Let X1 and X2 be random variables with CDF FX. Let Y be a random variable with PDF fY. Let X1⊥Y and X2⊥Y.
I know that $$P(X_1
But how to find P(X1>Y,X2>Y)? Will it be $1-P(X_1
Answer
Assuming the probability mass for X1=Y is zero, we have that X1>Y is equivalent to $\neg (X_1
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
No comments:
Post a Comment