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Suppose we have $X,Y$, two independent standard normal random variables. How can we calculate $P(|\min(X,Y)|<1)$. I am still learning multivariables probability, and I also realize there are a lot of nice properties of two standard normal r.vs but I am not sure how to use them. |
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For any continuous distribution: $\Pr(|\min(X,Y)| \lt k) = \Pr(\min(X,Y) \gt -k) - \Pr(\min(X,Y) \ge k)$ $= \Pr(X \gt -k) \Pr(Y \gt -k) - \Pr(X \ge k) \Pr(Y \ge k)$ $ = (1- F(-k))^2- (1- F(k))^2 $. In the case of a distribution which is symmetric about $0$, this reduces to $F(k)^2- (1- F(k))^2= 2F(k)-1 = F(k)-F(-k) = \Pr(|X| \le k)$. which is your result. |
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I'm going to try to answer my own question. Basically, when you try to graph the inequality of $|\min(X,Y)|<1$, you will get a L-shape graph. And the area of the function can be calculated as the following $\begin{align} \operatorname{Area}(|\min(X,Y)|<1) &= \operatorname{Area}(-1<X<1 \text{ and } Y >-1) + \operatorname{Area}(-1<Y<1 \text { and } X>1)\\ &=\operatorname{Area}(-1<X<1 \text{ and } Y>-1) + \operatorname{Area}(-1<X<1 \text { and } Y>1)\\ &=\operatorname{Area}(-1<X<1) \end{align}$ It is like rotating the lower right piece of that L-shape graph 90 degrees clockwise. Then the probability of $P(|\min(X,Y)|<1)$ can be easily calculated. $P(|\min(X,Y)|<1) = P(-1<X<1) = \Phi(1)-\Phi(-1)$ |
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