Assume the $1 \times N$ vector


$\mathbf X = [X\_1, X\_2, \ldots , X\_N]$


contains i.i.d. normal samples such that $\mathbf X$ has a multivariate normal distribution. Now assume another random variable $Y$ also has a normal distribution, independent from the samples in $\mathbf X$, such that $Y$ has a mean and variance not necessarily equal to that of $\mathbf X$.


What is the probability that $Y$ will exceed all the values in $\mathbf X$ concurrently (or jointly)? In other words what is the probability that the value of $Y$ will be the maximum value of $[X\_1, X\_2, \ldots , X\_N, Y]$


**Possible approach?**


In the simplest case $N=1$, the problem reduces to 


$p(Y>X\_1)$ = $p(Y-X\_1>0)$


which can be calculated by convolution of $p(Y)$ and $p(-X\_1)$ and integrating the resulting PDF from $0$ to $\infty$. When $Y$ is also i.i.d. w.r.t the samples of $\mathbf X$, the resulting PDF is zero-mean, such that $p(Y>X\_1)$ = 0.5. This makes intuitively sense, since the probability that one RV will exceed another i.i.d. RV is 0.5. The probability


$p(Y>X\_1, Y>X\_2, \ldots, Y>X\_N )$


should therefore equal $1/(N+1)$ if $Y$ is i.i.d. w.r.t. $\mathbf X$. When $N>1$, the problem does not seem straightforward (maybe it is?).



Probability that one RV will exceed many others
Assume without loss of generality that each $X\_i$ is standard normal and that $Y$ is normal with mean $\mu$ and variance $\sigma^2$. By definition, for every $y$,
$$
P[y\gt\max(X\_1,\ldots,X\_N)]=\Phi(y)^N,
$$
hence, by the independence of $Y$ from $(X\_i)\_i$,
$$
P[Y\gt\max(X\_1,\ldots,X\_N)]=E[\Phi(Y)^N]=\int\_\mathbb R\Phi(\mu+\sigma x)^N\varphi(x)\mathrm dx,
$$
where
$$
\varphi(x)=\frac1{\sqrt{2\pi}}\mathrm e^{-x^2/2},\qquad\Phi(x)=\int\_{-\infty}^x\varphi(z)\mathrm dz.
$$
Except when $\mu=0$, $\sigma^2=1$, I see no simplification. An equivalent formula is
$$
P[Y\gt\max(X\_1,\ldots,X\_N)]=\int\_\mathbb R\Phi(x)^N\varphi\left(\frac{x-\mu}\sigma\right)\frac{\mathrm dx}\sigma.
$$


