Let \( (X, Y) \) have the joint probability density function
\[
f_{X,Y}(x,y) =
\begin{cases}
\frac{4}{(x + y)^3}, & x>1, y>1 \\
0, & \text{otherwise}
\end{cases}
\]
Then which one of the following statements is NOT true?
\[ f_{X,Y}(x,y) = \begin{cases} \frac{4}{(x + y)^3}, & x>1, y>1 \\ 0, & \text{otherwise} \end{cases} \]
Then which one of the following statements is NOT true?Show Hint
- The probability density function of \( X + Y \) is \[ f_{X+Y}(z) = \frac{4}{z^3}(z - 2), \, z>2, \, \text{otherwise}. \]
- \( P(X + Y>4) = \frac{3}{4} \)
- \( E(X + Y) = 4 \log 2 \)
- \( E(Y | X = 2) = 4 \)
The Correct Option is C
Solution and Explanation
We are given the joint probability density function \( f_{X,Y}(x,y) \), and we need to evaluate the provided statements.
Step 1: Find the probability density function of \( X + Y \).
The given joint probability density function for \( X \) and \( Y \) can be used to find the probability density function of \( X + Y \), denoted by \( f_{X+Y}(z) \). The correct form of \( f_{X+Y}(z) \) is: \[ f_{X+Y}(z) = \frac{4}{z^3}(z - 2), \, z>2. \] This corresponds to option (A), so it is true.
Step 2: Evaluate \( P(X + Y>4) \).
To find \( P(X + Y>4) \), we integrate the probability density function \( f_{X+Y}(z) \) from 4 to infinity: \[ P(X + Y>4) = \int_4^\infty \frac{4}{z^3}(z - 2) \, dz = \frac{3}{4}. \] So, option (B) is also true.
Step 3: Evaluate \( E(X + Y) \).
The expected value of \( X + Y \) is given by: \[ E(X + Y) = \int_2^\infty z f_{X+Y}(z) \, dz. \] Using the correct form of the density function, the expected value \( E(X + Y) \) does not equal \( 4 \log 2 \), and this makes option (C) false.
Final Answer: \[ \boxed{\text{(C) } E(X + Y) = 4 \log 2 \text{ is NOT true}}. \]
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