Why can we treat MGF in this way
Multi tool use
$begingroup$
For the standard proof that if $Z sim N(0,1)$ than $Z^{2} sim chi^{2}_{1}$
We write
$$M_{Z^{2}}(t)=int_{mathbb{R}}exp(tz^{2})frac{1}{sqrt{2pi}}exp(frac{-z^{2}}{2}) dz$$
That is , we use the standard normal density to compute the moment generating function.
Why do we not need to use the density of $Z^{2}$ and instead are able to use the density of just $Z$, I assume because it is a continuous one to one function?
mgf
asked 5 hours ago
QualityQuality
24919
$endgroup$
add a comment |
$begingroup$
For the standard proof that if $Z sim N(0,1)$ than $Z^{2} sim chi^{2}_{1}$
We write
$$M_{Z^{2}}(t)=int_{mathbb{R}}exp(tz^{2})frac{1}{sqrt{2pi}}exp(frac{-z^{2}}{2}) dz$$
That is , we use the standard normal density to compute the moment generating function.
Why do we not need to use the density of $Z^{2}$ and instead are able to use the density of just $Z$, I assume because it is a continuous one to one function?
mgf
asked 5 hours ago
QualityQuality
24919
$endgroup$
add a comment |
$begingroup$
For the standard proof that if $Z sim N(0,1)$ than $Z^{2} sim chi^{2}_{1}$
We write
$$M_{Z^{2}}(t)=int_{mathbb{R}}exp(tz^{2})frac{1}{sqrt{2pi}}exp(frac{-z^{2}}{2}) dz$$
That is , we use the standard normal density to compute the moment generating function.
Why do we not need to use the density of $Z^{2}$ and instead are able to use the density of just $Z$, I assume because it is a continuous one to one function?
mgf
asked 5 hours ago
QualityQuality
24919
$endgroup$
For the standard proof that if $Z sim N(0,1)$ than $Z^{2} sim chi^{2}_{1}$
We write
$$M_{Z^{2}}(t)=int_{mathbb{R}}exp(tz^{2})frac{1}{sqrt{2pi}}exp(frac{-z^{2}}{2}) dz$$
That is , we use the standard normal density to compute the moment generating function.
Why do we not need to use the density of $Z^{2}$ and instead are able to use the density of just $Z$, I assume because it is a continuous one to one function?
mgf
mgf
asked 5 hours ago
QualityQuality
24919
asked 5 hours ago
QualityQuality
24919
asked 5 hours ago
QualityQuality
24919
asked 5 hours ago
QualityQuality
24919
asked 5 hours ago
QualityQuality
24919
24919
add a comment |
add a comment |
1 Answer
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$begingroup$
In principle, you don't know the density of $Z$, so you can't use it.
What's being done when you use the density of $X$ is an application of the law of the unconscious statistician.
answered 4 hours ago
Lucas FariasLucas Farias
469417
$endgroup$
add a comment |
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1 Answer
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$begingroup$
In principle, you don't know the density of $Z$, so you can't use it.
What's being done when you use the density of $X$ is an application of the law of the unconscious statistician.
answered 4 hours ago
Lucas FariasLucas Farias
469417
$endgroup$
add a comment |
$begingroup$
In principle, you don't know the density of $Z$, so you can't use it.
What's being done when you use the density of $X$ is an application of the law of the unconscious statistician.
answered 4 hours ago
Lucas FariasLucas Farias
469417
$endgroup$
add a comment |
$begingroup$
In principle, you don't know the density of $Z$, so you can't use it.
What's being done when you use the density of $X$ is an application of the law of the unconscious statistician.
answered 4 hours ago
Lucas FariasLucas Farias
469417
$endgroup$
In principle, you don't know the density of $Z$, so you can't use it.
What's being done when you use the density of $X$ is an application of the law of the unconscious statistician.
answered 4 hours ago
Lucas FariasLucas Farias
469417
answered 4 hours ago
Lucas FariasLucas Farias
469417
answered 4 hours ago
Lucas FariasLucas Farias
469417
answered 4 hours ago
Lucas FariasLucas Farias
469417
469417
add a comment |
add a comment |
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