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How can $sigma^2$ be derived as a function of $mu$ in a Gaussian pdf?
Prove Variance of a normal distribution is (sigma)^2 (using its moment generating function)Values of the PDF function get extremely highFor a CRV, X, if the pdf is given, can you find the pdf of a Y?What should be the probability density function (PDF) of following vector?PDF of a Gaussian with variance as a function of it's meanDerive PDF for general multivariate GaussianJoint Gaussian PDF Change of CoordinatesQuestions Leading From Application of Orthogonal Change of Coordinates to Transform a General Gaussian PDFMoment Function of Gaussian ProcessExpected value of Normal Lognormal Mixture
$begingroup$
I have a Gaussian pdf defined as
$$f_X(x) =frac1sigmasqrt2piexpleft-frac(x-mu)^22sigma^2right$$
whose $mu = fracd^26D$, where $d$ is distance parameter and $D$ is the diffusion coefficient.
How can $sigma^2$ be derived from $mu$? And will it be a function of $d$ and $D$ as well?
This is what I have come up with.
To my understanding, $$sigma^2=Var(X)$$ $$=E(X^2)-E(X)^2$$
where $E(X)^2 = mu^2 = (fracd^26D)^2$.
How can I proceed from here? How can $=E(X^2)$ be obtained?
or Is this a wrong approach? If so, what should I do?
probability random-variables normal-distribution variance expected-value
asked Mar 15 at 4:55
nashynashnashynash
1168
$endgroup$
add a comment |
$begingroup$
I have a Gaussian pdf defined as
$$f_X(x) =frac1sigmasqrt2piexpleft-frac(x-mu)^22sigma^2right$$
whose $mu = fracd^26D$, where $d$ is distance parameter and $D$ is the diffusion coefficient.
How can $sigma^2$ be derived from $mu$? And will it be a function of $d$ and $D$ as well?
This is what I have come up with.
To my understanding, $$sigma^2=Var(X)$$ $$=E(X^2)-E(X)^2$$
where $E(X)^2 = mu^2 = (fracd^26D)^2$.
How can I proceed from here? How can $=E(X^2)$ be obtained?
or Is this a wrong approach? If so, what should I do?
probability random-variables normal-distribution variance expected-value
asked Mar 15 at 4:55
nashynashnashynash
1168
$endgroup$
$begingroup$
I think using the definition instead of the shortcut will get you somewhere.
$endgroup$
– Sean Roberson
Mar 15 at 4:58
$begingroup$
@SeanRoberson Sorry, but I did not quite understand what you mean.
$endgroup$
– nashynash
Mar 15 at 5:21
$begingroup$
@SeanRoberson Do you mean to use the definition of expectation $E(X)=sumxp(x)$, where $p(x)$ is the PDF?
$endgroup$
– nashynash
Mar 15 at 8:19
$begingroup$
No. I meant $V(X) = E(X-mu)^2$.
$endgroup$
– Sean Roberson
Mar 15 at 8:21
$begingroup$
@SeanRoberson So then $V(X)= Ebig(frac1sigmasqrt2piexpleft-frac(x-mu)^22sigma^2right - mubig)^2$?
$endgroup$
– nashynash
Mar 15 at 8:30
add a comment |
$begingroup$
I have a Gaussian pdf defined as
$$f_X(x) =frac1sigmasqrt2piexpleft-frac(x-mu)^22sigma^2right$$
whose $mu = fracd^26D$, where $d$ is distance parameter and $D$ is the diffusion coefficient.
How can $sigma^2$ be derived from $mu$? And will it be a function of $d$ and $D$ as well?
This is what I have come up with.
To my understanding, $$sigma^2=Var(X)$$ $$=E(X^2)-E(X)^2$$
where $E(X)^2 = mu^2 = (fracd^26D)^2$.
How can I proceed from here? How can $=E(X^2)$ be obtained?
or Is this a wrong approach? If so, what should I do?
probability random-variables normal-distribution variance expected-value
asked Mar 15 at 4:55
nashynashnashynash
1168
$endgroup$
I have a Gaussian pdf defined as
$$f_X(x) =frac1sigmasqrt2piexpleft-frac(x-mu)^22sigma^2right$$
whose $mu = fracd^26D$, where $d$ is distance parameter and $D$ is the diffusion coefficient.
How can $sigma^2$ be derived from $mu$? And will it be a function of $d$ and $D$ as well?
This is what I have come up with.
To my understanding, $$sigma^2=Var(X)$$ $$=E(X^2)-E(X)^2$$
where $E(X)^2 = mu^2 = (fracd^26D)^2$.
How can I proceed from here? How can $=E(X^2)$ be obtained?
or Is this a wrong approach? If so, what should I do?
probability random-variables normal-distribution variance expected-value
probability random-variables normal-distribution variance expected-value
asked Mar 15 at 4:55
nashynashnashynash
1168
asked Mar 15 at 4:55
nashynashnashynash
1168
edited Mar 15 at 4:59
nashynash
asked Mar 15 at 4:55
nashynashnashynash
1168
asked Mar 15 at 4:55
nashynashnashynash
1168
asked Mar 15 at 4:55
nashynashnashynash
1168
1168
$begingroup$
I think using the definition instead of the shortcut will get you somewhere.
$endgroup$
– Sean Roberson
Mar 15 at 4:58
$begingroup$
@SeanRoberson Sorry, but I did not quite understand what you mean.
$endgroup$
– nashynash
Mar 15 at 5:21
$begingroup$
@SeanRoberson Do you mean to use the definition of expectation $E(X)=sumxp(x)$, where $p(x)$ is the PDF?
$endgroup$
– nashynash
Mar 15 at 8:19
$begingroup$
No. I meant $V(X) = E(X-mu)^2$.
$endgroup$
– Sean Roberson
Mar 15 at 8:21
$begingroup$
@SeanRoberson So then $V(X)= Ebig(frac1sigmasqrt2piexpleft-frac(x-mu)^22sigma^2right - mubig)^2$?
$endgroup$
– nashynash
Mar 15 at 8:30
add a comment |
$begingroup$
I think using the definition instead of the shortcut will get you somewhere.
$endgroup$
– Sean Roberson
Mar 15 at 4:58
$begingroup$
@SeanRoberson Sorry, but I did not quite understand what you mean.
$endgroup$
– nashynash
Mar 15 at 5:21
$begingroup$
@SeanRoberson Do you mean to use the definition of expectation $E(X)=sumxp(x)$, where $p(x)$ is the PDF?
$endgroup$
– nashynash
Mar 15 at 8:19
$begingroup$
No. I meant $V(X) = E(X-mu)^2$.
$endgroup$
– Sean Roberson
Mar 15 at 8:21
$begingroup$
@SeanRoberson So then $V(X)= Ebig(frac1sigmasqrt2piexpleft-frac(x-mu)^22sigma^2right - mubig)^2$?
$endgroup$
– nashynash
Mar 15 at 8:30
$begingroup$
I think using the definition instead of the shortcut will get you somewhere.
$endgroup$
– Sean Roberson
Mar 15 at 4:58
$begingroup$
I think using the definition instead of the shortcut will get you somewhere.
$endgroup$
– Sean Roberson
Mar 15 at 4:58
$begingroup$
@SeanRoberson Sorry, but I did not quite understand what you mean.
$endgroup$
– nashynash
Mar 15 at 5:21
$begingroup$
@SeanRoberson Sorry, but I did not quite understand what you mean.
$endgroup$
– nashynash
Mar 15 at 5:21
$begingroup$
@SeanRoberson Do you mean to use the definition of expectation $E(X)=sumxp(x)$, where $p(x)$ is the PDF?
$endgroup$
– nashynash
Mar 15 at 8:19
$begingroup$
@SeanRoberson Do you mean to use the definition of expectation $E(X)=sumxp(x)$, where $p(x)$ is the PDF?
$endgroup$
– nashynash
Mar 15 at 8:19
$begingroup$
No. I meant $V(X) = E(X-mu)^2$.
$endgroup$
– Sean Roberson
Mar 15 at 8:21
$begingroup$
No. I meant $V(X) = E(X-mu)^2$.
$endgroup$
– Sean Roberson
Mar 15 at 8:21
$begingroup$
@SeanRoberson So then $V(X)= Ebig(frac1sigmasqrt2piexpleft-frac(x-mu)^22sigma^2right - mubig)^2$?
$endgroup$
– nashynash
Mar 15 at 8:30
$begingroup$
@SeanRoberson So then $V(X)= Ebig(frac1sigmasqrt2piexpleft-frac(x-mu)^22sigma^2right - mubig)^2$?
$endgroup$
– nashynash
Mar 15 at 8:30
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The mean and variance parameters in a normal distribution are both free parameters, so setting the mean parameter does not impose any restrictions on the variance parameter. Hence, without more information on any outside restrictions, there is no way to "derive" the variance from the mean. It appears in your question that you have a formula for the mean parameter that comes from exogenous variables. Unless you also have a formula for the variance from these exogenous variables (which is not specified), you are out of luck.
answered Mar 15 at 5:00
BenBen
1,840215
$endgroup$
$begingroup$
Yes, the mean was obtained by differentiating another pdf and set to 0.
$endgroup$
– nashynash
Mar 15 at 5:07
add a comment |
Your Answer
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The mean and variance parameters in a normal distribution are both free parameters, so setting the mean parameter does not impose any restrictions on the variance parameter. Hence, without more information on any outside restrictions, there is no way to "derive" the variance from the mean. It appears in your question that you have a formula for the mean parameter that comes from exogenous variables. Unless you also have a formula for the variance from these exogenous variables (which is not specified), you are out of luck.
answered Mar 15 at 5:00
BenBen
1,840215
$endgroup$
$begingroup$
Yes, the mean was obtained by differentiating another pdf and set to 0.
$endgroup$
– nashynash
Mar 15 at 5:07
add a comment |
$begingroup$
The mean and variance parameters in a normal distribution are both free parameters, so setting the mean parameter does not impose any restrictions on the variance parameter. Hence, without more information on any outside restrictions, there is no way to "derive" the variance from the mean. It appears in your question that you have a formula for the mean parameter that comes from exogenous variables. Unless you also have a formula for the variance from these exogenous variables (which is not specified), you are out of luck.
answered Mar 15 at 5:00
BenBen
1,840215
$endgroup$
$begingroup$
Yes, the mean was obtained by differentiating another pdf and set to 0.
$endgroup$
– nashynash
Mar 15 at 5:07
add a comment |
$begingroup$
The mean and variance parameters in a normal distribution are both free parameters, so setting the mean parameter does not impose any restrictions on the variance parameter. Hence, without more information on any outside restrictions, there is no way to "derive" the variance from the mean. It appears in your question that you have a formula for the mean parameter that comes from exogenous variables. Unless you also have a formula for the variance from these exogenous variables (which is not specified), you are out of luck.
answered Mar 15 at 5:00
BenBen
1,840215
$endgroup$
The mean and variance parameters in a normal distribution are both free parameters, so setting the mean parameter does not impose any restrictions on the variance parameter. Hence, without more information on any outside restrictions, there is no way to "derive" the variance from the mean. It appears in your question that you have a formula for the mean parameter that comes from exogenous variables. Unless you also have a formula for the variance from these exogenous variables (which is not specified), you are out of luck.
answered Mar 15 at 5:00
BenBen
1,840215
answered Mar 15 at 5:00
BenBen
1,840215
answered Mar 15 at 5:00
BenBen
1,840215
answered Mar 15 at 5:00
BenBen
1,840215
1,840215
$begingroup$
Yes, the mean was obtained by differentiating another pdf and set to 0.
$endgroup$
– nashynash
Mar 15 at 5:07
add a comment |
$begingroup$
Yes, the mean was obtained by differentiating another pdf and set to 0.
$endgroup$
– nashynash
Mar 15 at 5:07
$begingroup$
Yes, the mean was obtained by differentiating another pdf and set to 0.
$endgroup$
– nashynash
Mar 15 at 5:07
$begingroup$
Yes, the mean was obtained by differentiating another pdf and set to 0.
$endgroup$
– nashynash
Mar 15 at 5:07
add a comment |
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$begingroup$
I think using the definition instead of the shortcut will get you somewhere.
$endgroup$
– Sean Roberson
Mar 15 at 4:58
$begingroup$
@SeanRoberson Sorry, but I did not quite understand what you mean.
$endgroup$
– nashynash
Mar 15 at 5:21
$begingroup$
@SeanRoberson Do you mean to use the definition of expectation $E(X)=sumxp(x)$, where $p(x)$ is the PDF?
$endgroup$
– nashynash
Mar 15 at 8:19
$begingroup$
No. I meant $V(X) = E(X-mu)^2$.
$endgroup$
– Sean Roberson
Mar 15 at 8:21
$begingroup$
@SeanRoberson So then $V(X)= Ebig(frac1sigmasqrt2piexpleft-frac(x-mu)^22sigma^2right - mubig)^2$?
$endgroup$
– nashynash
Mar 15 at 8:30