Wednesday, 3 April 2019

probability - Finding P(X1>Y,X2>Y)




Let X1 and X2 be random variables with CDF FX. Let Y be a random variable with PDF fY. Let X1Y and X2Y.



I know that $$P(X_1



But how to find P(X1>Y,X2>Y)? Will it be $1-P(X_1? Or $(1-P(X_1? Or something completely different?


Answer



Assuming the probability mass for X1=Y is zero, we have that X1>Y is equivalent to $\neg (X_1. Distributing a negation over a conjunction gives a disjunction: $$1-P((X_1 $$P( \neg ((X_1 $$ P( \neg (X_1 P((X1>Y)(X2>Y)). But you want the probability P((X1>Y)(X2>Y)). So your first guess doesn't work.



Your second guess doesn't work because by finding the probability of the two conditions separately and multiplying them, you're treating them as independent. But clearly they are dependent; if X1>Y, then it's more likely that Y is small, which makes it more likely that X2>Y.




To find the probability, just reverse the inequalities in the integrals. For instance, instead of $\int_{-\infty}^{\infty}P(X_1, take P(X1>y,X2>y|Y=y)fY(y)dy


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