Mathematics, 19.12.2019 19:31 jambunny26
Let (x, y) be a uniformly random point in the triangle in the plane with vertices (0,0), (0, , 0). find cov(x, y ) and corr(x, y ).
Answers: 2
, so that
is
![\mathrm{Cov}[X,Y]=E[(X-E[X])(Y-E[Y])]=E[XY]-E[X]E[Y]](/tpl/images/0426/3229/6a545.png)
![E[X]=\displaystyle\iint_{\Bbb R^2}xf_{X,Y}(x,y)\,\mathrm dA=\frac12\iint_Tx\,\mathrm dA=\frac12\int_0^1\int_0^{1-x}x\,\mathrm dy\,\mathrm dx=\frac1{12}](/tpl/images/0426/3229/b0475.png)
![E[Y]=\displaystyle\iint_{\Bbb R^2}yf_{X,Y}(x,y)\,\mathrm dA=\frac1{12}](/tpl/images/0426/3229/13c50.png)
![E[XY]=\displaystyle\iint_{\Bbb R^2}xyf_{X,Y}(x,y)\,\mathrm dA=\frac1{48}](/tpl/images/0426/3229/a4ff2.png)
![\mathrm{Cov}[X,Y]=\dfrac1{48}-\dfrac1{12^2}=\boxed{\dfrac1{72}}](/tpl/images/0426/3229/ffbbf.png)
![\mathrm{Corr}[X,Y]=\dfrac{\mathrm{Cov}[X,Y]}{\sqrt{\mathrm{Var}[X]\mathrm{Var}[Y]}}](/tpl/images/0426/3229/f66d2.png)
,![\mathrm{Var}[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2](/tpl/images/0426/3229/14d23.png)
![E[X^2]=\displaystyle\iint_{\Bbb R^2}x^2f_{X,Y}(x,y)\,\mathrm dA=\frac1{24}](/tpl/images/0426/3229/3c865.png)
![E[Y^2]=\displaystyle\iint_{\Bbb R^2}y^2f_{X,Y}(x,y)\,\mathrm dA=\frac1{24}](/tpl/images/0426/3229/1ce12.png)
![\mathrm{Corr}[X,Y]=\dfrac{\frac1{72}}{\sqrt{\left(\frac1{24}-\frac1{12^2}\right)^2}}=\boxed{\dfrac25}](/tpl/images/0426/3229/b6672.png)
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Let (x, y) be a uniformly random point in the triangle in the plane with vertices (0,0), (0, , 0). f...
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