Interview problems: Gaussian

Problems:
(1) X, Y are i.i.d. standard normal $\mathcal{N}(0, 1)$. What is $\mathbb{P}(X >3Y| X > 0)$?

(2) X, Y, Z are standard normal $\mathcal{N}(0, 1)$ with $\rho_{X, Y}=\rho_{X, Z}=\rho_{Y, Z}=0.5$. What is $\mathbb{P}(X+Y>Z+2)$?

(3) X, Y are i.i.d. standard normal $\mathcal{N}(0, 1)$. What is $\mathbb{E}[X|X+Y=1]$?





Solution:

(1) (X, Y) joint pdf is rotation invariant. Therefore, as shown in the figure, $$\mathbb{P}(X >3Y| X > 0)=\frac{\theta}{\pi} = \frac{1}{2} + \frac{\arctan 1/3}{\pi}.$$
Similarly, we can compute, for example, $\mathbb{P}(X >5Y)=1/2$ or $\mathbb{P}(X>0| X+Y>0)=3/4$.

(2) Let $W\equiv X+Y-Z$, $\text{Var}(W)=\text{Cov}(X+Y-Z, X+Y-Z)=2$. As a result, $W\sim \mathcal{N}(0, 2)$ and $\mathbb{P}(W>2)=1-\Phi(\sqrt{2})$.

(3) $\mathbb{E}[X|X+Y=1]=\mathbb{E}[X+Y-Y|X+Y=1]=1-\mathbb{E}[Y|X+Y=1]=1-\mathbb{E}[X|X+Y=1]$, and thus $\mathbb{E}[X|X+Y=1]=1/2$.

For a more general case, such conditional expectation can be solved by two properties of Gaussian: linear combination of Gaussians is still Gaussian, and zero correlation of two Gaussians means independence. The calculation is as follows:  
Let $Z=X+Y$, find $a$ such that $\text{Cov}(X+aZ, Z)=0$. Then, $$\mathbb{E}[X|Z=b]=\mathbb{E}[X+aZ-aZ|Z=b]=\mathbb{E}[X+aZ]-ab=-ab\,.$$
By $\text{Cov}(X+aZ, Z)=\text{Cov}((1+a)X+aY, X+Y)=2a+1=0$, we have $a=-1/2$ and thus $\mathbb{E}[X|Z=1]=1/2$.
Location: 美国加利福尼亚州帕罗奥图

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