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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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    Further hint on the L-shaped region. It can be partitioned into two rectangular regions and you can find the probability that $(X,Y)$ belongs in each, and add. The two probabilities are different. Or you can use express the region as $A\cup B$ where $P(A) = P(B)$ and use $P(A\cup B) = P(A) + P(B) - P(A\cap B)$.2011-11-16

2 Answers 2

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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}(-1-1) + \operatorname{Area}(-11)\\ &=\operatorname{Area}(-1-1) + \operatorname{Area}(-11)\\ &=\operatorname{Area}(-1

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

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    Your intuition is correct but you should be careful about how you write up the solution, since the _areas_ of the regions are not finite. Write the same thing in terms of the _probabilities_ (and add another line to go from P\{-1 < X < 1, Y > 1\} to P\{-1 < X < 1, Y < -1\} so that P\{-1 < X < 1, Y > -1\}+ P\{-1 < X < 1, Y < -1\} = P\{-1 < X < 1\} is immediately obvious, and I will be glad to upvote your answer.2011-11-17