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The random variables X and Y are jointly continuous, with a joint PDF of the form

f X,Y (x,y) = {c x y, if 0 ≤ x ≤ y ≤1, 0, otherwise,

where c is a normalizing constant

a) Is it true that f X given Y (2 given 0.5) is equal to zero?

Select an option

b) Is it true that f X given Y (0.5 given 2) is equal to zero?

Select an option

1 answer

To analyze the joint PDF \( f_{X,Y}(x,y) \) given by

\[ f_{X,Y}(x,y) = \begin{cases} cxy & \text{if } 0 \leq x \leq y \leq 1 \ 0 & \text{otherwise} \end{cases} \]

we first need to determine the normalizing constant \( c \) and the support of the random variables \( X \) and \( Y \) defined in this joint PDF.

a) Evaluating \( f_{X|Y}(2 | 0.5) \)

  1. Understanding the condition: The conditional PDF \( f_{X|Y}(x|y) \) is only defined when the joint PDF \( f_{X,Y}(x,y) \) is non-zero.
  2. Conditions for \( y = 0.5 \): For \( Y = 0.5 \), we look at the joint PDF: We need to find \( x \) such that \( 0 \leq x \leq 0.5 \leq 1 \). However, when we consider \( x = 2 \), we see that \( 2 \) does not satisfy the condition \( 0 \leq x \leq 0.5 \).
  3. Result: Therefore, since \( x = 2 \) is not in the support for \( Y = 0.5 \), we conclude that

\[ f_{X|Y}(2|0.5) = 0 \]

Thus, the statement in part (a) is true.

b) Evaluating \( f_{X|Y}(0.5 | 2) \)

  1. Understanding the condition: As before, we must consider the support of \( Y \).
  2. Conditions for \( y = 2 \): The condition states that \( 0 \leq x \leq y \leq 1 \). However, \( y = 2 \) does not satisfy \( y \leq 1 \) since 2 is outside this range.
  3. Result: Therefore, since \( y = 2 \) is outside the valid support for the joint PDF, we conclude that

\[ f_{X|Y}(0.5|2) = 0 \]

Thus, the statement in part (b) is also true.

Conclusion

  • a) True: \( f_{X|Y}(2 | 0.5) = 0 \)
  • b) True: \( f_{X|Y}(0.5 | 2) = 0 \)