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, $f(x,y)=begin{cases}frac{1}{area(D)}=frac{2}{3}&,x,y in D\ 0&,x,y not in Dend{cases}$.
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
asked Nov 19 at 23:03
dxdydz
1869
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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, $f(x,y)=begin{cases}frac{1}{area(D)}=frac{2}{3}&,x,y in D\ 0&,x,y not in Dend{cases}$.
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
asked Nov 19 at 23:03
dxdydz
1869
1
Just find the geometric centre. (The centroid, or centre of area).
– Graham Kemp
Nov 19 at 23:35
add a comment |
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, $f(x,y)=begin{cases}frac{1}{area(D)}=frac{2}{3}&,x,y in D\ 0&,x,y not in Dend{cases}$.
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
asked Nov 19 at 23:03
dxdydz
1869
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, $f(x,y)=begin{cases}frac{1}{area(D)}=frac{2}{3}&,x,y in D\ 0&,x,y not in Dend{cases}$.
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
asked Nov 19 at 23:03
dxdydz
1869
asked Nov 19 at 23:03
dxdydz
1869
asked Nov 19 at 23:03
dxdydz
1869
asked Nov 19 at 23:03
dxdydz
1869
asked Nov 19 at 23:03
dxdydz
1869
1869
1
Just find the geometric centre. (The centroid, or centre of area).
– Graham Kemp
Nov 19 at 23:35
add a comment |
1
Just find the geometric centre. (The centroid, or centre of area).
– Graham Kemp
Nov 19 at 23:35
1
1
Just find the geometric centre. (The centroid, or centre of area).
– Graham Kemp
Nov 19 at 23:35
Just find the geometric centre. (The centroid, or centre of area).
– Graham Kemp
Nov 19 at 23:35
add a comment |
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Just find the geometric centre. (The centroid, or centre of area).
– Graham Kemp
Nov 19 at 23:35