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Find $Cov(X,Y)$

My work:

I found that $E(X) = .75$ and $E(Y)=.50$ $E(XY) = \int_0^1{\int_0^1} xyf(x,y) dydx = {\int_0^1}\int_0^16xy^2 dy dx = \int_0^1 2x dx = 1$

so $Cov(X,Y) = E(XY) - E(X)\cdot E(Y) = 1 - .75\cdot.5= \frac58$

But my book says that the correct answer is $\cfrac{1}{40}$

Where did I go wrong? I would appreciate any tips.

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    Your joint pdf is $f(x,y)=6y$ when $0<y<x<1$. Presumably for $y > x$, $f(x,y) = 0$, so $\int_0^1 yf(x,y)dy = \int_0^x yf(x,y) dy$. – player3236 Nov 02 '20 at 05:28

2 Answers2

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$$ E(XY) = \int_0^1 \int_0^1 6xy^2 \mathbb{1}_{0 < y<x<1}dx dy = \int_0^1 \int_y^1 6xy^2dx dy= \frac{2}{5} $$

WeakLearner
  • 5,982
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$EXY=\int_0^{1}\int_0^{x} xy (6y)dydx$. This evaluates to $\frac 2 5$ so $cov(X,Y)=\frac 2 5-(0.75)(0.5)=0.025$

Kavi Rama Murthy
  • 311,013
  • I see so the first integral is from 0 to x because y is limited by x. But the second integral is from 0 to 1 because there is no constraint on x (besides 0 to 1) because x is always greater then y. Is my understanding of this correct? –  Nov 02 '20 at 05:56
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    @ Yes, It is correct. @BadwaterBarry – Kavi Rama Murthy Nov 02 '20 at 05:57
  • Thank you very much. –  Nov 02 '20 at 06:11