Expected value of X using the geometry of D and the uniformity of the joint distribution.











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I've asked this question before, but didn't get the answer I wanted.
Let $(X, Y)$ be a uniformly distributed random point on the quadrilateral $D$ with vertices $(0,0), (2,0), (1,1) text{ and } (0,1).$



(b) Find E[X] and E[Y].



So, $f(x,y)=begin{cases}frac{1}{area(D)}=frac{2}{3}&,x,y in D\ 0&,x,y not in Dend{cases}$.



enter image description here
I've computed the expectations of X and Y using the marginal density functions, but the last time I asked this question, there was a comment saying that its possible to calculate the expectations using the geometric meaning of D and the uniformity, but I still can't get it. Can somebody explain how to do that? Thanks.






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    Just find the geometric centre. (The centroid, or centre of area).
    – Graham Kemp
    2 days ago

















up vote
-1
down vote

favorite












I've asked this question before, but didn't get the answer I wanted.
Let $(X, Y)$ be a uniformly distributed random point on the quadrilateral $D$ with vertices $(0,0), (2,0), (1,1) text{ and } (0,1).$



(b) Find E[X] and E[Y].



So, $f(x,y)=begin{cases}frac{1}{area(D)}=frac{2}{3}&,x,y in D\ 0&,x,y not in Dend{cases}$.



enter image description here
I've computed the expectations of X and Y using the marginal density functions, but the last time I asked this question, there was a comment saying that its possible to calculate the expectations using the geometric meaning of D and the uniformity, but I still can't get it. Can somebody explain how to do that? Thanks.






probability







share|cite|improve this question


















  • 1




    Just find the geometric centre. (The centroid, or centre of area).
    – Graham Kemp
    2 days ago















up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











I've asked this question before, but didn't get the answer I wanted.
Let $(X, Y)$ be a uniformly distributed random point on the quadrilateral $D$ with vertices $(0,0), (2,0), (1,1) text{ and } (0,1).$



(b) Find E[X] and E[Y].



So, $f(x,y)=begin{cases}frac{1}{area(D)}=frac{2}{3}&,x,y in D\ 0&,x,y not in Dend{cases}$.



enter image description here
I've computed the expectations of X and Y using the marginal density functions, but the last time I asked this question, there was a comment saying that its possible to calculate the expectations using the geometric meaning of D and the uniformity, but I still can't get it. Can somebody explain how to do that? Thanks.






probability







share|cite|improve this question













I've asked this question before, but didn't get the answer I wanted.
Let $(X, Y)$ be a uniformly distributed random point on the quadrilateral $D$ with vertices $(0,0), (2,0), (1,1) text{ and } (0,1).$



(b) Find E[X] and E[Y].



So, $f(x,y)=begin{cases}frac{1}{area(D)}=frac{2}{3}&,x,y in D\ 0&,x,y not in Dend{cases}$.



enter image description here
I've computed the expectations of X and Y using the marginal density functions, but the last time I asked this question, there was a comment saying that its possible to calculate the expectations using the geometric meaning of D and the uniformity, but I still can't get it. Can somebody explain how to do that? Thanks.






probability




probability






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 2 days ago









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999




999








  • 1




    Just find the geometric centre. (The centroid, or centre of area).
    – Graham Kemp
    2 days ago
















  • 1




    Just find the geometric centre. (The centroid, or centre of area).
    – Graham Kemp
    2 days ago










1




1




Just find the geometric centre. (The centroid, or centre of area).
– Graham Kemp
2 days ago






Just find the geometric centre. (The centroid, or centre of area).
– Graham Kemp
2 days ago

















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