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$Z$ ~ $N(0, 1)$. Find $operatornameVar(Z)$
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$Z$ ~ $N(0, 1)$. Find $operatornameVar(Z)$
We know:
$E(Z) = 0$
$operatornameVar(Z) = E(Z^2) - E(Z)^2 = E(Z^2) - 0^2 = E(Z^2)$
Thus, we need:
$E(Z^2) = int_-infty^infty z^2 frac1sqrt2pie^-1/2(z^2)dz$
Using int by parts, $u = z, du = dz$, then $dv = z frac1sqrt2pie^-1/2z^2dz$, $v = -frac1sqrt2pie^-1/2z^2$
Then we have
$$left[-fraczsqrt2pie^-1/2z^2 right]_-infty^infty + frac1sqrt2piint_-infty^inftye^-1/2z^2dz$$
I'm aware that $left[-fraczsqrt2pie^-1/2z^2 right]_-infty^infty = (0-0) = 0$
No clue how to proceed.
probability expectation
edited Aug 6 at 21:52
Bernard
110k635103
asked Aug 6 at 21:22
Bas bas
39611
add a comment |Â
up vote
1
down vote
favorite
$Z$ ~ $N(0, 1)$. Find $operatornameVar(Z)$
We know:
$E(Z) = 0$
$operatornameVar(Z) = E(Z^2) - E(Z)^2 = E(Z^2) - 0^2 = E(Z^2)$
Thus, we need:
$E(Z^2) = int_-infty^infty z^2 frac1sqrt2pie^-1/2(z^2)dz$
Using int by parts, $u = z, du = dz$, then $dv = z frac1sqrt2pie^-1/2z^2dz$, $v = -frac1sqrt2pie^-1/2z^2$
Then we have
$$left[-fraczsqrt2pie^-1/2z^2 right]_-infty^infty + frac1sqrt2piint_-infty^inftye^-1/2z^2dz$$
I'm aware that $left[-fraczsqrt2pie^-1/2z^2 right]_-infty^infty = (0-0) = 0$
No clue how to proceed.
probability expectation
edited Aug 6 at 21:52
Bernard
110k635103
asked Aug 6 at 21:22
Bas bas
39611
2
What does the $1$ in $N(0,1)$ mean to you?
– Arthur
Aug 6 at 21:28
By definition $N(u, sigma^2)$, so $sigma^2 = 1$
– Bas bas
Aug 6 at 21:29
1
So you got it, right? You know that $sigma^2$ is another notation for Var($Z$)?
– Arnaud Mortier
Aug 6 at 21:50
Yeah, I understand now thank you.
– Bas bas
Aug 6 at 22:04
1
Yes, but the point of the exercise is to show that the formula for PDF of $N(0,1)$ does have variance $1$.
– Robert Israel
Aug 6 at 23:40
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$Z$ ~ $N(0, 1)$. Find $operatornameVar(Z)$
We know:
$E(Z) = 0$
$operatornameVar(Z) = E(Z^2) - E(Z)^2 = E(Z^2) - 0^2 = E(Z^2)$
Thus, we need:
$E(Z^2) = int_-infty^infty z^2 frac1sqrt2pie^-1/2(z^2)dz$
Using int by parts, $u = z, du = dz$, then $dv = z frac1sqrt2pie^-1/2z^2dz$, $v = -frac1sqrt2pie^-1/2z^2$
Then we have
$$left[-fraczsqrt2pie^-1/2z^2 right]_-infty^infty + frac1sqrt2piint_-infty^inftye^-1/2z^2dz$$
I'm aware that $left[-fraczsqrt2pie^-1/2z^2 right]_-infty^infty = (0-0) = 0$
No clue how to proceed.
probability expectation
edited Aug 6 at 21:52
Bernard
110k635103
asked Aug 6 at 21:22
Bas bas
39611
$Z$ ~ $N(0, 1)$. Find $operatornameVar(Z)$
We know:
$E(Z) = 0$
$operatornameVar(Z) = E(Z^2) - E(Z)^2 = E(Z^2) - 0^2 = E(Z^2)$
Thus, we need:
$E(Z^2) = int_-infty^infty z^2 frac1sqrt2pie^-1/2(z^2)dz$
Using int by parts, $u = z, du = dz$, then $dv = z frac1sqrt2pie^-1/2z^2dz$, $v = -frac1sqrt2pie^-1/2z^2$
Then we have
$$left[-fraczsqrt2pie^-1/2z^2 right]_-infty^infty + frac1sqrt2piint_-infty^inftye^-1/2z^2dz$$
I'm aware that $left[-fraczsqrt2pie^-1/2z^2 right]_-infty^infty = (0-0) = 0$
No clue how to proceed.
probability expectation
edited Aug 6 at 21:52
Bernard
110k635103
asked Aug 6 at 21:22
Bas bas
39611
edited Aug 6 at 21:52
Bernard
110k635103
edited Aug 6 at 21:52
Bernard
110k635103
edited Aug 6 at 21:52
Bernard
110k635103
110k635103
asked Aug 6 at 21:22
Bas bas
39611
asked Aug 6 at 21:22
Bas bas
39611
asked Aug 6 at 21:22
Bas bas
39611
39611
2
What does the $1$ in $N(0,1)$ mean to you?
– Arthur
Aug 6 at 21:28
By definition $N(u, sigma^2)$, so $sigma^2 = 1$
– Bas bas
Aug 6 at 21:29
1
So you got it, right? You know that $sigma^2$ is another notation for Var($Z$)?
– Arnaud Mortier
Aug 6 at 21:50
Yeah, I understand now thank you.
– Bas bas
Aug 6 at 22:04
1
Yes, but the point of the exercise is to show that the formula for PDF of $N(0,1)$ does have variance $1$.
– Robert Israel
Aug 6 at 23:40
add a comment |Â
2
What does the $1$ in $N(0,1)$ mean to you?
– Arthur
Aug 6 at 21:28
By definition $N(u, sigma^2)$, so $sigma^2 = 1$
– Bas bas
Aug 6 at 21:29
1
So you got it, right? You know that $sigma^2$ is another notation for Var($Z$)?
– Arnaud Mortier
Aug 6 at 21:50
Yeah, I understand now thank you.
– Bas bas
Aug 6 at 22:04
1
Yes, but the point of the exercise is to show that the formula for PDF of $N(0,1)$ does have variance $1$.
– Robert Israel
Aug 6 at 23:40
2
2
What does the $1$ in $N(0,1)$ mean to you?
– Arthur
Aug 6 at 21:28
What does the $1$ in $N(0,1)$ mean to you?
– Arthur
Aug 6 at 21:28
By definition $N(u, sigma^2)$, so $sigma^2 = 1$
– Bas bas
Aug 6 at 21:29
By definition $N(u, sigma^2)$, so $sigma^2 = 1$
– Bas bas
Aug 6 at 21:29
1
1
So you got it, right? You know that $sigma^2$ is another notation for Var($Z$)?
– Arnaud Mortier
Aug 6 at 21:50
So you got it, right? You know that $sigma^2$ is another notation for Var($Z$)?
– Arnaud Mortier
Aug 6 at 21:50
Yeah, I understand now thank you.
– Bas bas
Aug 6 at 22:04
Yeah, I understand now thank you.
– Bas bas
Aug 6 at 22:04
1
1
Yes, but the point of the exercise is to show that the formula for PDF of $N(0,1)$ does have variance $1$.
– Robert Israel
Aug 6 at 23:40
Yes, but the point of the exercise is to show that the formula for PDF of $N(0,1)$ does have variance $1$.
– Robert Israel
Aug 6 at 23:40
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
The standard method for evaluating $int_-infty^inftye^-frac12z^2dz$ isn't an obvious one, but one worth remembering. It involves squaring it and turning it into a double integral.
First note that
$$int_-infty^inftye^-frac12z^2dz=2int_0^inftye^-frac12z^2dz$$
as $e^-frac12z^2$ is an even function.
Let $I:=int_0^inftye^-frac12z^2dz$. Also note that "$z$" is just a dummy variable, so $I=int_0^inftye^-frac12x^2dx=int_0^inftye^-frac12y^2dy$.
$$
beginalign
I^2 &= int_0^inftye^-frac12x^2dx int_0^inftye^-frac12y^2dy
\ &= int_0^inftyint_0^inftye^-frac12x^2e^-frac12y^2dxdy
\ &= int_0^inftyint_0^inftye^-frac12(x^2+y^2)dxdy
\ &= int_0^fracpi2int_0^inftye^-frac12r^2rdrdtheta
\ &= fracpi2int_0^inftyre^-frac12r^2dr
\ &= fracpi2 left[ -e^-frac12r^2 right]_0^infty
\ &= fracpi2
endalign
$$
... and hence
$$ I = fracsqrt2pi2 $$
which, together with your working, gives:
$$
beginalign
mathrmVar(Z) &= E(Z^2)
\ &= frac1sqrt2piint_-infty^inftye^-frac12z^2dz
\ &= frac2sqrt2pi I
\ &= 1
endalign
$$
as required!
answered Aug 6 at 22:08
Malkin
1,395523
add a comment |Â
up vote
1
down vote
Using Fubini's theorem you can compute $(int e^-x^2)^2$=$pi$ and in the same way that $int e^-fracx^2alpha=sqrtalphapi$, which gives you the result you were looking for.
answered Aug 6 at 21:44
Leo Hö
195
add a comment |Â
up vote
-1
down vote
The answer is by definition of the law 1.
Moreover, you are trying to find the variance of a normal law knowing only the mean. Do you think it is even possible ? You could have any positive value for the variance.
answered Aug 6 at 21:32
Pjonin
3206
It seems you think that no meaningful Question has been posed, or at least that any value could be assigned. It might be a narrow reading of the Question (and perhaps an Answer could outline what additional information is needed for a solution), but merely pointing out a shortcoming would be better executed as a Comment on the the Question than posted as an Answer.
– hardmath
Aug 6 at 21:57
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The standard method for evaluating $int_-infty^inftye^-frac12z^2dz$ isn't an obvious one, but one worth remembering. It involves squaring it and turning it into a double integral.
First note that
$$int_-infty^inftye^-frac12z^2dz=2int_0^inftye^-frac12z^2dz$$
as $e^-frac12z^2$ is an even function.
Let $I:=int_0^inftye^-frac12z^2dz$. Also note that "$z$" is just a dummy variable, so $I=int_0^inftye^-frac12x^2dx=int_0^inftye^-frac12y^2dy$.
$$
beginalign
I^2 &= int_0^inftye^-frac12x^2dx int_0^inftye^-frac12y^2dy
\ &= int_0^inftyint_0^inftye^-frac12x^2e^-frac12y^2dxdy
\ &= int_0^inftyint_0^inftye^-frac12(x^2+y^2)dxdy
\ &= int_0^fracpi2int_0^inftye^-frac12r^2rdrdtheta
\ &= fracpi2int_0^inftyre^-frac12r^2dr
\ &= fracpi2 left[ -e^-frac12r^2 right]_0^infty
\ &= fracpi2
endalign
$$
... and hence
$$ I = fracsqrt2pi2 $$
which, together with your working, gives:
$$
beginalign
mathrmVar(Z) &= E(Z^2)
\ &= frac1sqrt2piint_-infty^inftye^-frac12z^2dz
\ &= frac2sqrt2pi I
\ &= 1
endalign
$$
as required!
answered Aug 6 at 22:08
Malkin
1,395523
add a comment |Â
up vote
1
down vote
accepted
The standard method for evaluating $int_-infty^inftye^-frac12z^2dz$ isn't an obvious one, but one worth remembering. It involves squaring it and turning it into a double integral.
First note that
$$int_-infty^inftye^-frac12z^2dz=2int_0^inftye^-frac12z^2dz$$
as $e^-frac12z^2$ is an even function.
Let $I:=int_0^inftye^-frac12z^2dz$. Also note that "$z$" is just a dummy variable, so $I=int_0^inftye^-frac12x^2dx=int_0^inftye^-frac12y^2dy$.
$$
beginalign
I^2 &= int_0^inftye^-frac12x^2dx int_0^inftye^-frac12y^2dy
\ &= int_0^inftyint_0^inftye^-frac12x^2e^-frac12y^2dxdy
\ &= int_0^inftyint_0^inftye^-frac12(x^2+y^2)dxdy
\ &= int_0^fracpi2int_0^inftye^-frac12r^2rdrdtheta
\ &= fracpi2int_0^inftyre^-frac12r^2dr
\ &= fracpi2 left[ -e^-frac12r^2 right]_0^infty
\ &= fracpi2
endalign
$$
... and hence
$$ I = fracsqrt2pi2 $$
which, together with your working, gives:
$$
beginalign
mathrmVar(Z) &= E(Z^2)
\ &= frac1sqrt2piint_-infty^inftye^-frac12z^2dz
\ &= frac2sqrt2pi I
\ &= 1
endalign
$$
as required!
answered Aug 6 at 22:08
Malkin
1,395523
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The standard method for evaluating $int_-infty^inftye^-frac12z^2dz$ isn't an obvious one, but one worth remembering. It involves squaring it and turning it into a double integral.
First note that
$$int_-infty^inftye^-frac12z^2dz=2int_0^inftye^-frac12z^2dz$$
as $e^-frac12z^2$ is an even function.
Let $I:=int_0^inftye^-frac12z^2dz$. Also note that "$z$" is just a dummy variable, so $I=int_0^inftye^-frac12x^2dx=int_0^inftye^-frac12y^2dy$.
$$
beginalign
I^2 &= int_0^inftye^-frac12x^2dx int_0^inftye^-frac12y^2dy
\ &= int_0^inftyint_0^inftye^-frac12x^2e^-frac12y^2dxdy
\ &= int_0^inftyint_0^inftye^-frac12(x^2+y^2)dxdy
\ &= int_0^fracpi2int_0^inftye^-frac12r^2rdrdtheta
\ &= fracpi2int_0^inftyre^-frac12r^2dr
\ &= fracpi2 left[ -e^-frac12r^2 right]_0^infty
\ &= fracpi2
endalign
$$
... and hence
$$ I = fracsqrt2pi2 $$
which, together with your working, gives:
$$
beginalign
mathrmVar(Z) &= E(Z^2)
\ &= frac1sqrt2piint_-infty^inftye^-frac12z^2dz
\ &= frac2sqrt2pi I
\ &= 1
endalign
$$
as required!
answered Aug 6 at 22:08
Malkin
1,395523
The standard method for evaluating $int_-infty^inftye^-frac12z^2dz$ isn't an obvious one, but one worth remembering. It involves squaring it and turning it into a double integral.
First note that
$$int_-infty^inftye^-frac12z^2dz=2int_0^inftye^-frac12z^2dz$$
as $e^-frac12z^2$ is an even function.
Let $I:=int_0^inftye^-frac12z^2dz$. Also note that "$z$" is just a dummy variable, so $I=int_0^inftye^-frac12x^2dx=int_0^inftye^-frac12y^2dy$.
$$
beginalign
I^2 &= int_0^inftye^-frac12x^2dx int_0^inftye^-frac12y^2dy
\ &= int_0^inftyint_0^inftye^-frac12x^2e^-frac12y^2dxdy
\ &= int_0^inftyint_0^inftye^-frac12(x^2+y^2)dxdy
\ &= int_0^fracpi2int_0^inftye^-frac12r^2rdrdtheta
\ &= fracpi2int_0^inftyre^-frac12r^2dr
\ &= fracpi2 left[ -e^-frac12r^2 right]_0^infty
\ &= fracpi2
endalign
$$
... and hence
$$ I = fracsqrt2pi2 $$
which, together with your working, gives:
$$
beginalign
mathrmVar(Z) &= E(Z^2)
\ &= frac1sqrt2piint_-infty^inftye^-frac12z^2dz
\ &= frac2sqrt2pi I
\ &= 1
endalign
$$
as required!
answered Aug 6 at 22:08
Malkin
1,395523
answered Aug 6 at 22:08
Malkin
1,395523
answered Aug 6 at 22:08
Malkin
1,395523
answered Aug 6 at 22:08
Malkin
1,395523
1,395523
add a comment |Â
add a comment |Â
up vote
1
down vote
Using Fubini's theorem you can compute $(int e^-x^2)^2$=$pi$ and in the same way that $int e^-fracx^2alpha=sqrtalphapi$, which gives you the result you were looking for.
answered Aug 6 at 21:44
Leo Hö
195
add a comment |Â
up vote
1
down vote
Using Fubini's theorem you can compute $(int e^-x^2)^2$=$pi$ and in the same way that $int e^-fracx^2alpha=sqrtalphapi$, which gives you the result you were looking for.
answered Aug 6 at 21:44
Leo Hö
195
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Using Fubini's theorem you can compute $(int e^-x^2)^2$=$pi$ and in the same way that $int e^-fracx^2alpha=sqrtalphapi$, which gives you the result you were looking for.
answered Aug 6 at 21:44
Leo Hö
195
Using Fubini's theorem you can compute $(int e^-x^2)^2$=$pi$ and in the same way that $int e^-fracx^2alpha=sqrtalphapi$, which gives you the result you were looking for.
answered Aug 6 at 21:44
Leo Hö
195
answered Aug 6 at 21:44
Leo Hö
195
answered Aug 6 at 21:44
Leo Hö
195
answered Aug 6 at 21:44
Leo Hö
195
195
add a comment |Â
add a comment |Â
up vote
-1
down vote
The answer is by definition of the law 1.
Moreover, you are trying to find the variance of a normal law knowing only the mean. Do you think it is even possible ? You could have any positive value for the variance.
answered Aug 6 at 21:32
Pjonin
3206
It seems you think that no meaningful Question has been posed, or at least that any value could be assigned. It might be a narrow reading of the Question (and perhaps an Answer could outline what additional information is needed for a solution), but merely pointing out a shortcoming would be better executed as a Comment on the the Question than posted as an Answer.
– hardmath
Aug 6 at 21:57
add a comment |Â
up vote
-1
down vote
The answer is by definition of the law 1.
Moreover, you are trying to find the variance of a normal law knowing only the mean. Do you think it is even possible ? You could have any positive value for the variance.
answered Aug 6 at 21:32
Pjonin
3206
It seems you think that no meaningful Question has been posed, or at least that any value could be assigned. It might be a narrow reading of the Question (and perhaps an Answer could outline what additional information is needed for a solution), but merely pointing out a shortcoming would be better executed as a Comment on the the Question than posted as an Answer.
– hardmath
Aug 6 at 21:57
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
The answer is by definition of the law 1.
Moreover, you are trying to find the variance of a normal law knowing only the mean. Do you think it is even possible ? You could have any positive value for the variance.
answered Aug 6 at 21:32
Pjonin
3206
The answer is by definition of the law 1.
Moreover, you are trying to find the variance of a normal law knowing only the mean. Do you think it is even possible ? You could have any positive value for the variance.
answered Aug 6 at 21:32
Pjonin
3206
answered Aug 6 at 21:32
Pjonin
3206
answered Aug 6 at 21:32
Pjonin
3206
answered Aug 6 at 21:32
Pjonin
3206
3206
It seems you think that no meaningful Question has been posed, or at least that any value could be assigned. It might be a narrow reading of the Question (and perhaps an Answer could outline what additional information is needed for a solution), but merely pointing out a shortcoming would be better executed as a Comment on the the Question than posted as an Answer.
– hardmath
Aug 6 at 21:57
add a comment |Â
It seems you think that no meaningful Question has been posed, or at least that any value could be assigned. It might be a narrow reading of the Question (and perhaps an Answer could outline what additional information is needed for a solution), but merely pointing out a shortcoming would be better executed as a Comment on the the Question than posted as an Answer.
– hardmath
Aug 6 at 21:57
It seems you think that no meaningful Question has been posed, or at least that any value could be assigned. It might be a narrow reading of the Question (and perhaps an Answer could outline what additional information is needed for a solution), but merely pointing out a shortcoming would be better executed as a Comment on the the Question than posted as an Answer.
– hardmath
Aug 6 at 21:57
It seems you think that no meaningful Question has been posed, or at least that any value could be assigned. It might be a narrow reading of the Question (and perhaps an Answer could outline what additional information is needed for a solution), but merely pointing out a shortcoming would be better executed as a Comment on the the Question than posted as an Answer.
– hardmath
Aug 6 at 21:57
add a comment |Â
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2
What does the $1$ in $N(0,1)$ mean to you?
– Arthur
Aug 6 at 21:28
By definition $N(u, sigma^2)$, so $sigma^2 = 1$
– Bas bas
Aug 6 at 21:29
1
So you got it, right? You know that $sigma^2$ is another notation for Var($Z$)?
– Arnaud Mortier
Aug 6 at 21:50
Yeah, I understand now thank you.
– Bas bas
Aug 6 at 22:04
1
Yes, but the point of the exercise is to show that the formula for PDF of $N(0,1)$ does have variance $1$.
– Robert Israel
Aug 6 at 23:40