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Suppose we have $X,Y$, two independent standard normal random variables. How can we calculate


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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Hint: graph the relevant zone in the XY plane – leonbloy Nov 16 '11 at 20:06
Draw a sketch of the plane with coordinate axes $x$ and $y$, and mark on it the region described as $$\{(x,y) \colon |\min(x,y) | < 1\}.$$ Then think of probabilities, not before you have finished the sketch. – Dilip Sarwate Nov 16 '11 at 20:08
I did. And I got a L-shape graph but I don't know how to start from here. – geraldgreen Nov 16 '11 at 20:09
Hint: $\mathbb P(|\min(X,Y)| < 1) = \mathbb P( \min(X,Y) > -1 ) - \mathbb P( \min(X,Y) \geq 1 )$. Can you continue from there? – cardinal Nov 16 '11 at 20:09
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 $P(A\cup B) = P(A) + P(B) - P(A\cap B)$. – Dilip Sarwate Nov 16 '11 at 23:52
up vote 1 down vote accepted

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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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. – Dilip Sarwate Nov 17 '11 at 2:07

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