Mixing
An Integration problem of Bivariate Normal Distribution
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I'm currently solving previous years problems for my M.Sc Statistics entrance exam on IIT (India).
Problem->
Find the value of
$$frac{1}{2pi}int_{-infty}^{infty}int_{-infty}^y e^{frac{-1}{2}(x^2 + y^2)}mathop{dx} mathop{dy}$$
This is a PDF of Bivariate Normal(0,0,1,1,cor=0)
Now I can't find an right approach to solve it, I tried integrating but it became messed up as after using the integration by parts form. Thank you for your time :-)
integration
edited Jan 23 at 9:45
Paras Khosla
1,826219
asked Jan 23 at 7:42
user194259user194259
305
$endgroup$
add a comment |
$begingroup$
I'm currently solving previous years problems for my M.Sc Statistics entrance exam on IIT (India).
Problem->
Find the value of
$$frac{1}{2pi}int_{-infty}^{infty}int_{-infty}^y e^{frac{-1}{2}(x^2 + y^2)}mathop{dx} mathop{dy}$$
This is a PDF of Bivariate Normal(0,0,1,1,cor=0)
Now I can't find an right approach to solve it, I tried integrating but it became messed up as after using the integration by parts form. Thank you for your time :-)
integration
edited Jan 23 at 9:45
Paras Khosla
1,826219
asked Jan 23 at 7:42
user194259user194259
305
$endgroup$
$begingroup$
If you can use that this is circularly symmetric and a distribution, then you may exploit the symmetry, right? A distribution adds up to 1, it is divided into two symmetrical parts, each part shall add up to 0.5? If you wish to deal with the double integral typical approach would be making the exponent square.
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– keoxkeox
Jan 23 at 8:27
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This has been asked before but I cannot find the question.
$endgroup$
– StubbornAtom
Jan 23 at 8:38
$begingroup$
Here's one: math.stackexchange.com/questions/2675055/….
$endgroup$
– StubbornAtom
Jan 23 at 10:51
add a comment |
$begingroup$
I'm currently solving previous years problems for my M.Sc Statistics entrance exam on IIT (India).
Problem->
Find the value of
$$frac{1}{2pi}int_{-infty}^{infty}int_{-infty}^y e^{frac{-1}{2}(x^2 + y^2)}mathop{dx} mathop{dy}$$
This is a PDF of Bivariate Normal(0,0,1,1,cor=0)
Now I can't find an right approach to solve it, I tried integrating but it became messed up as after using the integration by parts form. Thank you for your time :-)
integration
edited Jan 23 at 9:45
Paras Khosla
1,826219
asked Jan 23 at 7:42
user194259user194259
305
$endgroup$
I'm currently solving previous years problems for my M.Sc Statistics entrance exam on IIT (India).
Problem->
Find the value of
$$frac{1}{2pi}int_{-infty}^{infty}int_{-infty}^y e^{frac{-1}{2}(x^2 + y^2)}mathop{dx} mathop{dy}$$
This is a PDF of Bivariate Normal(0,0,1,1,cor=0)
Now I can't find an right approach to solve it, I tried integrating but it became messed up as after using the integration by parts form. Thank you for your time :-)
integration
integration
edited Jan 23 at 9:45
Paras Khosla
1,826219
asked Jan 23 at 7:42
user194259user194259
305
edited Jan 23 at 9:45
Paras Khosla
1,826219
asked Jan 23 at 7:42
user194259user194259
305
edited Jan 23 at 9:45
Paras Khosla
1,826219
edited Jan 23 at 9:45
Paras Khosla
1,826219
edited Jan 23 at 9:45
Paras Khosla
1,826219
1,826219
asked Jan 23 at 7:42
user194259user194259
305
asked Jan 23 at 7:42
user194259user194259
305
asked Jan 23 at 7:42
user194259user194259
305
305
$begingroup$
If you can use that this is circularly symmetric and a distribution, then you may exploit the symmetry, right? A distribution adds up to 1, it is divided into two symmetrical parts, each part shall add up to 0.5? If you wish to deal with the double integral typical approach would be making the exponent square.
$endgroup$
– keoxkeox
Jan 23 at 8:27
$begingroup$
This has been asked before but I cannot find the question.
$endgroup$
– StubbornAtom
Jan 23 at 8:38
$begingroup$
Here's one: math.stackexchange.com/questions/2675055/….
$endgroup$
– StubbornAtom
Jan 23 at 10:51
add a comment |
$begingroup$
If you can use that this is circularly symmetric and a distribution, then you may exploit the symmetry, right? A distribution adds up to 1, it is divided into two symmetrical parts, each part shall add up to 0.5? If you wish to deal with the double integral typical approach would be making the exponent square.
$endgroup$
– keoxkeox
Jan 23 at 8:27
$begingroup$
This has been asked before but I cannot find the question.
$endgroup$
– StubbornAtom
Jan 23 at 8:38
$begingroup$
Here's one: math.stackexchange.com/questions/2675055/….
$endgroup$
– StubbornAtom
Jan 23 at 10:51
$begingroup$
If you can use that this is circularly symmetric and a distribution, then you may exploit the symmetry, right? A distribution adds up to 1, it is divided into two symmetrical parts, each part shall add up to 0.5? If you wish to deal with the double integral typical approach would be making the exponent square.
$endgroup$
– keoxkeox
Jan 23 at 8:27
$begingroup$
If you can use that this is circularly symmetric and a distribution, then you may exploit the symmetry, right? A distribution adds up to 1, it is divided into two symmetrical parts, each part shall add up to 0.5? If you wish to deal with the double integral typical approach would be making the exponent square.
$endgroup$
– keoxkeox
Jan 23 at 8:27
$begingroup$
This has been asked before but I cannot find the question.
$endgroup$
– StubbornAtom
Jan 23 at 8:38
$begingroup$
This has been asked before but I cannot find the question.
$endgroup$
– StubbornAtom
Jan 23 at 8:38
$begingroup$
Here's one: math.stackexchange.com/questions/2675055/….
$endgroup$
– StubbornAtom
Jan 23 at 10:51
$begingroup$
Here's one: math.stackexchange.com/questions/2675055/….
$endgroup$
– StubbornAtom
Jan 23 at 10:51
add a comment |
1 Answer
1
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Your integral is $int_{(x,,y)inBbb R^2land xle y}phi(x)phi(y)dxdy$, with $phi$ the $N(0,,1)$ pdf. Therefore, it is the probability $xle y$ subject to $x,,y$ being $N(0,,1)$ iids. In other words, $1/2$.
answered Jan 23 at 10:58
J.G.J.G.
29.8k22947
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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
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active
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$begingroup$
Your integral is $int_{(x,,y)inBbb R^2land xle y}phi(x)phi(y)dxdy$, with $phi$ the $N(0,,1)$ pdf. Therefore, it is the probability $xle y$ subject to $x,,y$ being $N(0,,1)$ iids. In other words, $1/2$.
answered Jan 23 at 10:58
J.G.J.G.
29.8k22947
$endgroup$
add a comment |
$begingroup$
Your integral is $int_{(x,,y)inBbb R^2land xle y}phi(x)phi(y)dxdy$, with $phi$ the $N(0,,1)$ pdf. Therefore, it is the probability $xle y$ subject to $x,,y$ being $N(0,,1)$ iids. In other words, $1/2$.
answered Jan 23 at 10:58
J.G.J.G.
29.8k22947
$endgroup$
add a comment |
$begingroup$
Your integral is $int_{(x,,y)inBbb R^2land xle y}phi(x)phi(y)dxdy$, with $phi$ the $N(0,,1)$ pdf. Therefore, it is the probability $xle y$ subject to $x,,y$ being $N(0,,1)$ iids. In other words, $1/2$.
answered Jan 23 at 10:58
J.G.J.G.
29.8k22947
$endgroup$
Your integral is $int_{(x,,y)inBbb R^2land xle y}phi(x)phi(y)dxdy$, with $phi$ the $N(0,,1)$ pdf. Therefore, it is the probability $xle y$ subject to $x,,y$ being $N(0,,1)$ iids. In other words, $1/2$.
answered Jan 23 at 10:58
J.G.J.G.
29.8k22947
answered Jan 23 at 10:58
J.G.J.G.
29.8k22947
answered Jan 23 at 10:58
J.G.J.G.
29.8k22947
answered Jan 23 at 10:58
J.G.J.G.
29.8k22947
29.8k22947
add a comment |
add a comment |
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If you can use that this is circularly symmetric and a distribution, then you may exploit the symmetry, right? A distribution adds up to 1, it is divided into two symmetrical parts, each part shall add up to 0.5? If you wish to deal with the double integral typical approach would be making the exponent square.
$endgroup$
– keoxkeox
Jan 23 at 8:27
$begingroup$
This has been asked before but I cannot find the question.
$endgroup$
– StubbornAtom
Jan 23 at 8:38
$begingroup$
Here's one: math.stackexchange.com/questions/2675055/….
$endgroup$
– StubbornAtom
Jan 23 at 10:51