The random variables X and Y are continuous. Is this enough information to determine the value of P(X² = e^(3Y))? Select an option The random variables X and Y are jointly continuous. Is this enough information to determine the value of P(X² = e^(3Y))? Select an option

Answer 1

To determine the probability \( P(X^2 = e^{3Y}) \) for continuous random variables \( X \) and \( Y \), we need to analyze how probabilities work for continuous variables.

1. If \( X \) and \( Y \) are independent continuous random variables: The probability of any specific value (or event) occurring for continuous random variables is zero. Therefore, \( P(X^2 = e^{3Y}) = 0 \).

2. If \( X \) and \( Y \) are jointly continuous random variables: The conclusion is essentially the same. Joint continuous distributions imply that the probability of any specific event (like \( X^2 = e^{3Y} \)) also equals zero. So, in this case, \( P(X^2 = e^{3Y}) = 0 \) as well.

In summary, regardless of whether \( X \) and \( Y \) are independent or jointly continuous, the value of \( P(X^2 = e^{3Y}) \) is \( 0 \).

So the answer to your question is no, it is not enough information to give a non-zero probability, \( P(X^2 = e^{3Y}) = 0 \).