I am looking to solve the following integration. Given $Z_1,Z_2,\ldots,$ are all independent and standard gaussian variable, $Z_i = N(0,1)$.
For the first case, I am interested in the probability $P(Z_1+Z_2< \sqrt{2}\alpha , Z_2+Z_3<\sqrt{2}\alpha)$, or simply $$\frac{Z_1+Z_2}{\sqrt{2}} < \alpha$$ $$\frac{Z_2+Z_3}{\sqrt{2}} < \alpha$$
where $\alpha$ is a constant. I can solve this numerically. First question is there analytical solution to this?
Secondly, for higher dimensional case, what is the probability that $$\frac{Z_1+Z_2+Z_3}{\sqrt{3}} < \alpha$$ $$\frac{Z_2+Z_3+Z_4}{\sqrt{3}} < \alpha$$ $$\frac{Z_3+Z_4+Z_5}{\sqrt{3}} < \alpha$$
Thank you for your help.
Hint
If $U\sim N(\mu_U,\sigma_U)$ and $V\sim N(\mu_V,\sigma_V)$ then $$U + V \sim N(\mu_U + \mu_V,\; \sigma_U^2 + \sigma_V^2 + 2\sigma_{U,V})$$ therefore $$X=Z_1+Z_2 \sim N(0\,,\; 2)$$ and $$Y=Z_2+Z_3 \sim N(0\,,\; 2)$$ on the other hand $$\operatorname{Cov}(X,Y)=\operatorname{Cov}(Z_1+Z_2,Z_2+Z_3)\\ $$ thus $$\operatorname{Cov}(X,Y)=\operatorname{Cov}(Z_1,Z_2)+\operatorname{Cov}(Z_2,Z_2)+\operatorname{Cov}(Z_1,Z_3)+\operatorname{Cov}(Z_2,Z_3)=\operatorname{Var}(Z_2)=1$$ as a result $$\rho_{X,Y}=\frac{\operatorname{Cov}(X,Y)}{\sqrt{\operatorname{Var}(X)\operatorname{Var}(Y)}}=\frac{1}{2}$$ we have $$f_{X,Y}(x,y)=\frac{1}{2\sqrt{3}\pi}\exp\left(-\frac{x^2-xy+y^2}{3}\right)$$ For more details , see Multivariate normal distribution. Finally $$\mathbb{P}(X<\sqrt{2}\alpha\,,\,Y<\sqrt{2}\alpha)=\int_{-\infty}^{\sqrt{2}\alpha}\int_{-\infty}^{\sqrt{2}\alpha}f_{X,Y}(x,y)dxdy$$ Now set \begin{cases} x=\frac{\sqrt{2}}{2}(u-v)\\ y=\frac{\sqrt{2}}{2}(u+v) \end{cases}