Mixing
Calculating the probability of a vector of standard normal distributed variables.
$begingroup$
Let $X,Y$ be two i.i.d random variables with standard normal distribution which is the probability that the vector $(X,Y)$ to be in the second quadrant with distance to the origin greater than $2$?
What I think I am being asked is to calculate $$mathbb{P}(Xleq 0,Ygeq 0, X^2+Y^2leq 4)$$
How do I calculate that probability?
probability-distributions
asked Jan 16 at 23:51
John KeeperJohn Keeper
532315
$endgroup$
add a comment |
$begingroup$
Let $X,Y$ be two i.i.d random variables with standard normal distribution which is the probability that the vector $(X,Y)$ to be in the second quadrant with distance to the origin greater than $2$?
What I think I am being asked is to calculate $$mathbb{P}(Xleq 0,Ygeq 0, X^2+Y^2leq 4)$$
How do I calculate that probability?
probability-distributions
asked Jan 16 at 23:51
John KeeperJohn Keeper
532315
$endgroup$
$begingroup$
You have a joint probability of two independent variables so that should make it easier. Also, do you mean $X^2+Y^2gt 4$?
$endgroup$
– John Douma
Jan 16 at 23:58
add a comment |
$begingroup$
Let $X,Y$ be two i.i.d random variables with standard normal distribution which is the probability that the vector $(X,Y)$ to be in the second quadrant with distance to the origin greater than $2$?
What I think I am being asked is to calculate $$mathbb{P}(Xleq 0,Ygeq 0, X^2+Y^2leq 4)$$
How do I calculate that probability?
probability-distributions
asked Jan 16 at 23:51
John KeeperJohn Keeper
532315
$endgroup$
Let $X,Y$ be two i.i.d random variables with standard normal distribution which is the probability that the vector $(X,Y)$ to be in the second quadrant with distance to the origin greater than $2$?
What I think I am being asked is to calculate $$mathbb{P}(Xleq 0,Ygeq 0, X^2+Y^2leq 4)$$
How do I calculate that probability?
probability-distributions
probability-distributions
asked Jan 16 at 23:51
John KeeperJohn Keeper
532315
asked Jan 16 at 23:51
John KeeperJohn Keeper
532315
asked Jan 16 at 23:51
John KeeperJohn Keeper
532315
asked Jan 16 at 23:51
John KeeperJohn Keeper
532315
asked Jan 16 at 23:51
John KeeperJohn Keeper
532315
532315
$begingroup$
You have a joint probability of two independent variables so that should make it easier. Also, do you mean $X^2+Y^2gt 4$?
$endgroup$
– John Douma
Jan 16 at 23:58
add a comment |
$begingroup$
You have a joint probability of two independent variables so that should make it easier. Also, do you mean $X^2+Y^2gt 4$?
$endgroup$
– John Douma
Jan 16 at 23:58
$begingroup$
You have a joint probability of two independent variables so that should make it easier. Also, do you mean $X^2+Y^2gt 4$?
$endgroup$
– John Douma
Jan 16 at 23:58
$begingroup$
You have a joint probability of two independent variables so that should make it easier. Also, do you mean $X^2+Y^2gt 4$?
$endgroup$
– John Douma
Jan 16 at 23:58
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $R$ be the region corresponding to the set of points $(x,y)inmathbb{R^2}$ such that $xleq 0$, $ygeq 0$ and $x^2+y^2>4$. Then by independence the joint density of $(X,Y)$ is given by
$$
f(x,y)=frac{1}{2pi}expleft(-frac{1}{2}(x^2+y^2)right).
$$
Then the required probability is given by
$$
p=int_R f(x,y), dx, dy=int_{pi/2}^piint_{2}^inftyfrac{1}{2pi}expleft(-frac{1}{2}r^2right)r, dr,dtheta
$$
by changing to polar coordinates. You should be able to compute the integral.
answered Jan 17 at 0:03
Foobaz JohnFoobaz John
22.1k41452
$endgroup$
add a comment |
$begingroup$
Let $rho(x)$ be the usual standard normal density,
$$
rho(x)=frac 1{sqrt{2pi}}exp-frac{x^2}2 ,
$$
so we know the joint density for $(X,Y)$, it is $rho(x)rho(y)$ with respect to the standard Lebesgue mass $dx; dy$, so we have to calculate:
$$
begin{aligned}
p
&=
mathbb{P}(Xleq 0,Ygeq 0, X^2+Y^2color{red}{geq} 2^2)
\
&=
int_{substack{(x,y)inBbb R^2\x<0\y>0\x^2+y^2ge 2^2}}
rho(x); rho(y); dx; dy
\
&qquadtext{ change of variables }(x,y)=(rcos t,rsin t)
\
&=
int_{rge 2}
int_{tin(pi/2,pi)}
frac 1{2pi}
exp-frac {r^2}2
cdot
r; dr; dt
\
&=
frac pi 2
int_{rge 2}
frac 1{2pi}
exp-frac {r^2}2
cdot frac 12d(r^2)
\
&=
frac 14left[ -exp-frac{r^2}2 right]_{r=2}^infty
=frac 14exp(-2)approx 0.0338338208091532dots .
end{aligned}
$$
In such cases i usually also start a computer simulation:
sage: import numpy
sage: N = 10**8 # samples
sage: X = numpy.random.normal(size=N)
sage: Y = numpy.random.normal(size=N)
sage: V = (X^2+Y^2)[ (X<0) & (Y>0) ]
sage: nr_goodCases = len( V[ V>= 4 ] )
sage: nr_goodCases
3383058
sage: print "Proportion of good cases ~ %.8f" % (nr_goodCases / N)
Proportion of good cases ~ 0.03383058
this time...
answered Jan 17 at 0:49
dan_fuleadan_fulea
6,6381312
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $R$ be the region corresponding to the set of points $(x,y)inmathbb{R^2}$ such that $xleq 0$, $ygeq 0$ and $x^2+y^2>4$. Then by independence the joint density of $(X,Y)$ is given by
$$
f(x,y)=frac{1}{2pi}expleft(-frac{1}{2}(x^2+y^2)right).
$$
Then the required probability is given by
$$
p=int_R f(x,y), dx, dy=int_{pi/2}^piint_{2}^inftyfrac{1}{2pi}expleft(-frac{1}{2}r^2right)r, dr,dtheta
$$
by changing to polar coordinates. You should be able to compute the integral.
answered Jan 17 at 0:03
Foobaz JohnFoobaz John
22.1k41452
$endgroup$
add a comment |
$begingroup$
Let $R$ be the region corresponding to the set of points $(x,y)inmathbb{R^2}$ such that $xleq 0$, $ygeq 0$ and $x^2+y^2>4$. Then by independence the joint density of $(X,Y)$ is given by
$$
f(x,y)=frac{1}{2pi}expleft(-frac{1}{2}(x^2+y^2)right).
$$
Then the required probability is given by
$$
p=int_R f(x,y), dx, dy=int_{pi/2}^piint_{2}^inftyfrac{1}{2pi}expleft(-frac{1}{2}r^2right)r, dr,dtheta
$$
by changing to polar coordinates. You should be able to compute the integral.
answered Jan 17 at 0:03
Foobaz JohnFoobaz John
22.1k41452
$endgroup$
add a comment |
$begingroup$
Let $R$ be the region corresponding to the set of points $(x,y)inmathbb{R^2}$ such that $xleq 0$, $ygeq 0$ and $x^2+y^2>4$. Then by independence the joint density of $(X,Y)$ is given by
$$
f(x,y)=frac{1}{2pi}expleft(-frac{1}{2}(x^2+y^2)right).
$$
Then the required probability is given by
$$
p=int_R f(x,y), dx, dy=int_{pi/2}^piint_{2}^inftyfrac{1}{2pi}expleft(-frac{1}{2}r^2right)r, dr,dtheta
$$
by changing to polar coordinates. You should be able to compute the integral.
answered Jan 17 at 0:03
Foobaz JohnFoobaz John
22.1k41452
$endgroup$
Let $R$ be the region corresponding to the set of points $(x,y)inmathbb{R^2}$ such that $xleq 0$, $ygeq 0$ and $x^2+y^2>4$. Then by independence the joint density of $(X,Y)$ is given by
$$
f(x,y)=frac{1}{2pi}expleft(-frac{1}{2}(x^2+y^2)right).
$$
Then the required probability is given by
$$
p=int_R f(x,y), dx, dy=int_{pi/2}^piint_{2}^inftyfrac{1}{2pi}expleft(-frac{1}{2}r^2right)r, dr,dtheta
$$
by changing to polar coordinates. You should be able to compute the integral.
answered Jan 17 at 0:03
Foobaz JohnFoobaz John
22.1k41452
answered Jan 17 at 0:03
Foobaz JohnFoobaz John
22.1k41452
answered Jan 17 at 0:03
Foobaz JohnFoobaz John
22.1k41452
answered Jan 17 at 0:03
Foobaz JohnFoobaz John
22.1k41452
22.1k41452
add a comment |
add a comment |
$begingroup$
Let $rho(x)$ be the usual standard normal density,
$$
rho(x)=frac 1{sqrt{2pi}}exp-frac{x^2}2 ,
$$
so we know the joint density for $(X,Y)$, it is $rho(x)rho(y)$ with respect to the standard Lebesgue mass $dx; dy$, so we have to calculate:
$$
begin{aligned}
p
&=
mathbb{P}(Xleq 0,Ygeq 0, X^2+Y^2color{red}{geq} 2^2)
\
&=
int_{substack{(x,y)inBbb R^2\x<0\y>0\x^2+y^2ge 2^2}}
rho(x); rho(y); dx; dy
\
&qquadtext{ change of variables }(x,y)=(rcos t,rsin t)
\
&=
int_{rge 2}
int_{tin(pi/2,pi)}
frac 1{2pi}
exp-frac {r^2}2
cdot
r; dr; dt
\
&=
frac pi 2
int_{rge 2}
frac 1{2pi}
exp-frac {r^2}2
cdot frac 12d(r^2)
\
&=
frac 14left[ -exp-frac{r^2}2 right]_{r=2}^infty
=frac 14exp(-2)approx 0.0338338208091532dots .
end{aligned}
$$
In such cases i usually also start a computer simulation:
sage: import numpy
sage: N = 10**8 # samples
sage: X = numpy.random.normal(size=N)
sage: Y = numpy.random.normal(size=N)
sage: V = (X^2+Y^2)[ (X<0) & (Y>0) ]
sage: nr_goodCases = len( V[ V>= 4 ] )
sage: nr_goodCases
3383058
sage: print "Proportion of good cases ~ %.8f" % (nr_goodCases / N)
Proportion of good cases ~ 0.03383058
this time...
answered Jan 17 at 0:49
dan_fuleadan_fulea
6,6381312
$endgroup$
add a comment |
$begingroup$
Let $rho(x)$ be the usual standard normal density,
$$
rho(x)=frac 1{sqrt{2pi}}exp-frac{x^2}2 ,
$$
so we know the joint density for $(X,Y)$, it is $rho(x)rho(y)$ with respect to the standard Lebesgue mass $dx; dy$, so we have to calculate:
$$
begin{aligned}
p
&=
mathbb{P}(Xleq 0,Ygeq 0, X^2+Y^2color{red}{geq} 2^2)
\
&=
int_{substack{(x,y)inBbb R^2\x<0\y>0\x^2+y^2ge 2^2}}
rho(x); rho(y); dx; dy
\
&qquadtext{ change of variables }(x,y)=(rcos t,rsin t)
\
&=
int_{rge 2}
int_{tin(pi/2,pi)}
frac 1{2pi}
exp-frac {r^2}2
cdot
r; dr; dt
\
&=
frac pi 2
int_{rge 2}
frac 1{2pi}
exp-frac {r^2}2
cdot frac 12d(r^2)
\
&=
frac 14left[ -exp-frac{r^2}2 right]_{r=2}^infty
=frac 14exp(-2)approx 0.0338338208091532dots .
end{aligned}
$$
In such cases i usually also start a computer simulation:
sage: import numpy
sage: N = 10**8 # samples
sage: X = numpy.random.normal(size=N)
sage: Y = numpy.random.normal(size=N)
sage: V = (X^2+Y^2)[ (X<0) & (Y>0) ]
sage: nr_goodCases = len( V[ V>= 4 ] )
sage: nr_goodCases
3383058
sage: print "Proportion of good cases ~ %.8f" % (nr_goodCases / N)
Proportion of good cases ~ 0.03383058
this time...
answered Jan 17 at 0:49
dan_fuleadan_fulea
6,6381312
$endgroup$
add a comment |
$begingroup$
Let $rho(x)$ be the usual standard normal density,
$$
rho(x)=frac 1{sqrt{2pi}}exp-frac{x^2}2 ,
$$
so we know the joint density for $(X,Y)$, it is $rho(x)rho(y)$ with respect to the standard Lebesgue mass $dx; dy$, so we have to calculate:
$$
begin{aligned}
p
&=
mathbb{P}(Xleq 0,Ygeq 0, X^2+Y^2color{red}{geq} 2^2)
\
&=
int_{substack{(x,y)inBbb R^2\x<0\y>0\x^2+y^2ge 2^2}}
rho(x); rho(y); dx; dy
\
&qquadtext{ change of variables }(x,y)=(rcos t,rsin t)
\
&=
int_{rge 2}
int_{tin(pi/2,pi)}
frac 1{2pi}
exp-frac {r^2}2
cdot
r; dr; dt
\
&=
frac pi 2
int_{rge 2}
frac 1{2pi}
exp-frac {r^2}2
cdot frac 12d(r^2)
\
&=
frac 14left[ -exp-frac{r^2}2 right]_{r=2}^infty
=frac 14exp(-2)approx 0.0338338208091532dots .
end{aligned}
$$
In such cases i usually also start a computer simulation:
sage: import numpy
sage: N = 10**8 # samples
sage: X = numpy.random.normal(size=N)
sage: Y = numpy.random.normal(size=N)
sage: V = (X^2+Y^2)[ (X<0) & (Y>0) ]
sage: nr_goodCases = len( V[ V>= 4 ] )
sage: nr_goodCases
3383058
sage: print "Proportion of good cases ~ %.8f" % (nr_goodCases / N)
Proportion of good cases ~ 0.03383058
this time...
answered Jan 17 at 0:49
dan_fuleadan_fulea
6,6381312
$endgroup$
Let $rho(x)$ be the usual standard normal density,
$$
rho(x)=frac 1{sqrt{2pi}}exp-frac{x^2}2 ,
$$
so we know the joint density for $(X,Y)$, it is $rho(x)rho(y)$ with respect to the standard Lebesgue mass $dx; dy$, so we have to calculate:
$$
begin{aligned}
p
&=
mathbb{P}(Xleq 0,Ygeq 0, X^2+Y^2color{red}{geq} 2^2)
\
&=
int_{substack{(x,y)inBbb R^2\x<0\y>0\x^2+y^2ge 2^2}}
rho(x); rho(y); dx; dy
\
&qquadtext{ change of variables }(x,y)=(rcos t,rsin t)
\
&=
int_{rge 2}
int_{tin(pi/2,pi)}
frac 1{2pi}
exp-frac {r^2}2
cdot
r; dr; dt
\
&=
frac pi 2
int_{rge 2}
frac 1{2pi}
exp-frac {r^2}2
cdot frac 12d(r^2)
\
&=
frac 14left[ -exp-frac{r^2}2 right]_{r=2}^infty
=frac 14exp(-2)approx 0.0338338208091532dots .
end{aligned}
$$
In such cases i usually also start a computer simulation:
sage: import numpy
sage: N = 10**8 # samples
sage: X = numpy.random.normal(size=N)
sage: Y = numpy.random.normal(size=N)
sage: V = (X^2+Y^2)[ (X<0) & (Y>0) ]
sage: nr_goodCases = len( V[ V>= 4 ] )
sage: nr_goodCases
3383058
sage: print "Proportion of good cases ~ %.8f" % (nr_goodCases / N)
Proportion of good cases ~ 0.03383058
this time...
answered Jan 17 at 0:49
dan_fuleadan_fulea
6,6381312
answered Jan 17 at 0:49
dan_fuleadan_fulea
6,6381312
answered Jan 17 at 0:49
dan_fuleadan_fulea
6,6381312
answered Jan 17 at 0:49
dan_fuleadan_fulea
6,6381312
6,6381312
add a comment |
add a comment |
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$begingroup$
You have a joint probability of two independent variables so that should make it easier. Also, do you mean $X^2+Y^2gt 4$?
$endgroup$
– John Douma
Jan 16 at 23:58