Find the expectation of vertices
$begingroup$
Question: A point is chosen uniformly at random from the triangle in the plane with vertices (0,0), (1,0), (0,2). Let X be the x-coordinate of this point, and let Y be the y-coordinate of this point.
$$Compute: E(XY^2)$$
I know that $E(XY^2) = E(X)*E(Y^2)$ but I don't know how to compute these two numbers. Could anyone point me in the right direction?
probability uniform-distribution expected-value
edited Dec 4 '18 at 16:42
gt6989b
33.3k22452
asked Dec 4 '18 at 16:32
ZazmadzeZazmadze
12
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add a comment |
$begingroup$
Question: A point is chosen uniformly at random from the triangle in the plane with vertices (0,0), (1,0), (0,2). Let X be the x-coordinate of this point, and let Y be the y-coordinate of this point.
$$Compute: E(XY^2)$$
I know that $E(XY^2) = E(X)*E(Y^2)$ but I don't know how to compute these two numbers. Could anyone point me in the right direction?
probability uniform-distribution expected-value
edited Dec 4 '18 at 16:42
gt6989b
33.3k22452
asked Dec 4 '18 at 16:32
ZazmadzeZazmadze
12
$endgroup$
add a comment |
$begingroup$
Question: A point is chosen uniformly at random from the triangle in the plane with vertices (0,0), (1,0), (0,2). Let X be the x-coordinate of this point, and let Y be the y-coordinate of this point.
$$Compute: E(XY^2)$$
I know that $E(XY^2) = E(X)*E(Y^2)$ but I don't know how to compute these two numbers. Could anyone point me in the right direction?
probability uniform-distribution expected-value
edited Dec 4 '18 at 16:42
gt6989b
33.3k22452
asked Dec 4 '18 at 16:32
ZazmadzeZazmadze
12
$endgroup$
Question: A point is chosen uniformly at random from the triangle in the plane with vertices (0,0), (1,0), (0,2). Let X be the x-coordinate of this point, and let Y be the y-coordinate of this point.
$$Compute: E(XY^2)$$
I know that $E(XY^2) = E(X)*E(Y^2)$ but I don't know how to compute these two numbers. Could anyone point me in the right direction?
probability uniform-distribution expected-value
probability uniform-distribution expected-value
edited Dec 4 '18 at 16:42
gt6989b
33.3k22452
asked Dec 4 '18 at 16:32
ZazmadzeZazmadze
12
edited Dec 4 '18 at 16:42
gt6989b
33.3k22452
asked Dec 4 '18 at 16:32
ZazmadzeZazmadze
12
edited Dec 4 '18 at 16:42
gt6989b
33.3k22452
edited Dec 4 '18 at 16:42
gt6989b
33.3k22452
edited Dec 4 '18 at 16:42
gt6989b
33.3k22452
33.3k22452
asked Dec 4 '18 at 16:32
ZazmadzeZazmadze
12
asked Dec 4 '18 at 16:32
ZazmadzeZazmadze
12
asked Dec 4 '18 at 16:32
ZazmadzeZazmadze
12
12
add a comment |
add a comment |
1 Answer
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$begingroup$
Your $(X,Y)$ are distributed uniform over that triangle $T$. Can you write down their pdf $f(x,y)$? (Remember that $iint_T f(x,y) dxdy = 1.$)
After you do that, the result is
$$
mathbb{E}left[XY^2right] = iint_T xy^2 f(x,y)dxdy.
$$
answered Dec 4 '18 at 16:42
gt6989bgt6989b
33.3k22452
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add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your $(X,Y)$ are distributed uniform over that triangle $T$. Can you write down their pdf $f(x,y)$? (Remember that $iint_T f(x,y) dxdy = 1.$)
After you do that, the result is
$$
mathbb{E}left[XY^2right] = iint_T xy^2 f(x,y)dxdy.
$$
answered Dec 4 '18 at 16:42
gt6989bgt6989b
33.3k22452
$endgroup$
add a comment |
$begingroup$
Your $(X,Y)$ are distributed uniform over that triangle $T$. Can you write down their pdf $f(x,y)$? (Remember that $iint_T f(x,y) dxdy = 1.$)
After you do that, the result is
$$
mathbb{E}left[XY^2right] = iint_T xy^2 f(x,y)dxdy.
$$
answered Dec 4 '18 at 16:42
gt6989bgt6989b
33.3k22452
$endgroup$
add a comment |
$begingroup$
Your $(X,Y)$ are distributed uniform over that triangle $T$. Can you write down their pdf $f(x,y)$? (Remember that $iint_T f(x,y) dxdy = 1.$)
After you do that, the result is
$$
mathbb{E}left[XY^2right] = iint_T xy^2 f(x,y)dxdy.
$$
answered Dec 4 '18 at 16:42
gt6989bgt6989b
33.3k22452
$endgroup$
Your $(X,Y)$ are distributed uniform over that triangle $T$. Can you write down their pdf $f(x,y)$? (Remember that $iint_T f(x,y) dxdy = 1.$)
After you do that, the result is
$$
mathbb{E}left[XY^2right] = iint_T xy^2 f(x,y)dxdy.
$$
answered Dec 4 '18 at 16:42
gt6989bgt6989b
33.3k22452
answered Dec 4 '18 at 16:42
gt6989bgt6989b
33.3k22452
answered Dec 4 '18 at 16:42
gt6989bgt6989b
33.3k22452
answered Dec 4 '18 at 16:42
gt6989bgt6989b
33.3k22452
33.3k22452
add a comment |
add a comment |
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