Question
Download Solution PDFFor the joint density fxy(x, y) = x2 + Cy; 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, the value of constant C is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept used
We know that the probability of joint density \(\iint \)fxy(xy)dxdy
Calculation
\(\mathop \iint \limits_{x = 0y = 0}^{x = 1y = 1} \) (x2 + Cy)dxdy = 1
After integration we get
⇒ (x3/3)10 + C(y2/2)10 = 1
⇒ (1/3) + (1/2)C = 1
⇒ (1/2)C = 1 – 1/3
⇒ C(1/2) = 2/3
⇒ C = 4/3
∴ The value of C is 4/3
Last updated on Jul 17, 2025
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