What's the probability the cumulative average of multiple gaussian variables exceed a certain value?

Consider a number $n$ of independent, normally distributed variables $Y_1$, $Y_2$, …, $Y_n$.
Consider also the $n$ variables defined by the average of the last $h$ terms, $X_h = frac{sum^{n}_{k = n -h + 1} Y_k}{h}$ for $h in {1,.., n}$. The variables $sqrt{h}X_h$ are normally distributed.

I’d like to understand what is the probability of having $P{sqrt{h}X_h > T text{ or } sqrt{h-1}X_{h-1} > T text{ or .. or } X_1>T }$, in words having one or more of the $sqrt{h}X_h$ to output above a threshold value.

Do you have any reference or suggestions about this problem?
I can work out the $n = 2$ case through geometry working on the expression $P{Y_1 + Y_2 > sqrt{2}T , |, Y_1<T }$. However I soon get lost into integral hell when dimension increases.

Thank you

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EDIT:

I add my solution when $n = 2$. As remarked by @ConMan, the probability we are looking for is equal to one minus the probability for all $sqrt{h}X_h$ to output under threshold $T$, so:

$$ P = 1 – p(Y_1 + Y_2 < sqrt{2}T,|,Y_1< T)p(Y_1<T) $$

We can calculate the term $ p(Y_1 + Y_2 < sqrt{2}T,|,Y_1< T) $ geometrically (see picture), through:

$$ p(Y_1 + Y_2 < sqrt{2}T,|,Y_1< T) = int_{-infty}^{sqrt{2}T -T} text{d}x N(x) + int^{infty}_{sqrt{2}T -T}text{d}x N(x)Big[ 1 – int^{T}_{sqrt{2}T -x}text{d}y N(y)Big]=\
= 1 – int^{infty}_{sqrt{2}T -T}text{d}x N(x)Big( F(T) – F(sqrt{2} T – x) Big)=\
= 1 – Phi(T)big(1 – Phi(sqrt{2}T – T)big) + int^{infty}_{sqrt{2}T -T}text{d}x N(x) Phi(sqrt{2} T – x) $$

Naming $H(T)$ the last integral term, the probability we are looking for become:

$$P = 1 – Phi(T)Big( 1 -Phi(T)big(1-Phi(sqrt{2}T – T)big) + H(T) Big)$$

If $T = 3$, we find a probability $P sim 2.5%$ which implies a value larger than the probability $1 – Phi(T)$ by a factor $frac{P(T)}{1-Phi(T)} sim 1.823$.

Where we noted with $Phi$ and $N$ the normal partition and probability density function respectively.
This seems confirmed by simulations, in python:

import scipy.stats as sts
import numpy as np

N = 1000000
M = 2
host_matrix = np.zeros((N,M))

for n in range(N):
    arr = sts.norm(0).rvs(M)
    wind = np.cumsum(arr)/np.sqrt(np.arange(M) + 1)
    host_matrix[n] = wind

Here a routine for calculating the value of the formula above:

def fp_prb(thr):
    import scipy.integrate as integrate
    def hard_term_f(x, T):
        return sts.norm.pdf(x)*sts.norm.cdf(sqrt(2)*T - x)
    
    hard_term = integrate.quad(lambda x: hard_term_f(x,thr),thr*(sqrt(2) - 1),+np.inf)[0]
    F_t = F(thr)
    F_sqrt = F(thr*(sqrt(2)-1))
    
    prb = 1 - F_t*( 1 - F_t*(1 - F_sqrt) + hard_term )    
    return  prb

The observed frequency and probability ratio can be computed through:

thr = 3

no_fp = len(np.argwhere((host_matrix[:,0] < thr ) & (host_matrix[:,1] < thr )))
fp_freq =  1 - no_fp/N
norm_freq = 1 - sts.norm.cdf(thr)
print('FP frequency: {:.4f}'.format(fp_freq))
print('Normal detections frequency: {:.4f}'.format(norm_freq))
print('Ratio: {:.3f}'.format(fp_freq/norm_freq))

An example output:

FP frequency: 0.0025
Normal detections frequency: 0.0013
Ratio: 1.841