What is the probability that $X+Y$ and $X$ are both greater than $0$? $X$ and $Y$ are both normal standard gaussian.

Question

The question gives me:$X,Y\sim N(0,1)$ iid. I also know that $X+Y\sim N(0,2)$ because it is a sum of normal random variables. From my understanding, the sum and $X$ are not indepedent therefore the way to solve would be to rephrase it like this: $$P(X>0\text{ and }X+Y>0) = P(X+Y>0|X>0)*P(X>0).$$ I know that $P(X>0) = 0.5$ and that I can rewrite the first probability as: $P(Y>-x|X=x,x>0) = 1-P(Y<-x)$. But all this still depends on x, which I suppose should be removed somehow. I thought of integrating over all possible values of $X$ the cdf of the normal, but it seemed too many calculations for a simple question. How can I solve this in a simpler way?

Answer 1

There are exactly two "disjoint" ways this can happen:

• $X\geq 0$ and $Y\geq 0$
• $X\geq 0$ and $Y\leq 0$ and $|X|\geq |Y|$.

Notice that in each case, these are independent events.

Answer 2

Also $X$ and $-Y$ are iid with standard normal distribution so we can also go for finding the probability that $X$ and $X-Y$ are both positive. Not because that is essentially more easy, but only a bit more convenient.

We can write the probability of this event as: $$\int_0^{\infty}P(X-Y>0|X=x)\phi(x)dx=\int_0^{\infty}P(x-Y>0|X=x)\phi(x)dx=$$ $$\int_0^{\infty}P(x-Y>0)\phi(x)dx = \int_0^{\infty}P(Y<x)\phi(x)dx =\int_0^{\infty}\Phi(x)\phi(x)dx$$where $\phi(x)$ and $\Phi(x)$ are notations for PDF and CDF of standard normal distribution.

An antiderivative of $\Phi(x)\phi(x)$ is $\frac12\Phi(x)^2$ so the final answer is:$$\frac12\left(1^2-\left(\frac12\right)^2\right)=\frac38$$

Answer 3

Let $R_1 = \{(x,y) : x> 0, x+y > 0\}$,
$R_2 = \{(x,y) : x< 0, x+y < 0\}$,
$R_3 = \{(x,y) : x> 0, x+y < 0\}$,
$R_4 = \{(x,y) : x< 0, x+y > 0\}$.

I suggest that your draw these regions in the 2D plane. The probability you're looking for is $\mathbb P((X,Y)\in R_1)$.

By symmetry, $\mathbb P((X,Y)\in R_2) = \mathbb P((X,Y)\in R_1)$ and $\mathbb P((X,Y)\in R_4) = \mathbb P((X,Y)\in R_3)$

Note that $R_1$ is made up of 3 disjoint rotated copies of $R_3$, thus by rotational invariance of $(X,Y)$, $\mathbb P((X,Y)\in R_1) = 3\mathbb P((X,Y)\in R_3)$.

Finally, $$1=\sum_{i=1}^4 \mathbb P((X,Y)\in R_i) = \mathbb P((X,Y)\in R_1) + \mathbb P((X,Y)\in R_1) + \frac 13 \mathbb P((X,Y)\in R_1) + \frac 13 \mathbb P((X,Y)\in R_1),$$ thus $\mathbb P((X,Y)\in R_1) = \frac 38$.

Answer 4

If $X$ and $Y$ are independent and $N(0,1)$ then $X+iY =Re^{i\Theta}$ where $R$ and $\Theta$ are independent and $\Theta$ is uniform on $(-\pi,\pi].$ Hence $$\Pr(X>0,X+Y>0)=\Pr(X>0, Y>0)+\Pr (X>0, \ 0>Y>-X)$$$$=\Pr (-\pi/4<\Theta<\pi/2)=3/8.$$