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Consider that X and Y have the joint pdf $f(x,y) = (2/3)(x+1)$ for $0 < x < 1$ and $0 < y < 1$ and 0 otherwise. What is $P(X < 2Y < 3X)$? My book say the answer is 73/162

But I keep doing $\int_0^1 \int_{x/2}^{3x/2} f(x,y) \, dy \, dx$ and I am not getting 73/162

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1 Answer 1

up vote 7 down vote accepted

The region you need to integrate is the blue region as shown in the figure below. enter image description here

Hence, your integral should go as follows.

\begin{align} \int_0^{2/3} \int_{x/2}^{3x/2} f(x,y) \, dy \, dx + \int_{2/3}^1 \int_{x/2}^{1} f(x,y) \, dy \, dx & = \int_0^{2/3} \int_{x/2}^{3x/2} \dfrac23(x+1) \, dy \, dx\\ & + \int_{2/3}^1 \int_{x/2}^{1} \dfrac23 (x+1) \, dy \, dx\\ & = \int_0^{2/3} \dfrac23x(x+1) \, dx\\ & + \int_{2/3}^1 \dfrac23 (1-x/2)(x+1) \, dx\\ & = \dfrac23 \left(x^3/3+x^2/2 \right)_0^{2/3}\\ & + \dfrac23 \left(x^2/2+x - x^3/6 - x^2/4 \right)_{2/3}^1\\ & = \dfrac23 \left((2/3)^3/3+(2/3)^2/2 \right)\\ & + \dfrac23 \left(1/2+1 - 1/6 - 1/4 \right)\\ & - \dfrac23 \left((2/3)^2/2 + (2/3) \right)\\ & + \dfrac23 \left((2/3)^3/6 + (2/3)^2/4 \right) \\ & = \dfrac{73}{162} \end{align}

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Oh shoot, that graph is much much better than my quick and dirty plot on scrap paper. –  sidht Nov 2 '12 at 3:55
2  
@jak The graph was made using grapher (en.wikipedia.org/wiki/Grapher) software on mac. –  user17762 Nov 2 '12 at 4:03
    
I have Mathematica, so it okay. Thank you however –  sidht Nov 2 '12 at 4:08

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