbb R f_X(x)\sum_{k=1}^M\big(\int_{Z_k-x}f_N(s)\operatorname d s\big)^2\operatorname d x \end{align}$$