Can we write the probability that a random normal variable is greater than others $N$ normal variables as the product of cumulative density functions?

Question

For $Y_1 \sim N(\mu_y;\sigma_y^2)$ and $X_i, ..., X_N$ random variables distrubted as $N(\mu_i, \sigma_i^2)$, can we write $P(Y_1 \geq X_i \ \forall i \in \{2,...,N\})$ as $\prod_{i=1}^{N}F(z_i)$, where $F(\cdot)$ is the cumulative density function of a standard normal distribution and $z_i=\frac{\mu_y-\mu_i}{\sqrt{\sigma_y^2+\sigma_i^2}}$?

For the case $i=1$, I know it is easy to show that $P(Y_i \geq X_1)=F(\frac{\mu_y-\mu_i}{\sqrt{\sigma_y^2+\sigma_1^2}})$, but can we generalize to the $N$ case $P(Y_1 \geq X_i \ \forall i \in \{2,...,N\})=\prod_{i=1}^{N}F(z_i)$? In the case that $\mu_y=\mu_i \forall i$, it would be $P(Y_1 \geq X_i \ \forall i \in \{2,...,N\})=F(z_1)^N$.

My doubt emerges since it might appear that any change in the variance of $Y$ will not affect $P(Y_1 \geq X_i \ \forall i \in \{2,...,N\})$ as long as $Y$ has the same mean of any other $X_i$. While intuitively $Y$ would need high values to beat all the other distributions.

(It is not an assigment, just curiosity)

Answer 1

TL; DNR Version: The answer is not what the OP wants it to be, and the desired probability must be computed by numerical integration. A very special case of the problem posed here is this question and my answer discusses this special case in some detail.

Answer in more detail: The answer that the OP seeks cannot be determined in closed form and numerical integration is needed. Even for the special case when all the random variables are assumed to be independent, the events $\{X_1 \leq Y\},$ $X_2\leq Y\}, \cdots, \{X_n \leq Y\}$ are not independent events and so $$P(X_1 \leq Y, X_2\leq Y, \cdots, X_n \leq Y) \neq \bigcap_{i=1}^n P(X_i \leq Y).$$ The events are, however, conditionally independent given the value of $Y$, that is,

\begin{align}P\left(\bigcap \{X_i \leq Y \mid Y = \alpha\} \right) &= \prod_{i=1}^n P(X_i\leq \alpha)\\ &= \prod_{i=1}^n \Phi\left(\frac{\alpha-\mu_i}{\sigma_i}\right) \end{align} where $\Phi(\cdot)$ is the CDF of the standard normal random variable. It follows that $$P(\max X_i \leq Y) = \int_{-\infty}^\infty \prod_{i=1}^n \Phi\left(\frac{\alpha-\mu_i}{\sigma_i}\right)f_Y( \alpha) \, \mathrm d\alpha$$ where $f_Y(\cdot)$ is the pdf of $Y$. As noted in the TL; DNR version, even for the special case of $\mu_i =0, \sigma_i = \sigma ~ \forall i$ treated in this answer, there is no closed form expression for the value of the integral.

Answer 2

I dont think you can. The events $\{Y_1 \geq X_1\}$ and e.g. $\{Y_1 \geq X_2\}$ do not seem to be independent. I will try to explain how you can get the probability. This is probably not the only way, but what you can do is the following:

1. Rewritte the probability in terms of the maximum: \begin{align} \mathbb{P}(Y_1 \geq X_i, i \in \{1,...,N\}) = \mathbb{P}(Y_1 \geq \max(X_1,...,X_N)) \end{align}
2. Derive the distribution of $\max(X_1,...,X_N)$ (assuming they are independent). This is what eventually will factorize.
3. Write $\mathbb{P}(Y_1 \geq \max(X_1,...,X_N)) = \mathbb{E}_{Y_1}(\mathbb{P}(Y_1 \geq \max(X_1,...,X_N) | Y_1))$ and calculate the integral.