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Let $f_{X,Y}(x,y) = kxy$, for $0 ≤ x, y ≤ 1,$ otherwise $f_{X,Y}(x,y) = 0$.

(a) Determine $k$ such that $f_{X,Y}(x,y)$ is a PDF.

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    Reminder: A property of PDF is such that its integral over its domain equals *what*. Apply your calculus. I think you know what the *what* is.2012-08-12

1 Answers 1

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Selected from PDF at WikiPedia.

The probability density function is nonnegative everywhere, and its integral over the entire space is equal to one.

As implied, $\int_{R^2} f_{X,Y}(x,y)=1$. (Property of PDF)

Thus $\int_0^1\int_0^1 kxy dxdy=1$, with the rest being mere integration.

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    k = 4? :) Thanks so much!2012-08-12
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    @Panda Yep, that's the answer. Harder questions involve triangular or circular area of integration which will just complicate integration, though-- e.g. $0<=x<=y<=1$.2012-08-12