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What is distribution of normal plus minus a constant?
Hypothesis Testing of the normal distributionNormal and standard distributionScaling the normal distribution?Add Chi-Squared Distribution to Normal DistributionLog-normal distributionLog-Likelihood Ratio of signals with multivariate normal distribution.Convolution of normal distribution not equal to product with constant?Marginalization for normal distributionWhat's the distribution of a noncentral chi squared variable plus a constant?Normal distribution governed by a Bernoulli distribution
$begingroup$
$Ysim N(mu, sigma^2)$, then $Y+mu_0=N(mu+mu_0,sigma^2)$ and $nYsim N(mu, (nsigma)^2)$. What about $-Y+mu_0,$ will it be $N(mu+mu_0,(-sigma)^2)$
statistics probability-distributions normal-distribution
asked Dec 12 '17 at 17:29
CoolKidCoolKid
1,260829
$endgroup$
add a comment |
$begingroup$
$Ysim N(mu, sigma^2)$, then $Y+mu_0=N(mu+mu_0,sigma^2)$ and $nYsim N(mu, (nsigma)^2)$. What about $-Y+mu_0,$ will it be $N(mu+mu_0,(-sigma)^2)$
statistics probability-distributions normal-distribution
asked Dec 12 '17 at 17:29
CoolKidCoolKid
1,260829
$endgroup$
2
$begingroup$
Why not $-mu+mu_0$ ?
$endgroup$
– Yves Daoust
Dec 12 '17 at 17:32
$begingroup$
There's a typo. $$nYsim N(nmu, (nsigma)^2)$$ Take $n=-1$.
$endgroup$
– Abishanka Saha
Dec 12 '17 at 17:35
add a comment |
$begingroup$
$Ysim N(mu, sigma^2)$, then $Y+mu_0=N(mu+mu_0,sigma^2)$ and $nYsim N(mu, (nsigma)^2)$. What about $-Y+mu_0,$ will it be $N(mu+mu_0,(-sigma)^2)$
statistics probability-distributions normal-distribution
asked Dec 12 '17 at 17:29
CoolKidCoolKid
1,260829
$endgroup$
$Ysim N(mu, sigma^2)$, then $Y+mu_0=N(mu+mu_0,sigma^2)$ and $nYsim N(mu, (nsigma)^2)$. What about $-Y+mu_0,$ will it be $N(mu+mu_0,(-sigma)^2)$
statistics probability-distributions normal-distribution
statistics probability-distributions normal-distribution
asked Dec 12 '17 at 17:29
CoolKidCoolKid
1,260829
asked Dec 12 '17 at 17:29
CoolKidCoolKid
1,260829
asked Dec 12 '17 at 17:29
CoolKidCoolKid
1,260829
asked Dec 12 '17 at 17:29
CoolKidCoolKid
1,260829
asked Dec 12 '17 at 17:29
CoolKidCoolKid
1,260829
1,260829
2
$begingroup$
Why not $-mu+mu_0$ ?
$endgroup$
– Yves Daoust
Dec 12 '17 at 17:32
$begingroup$
There's a typo. $$nYsim N(nmu, (nsigma)^2)$$ Take $n=-1$.
$endgroup$
– Abishanka Saha
Dec 12 '17 at 17:35
add a comment |
2
$begingroup$
Why not $-mu+mu_0$ ?
$endgroup$
– Yves Daoust
Dec 12 '17 at 17:32
$begingroup$
There's a typo. $$nYsim N(nmu, (nsigma)^2)$$ Take $n=-1$.
$endgroup$
– Abishanka Saha
Dec 12 '17 at 17:35
2
2
$begingroup$
Why not $-mu+mu_0$ ?
$endgroup$
– Yves Daoust
Dec 12 '17 at 17:32
$begingroup$
Why not $-mu+mu_0$ ?
$endgroup$
– Yves Daoust
Dec 12 '17 at 17:32
$begingroup$
There's a typo. $$nYsim N(nmu, (nsigma)^2)$$ Take $n=-1$.
$endgroup$
– Abishanka Saha
Dec 12 '17 at 17:35
$begingroup$
There's a typo. $$nYsim N(nmu, (nsigma)^2)$$ Take $n=-1$.
$endgroup$
– Abishanka Saha
Dec 12 '17 at 17:35
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $Y∼N(mu,sigma^2)$ then $-Y∼N(-mu,sigma^2)$. Thus $−Y+mu_0~N(-mu,sigma^2)$
edited Mar 11 at 11:40
postmortes
2,16531222
answered Mar 11 at 10:01
Dinh HanksDinh Hanks
11
New contributor
Dinh Hanks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
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$begingroup$
If $Y∼N(mu,sigma^2)$ then $-Y∼N(-mu,sigma^2)$. Thus $−Y+mu_0~N(-mu,sigma^2)$
edited Mar 11 at 11:40
postmortes
2,16531222
answered Mar 11 at 10:01
Dinh HanksDinh Hanks
11
New contributor
Dinh Hanks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
If $Y∼N(mu,sigma^2)$ then $-Y∼N(-mu,sigma^2)$. Thus $−Y+mu_0~N(-mu,sigma^2)$
edited Mar 11 at 11:40
postmortes
2,16531222
answered Mar 11 at 10:01
Dinh HanksDinh Hanks
11
New contributor
Dinh Hanks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
If $Y∼N(mu,sigma^2)$ then $-Y∼N(-mu,sigma^2)$. Thus $−Y+mu_0~N(-mu,sigma^2)$
edited Mar 11 at 11:40
postmortes
2,16531222
answered Mar 11 at 10:01
Dinh HanksDinh Hanks
11
New contributor
Dinh Hanks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
If $Y∼N(mu,sigma^2)$ then $-Y∼N(-mu,sigma^2)$. Thus $−Y+mu_0~N(-mu,sigma^2)$
edited Mar 11 at 11:40
postmortes
2,16531222
answered Mar 11 at 10:01
Dinh HanksDinh Hanks
11
New contributor
Dinh Hanks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 11 at 11:40
postmortes
2,16531222
edited Mar 11 at 11:40
postmortes
2,16531222
edited Mar 11 at 11:40
postmortes
2,16531222
2,16531222
answered Mar 11 at 10:01
Dinh HanksDinh Hanks
11
New contributor
Dinh Hanks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Mar 11 at 10:01
Dinh HanksDinh Hanks
11
answered Mar 11 at 10:01
Dinh HanksDinh Hanks
11
11
New contributor
Dinh Hanks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Dinh Hanks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Dinh Hanks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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2
$begingroup$
Why not $-mu+mu_0$ ?
$endgroup$
– Yves Daoust
Dec 12 '17 at 17:32
$begingroup$
There's a typo. $$nYsim N(nmu, (nsigma)^2)$$ Take $n=-1$.
$endgroup$
– Abishanka Saha
Dec 12 '17 at 17:35