Mixing
Using change of variables to transform density functions
$begingroup$
I'm was working on some exercises on statistical inference and came across a question I could not solve. After a while I decided to take a look at the solution to hopefully understand the problem better.
Situation: Let $X_{1} dots X_{n}$ be independent and identically distributed random variables, each with the $N(mu,sigma^2)$ distribution. We have that $mu inmathbb{R}$ is a known parameter and $sigma^2$ is unknown. We consider $T=frac{sum^{n}_{i=i}(X_{i}-bar{X})^{2}}{n-1}$ and can use the fact that $frac{(n-1)}{sigma^2} Tsim Gamma(frac{n-1}{2},frac{1}{2})$.
My problem: In the solution is the following line:
Since $T=cY$, where $c=frac{sigma^2}{n-1}$ and $Ysim
Gamma(frac{n-1}{2},frac{1}{2})$, using a change of variables yields
that $f_{T}(x)=c^{-1}f_{Y}(x/c)$.
How does this change of variables work? How is it computed and why is it even necessary? Is it because the distribution of $T$ itself isn't explicitly given, but rather the distribution of $cT$? Any help is appreciated.
statistics normal-distribution statistical-inference change-of-variable gamma-distribution
asked Jan 3 at 15:52
S. CrimS. Crim
35612
$endgroup$
add a comment |
$begingroup$
I'm was working on some exercises on statistical inference and came across a question I could not solve. After a while I decided to take a look at the solution to hopefully understand the problem better.
Situation: Let $X_{1} dots X_{n}$ be independent and identically distributed random variables, each with the $N(mu,sigma^2)$ distribution. We have that $mu inmathbb{R}$ is a known parameter and $sigma^2$ is unknown. We consider $T=frac{sum^{n}_{i=i}(X_{i}-bar{X})^{2}}{n-1}$ and can use the fact that $frac{(n-1)}{sigma^2} Tsim Gamma(frac{n-1}{2},frac{1}{2})$.
My problem: In the solution is the following line:
Since $T=cY$, where $c=frac{sigma^2}{n-1}$ and $Ysim
Gamma(frac{n-1}{2},frac{1}{2})$, using a change of variables yields
that $f_{T}(x)=c^{-1}f_{Y}(x/c)$.
How does this change of variables work? How is it computed and why is it even necessary? Is it because the distribution of $T$ itself isn't explicitly given, but rather the distribution of $cT$? Any help is appreciated.
statistics normal-distribution statistical-inference change-of-variable gamma-distribution
asked Jan 3 at 15:52
S. CrimS. Crim
35612
$endgroup$
add a comment |
$begingroup$
I'm was working on some exercises on statistical inference and came across a question I could not solve. After a while I decided to take a look at the solution to hopefully understand the problem better.
Situation: Let $X_{1} dots X_{n}$ be independent and identically distributed random variables, each with the $N(mu,sigma^2)$ distribution. We have that $mu inmathbb{R}$ is a known parameter and $sigma^2$ is unknown. We consider $T=frac{sum^{n}_{i=i}(X_{i}-bar{X})^{2}}{n-1}$ and can use the fact that $frac{(n-1)}{sigma^2} Tsim Gamma(frac{n-1}{2},frac{1}{2})$.
My problem: In the solution is the following line:
Since $T=cY$, where $c=frac{sigma^2}{n-1}$ and $Ysim
Gamma(frac{n-1}{2},frac{1}{2})$, using a change of variables yields
that $f_{T}(x)=c^{-1}f_{Y}(x/c)$.
How does this change of variables work? How is it computed and why is it even necessary? Is it because the distribution of $T$ itself isn't explicitly given, but rather the distribution of $cT$? Any help is appreciated.
statistics normal-distribution statistical-inference change-of-variable gamma-distribution
asked Jan 3 at 15:52
S. CrimS. Crim
35612
$endgroup$
I'm was working on some exercises on statistical inference and came across a question I could not solve. After a while I decided to take a look at the solution to hopefully understand the problem better.
Situation: Let $X_{1} dots X_{n}$ be independent and identically distributed random variables, each with the $N(mu,sigma^2)$ distribution. We have that $mu inmathbb{R}$ is a known parameter and $sigma^2$ is unknown. We consider $T=frac{sum^{n}_{i=i}(X_{i}-bar{X})^{2}}{n-1}$ and can use the fact that $frac{(n-1)}{sigma^2} Tsim Gamma(frac{n-1}{2},frac{1}{2})$.
My problem: In the solution is the following line:
Since $T=cY$, where $c=frac{sigma^2}{n-1}$ and $Ysim
Gamma(frac{n-1}{2},frac{1}{2})$, using a change of variables yields
that $f_{T}(x)=c^{-1}f_{Y}(x/c)$.
How does this change of variables work? How is it computed and why is it even necessary? Is it because the distribution of $T$ itself isn't explicitly given, but rather the distribution of $cT$? Any help is appreciated.
statistics normal-distribution statistical-inference change-of-variable gamma-distribution
statistics normal-distribution statistical-inference change-of-variable gamma-distribution
asked Jan 3 at 15:52
S. CrimS. Crim
35612
asked Jan 3 at 15:52
S. CrimS. Crim
35612
edited Jan 3 at 15:59
S. Crim
asked Jan 3 at 15:52
S. CrimS. Crim
35612
asked Jan 3 at 15:52
S. CrimS. Crim
35612
asked Jan 3 at 15:52
S. CrimS. Crim
35612
35612
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can see this as a simple change of variable in an integral:
$$ P(T < t) = P(cY < t) = P(Y < frac{t}{c}) =int_{-infty}^{frac{t}{c}} f^Y(x)dx $$
define $ u = cx Rightarrow " du = c dx" iff dx "=" frac{1}{c}du.$
This yields
$$ P(T< t) = int_{-infty}^{frac{t}{c}} f^Y(x)dx = int_{-infty}^t frac{1}{c} f^Y(u/c) du .$$
And $c^{-1}f^Y(xc^{-1})$ is a probability density function for $T$.
answered Jan 3 at 16:09
DigitalisDigitalis
528216
$endgroup$
$begingroup$
The originial exercise was about computing the Fisher information based on $T$, i.e. $I_{T}(sigma^2)$. Is this step necessary because we only know the distribution of $cT$ and not $T$? Or is there another reason for having to change the variables?
$endgroup$
– S. Crim
Jan 3 at 16:16
$begingroup$
I'm afraid I'm not able to help you with that, I'm not very gifted when it comes to computing Fischer information. I'm also not familiar with the concept of calculating the fisher information based on a given statistic $T$. :(
$endgroup$
– Digitalis
Jan 3 at 16:55
$begingroup$
Fair enough, thanks for the help anyway!
$endgroup$
– S. Crim
Jan 3 at 17:16
$begingroup$
Hope I can revive this question. Shouldn't there be a $t/c$ as the upper boundary of your last integral in the last line? If not, why did it turn into a $t$?
$endgroup$
– S. Crim
2 days ago
1
$begingroup$
@S.Crim By using the substitution $u = cx$ then $u$ is a function of $x$ and we can write it as $u(x) = cx$. Then when $x$ approches $-infty$ then so does $u$ since $c > 0$. When $x = frac{t}{c}$ then $u = u(frac{t}{c}) = c cdot frac{t}{c} = t.$
$endgroup$
– Digitalis
yesterday
add a comment |
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1 Answer
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1 Answer
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active
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votes
$begingroup$
You can see this as a simple change of variable in an integral:
$$ P(T < t) = P(cY < t) = P(Y < frac{t}{c}) =int_{-infty}^{frac{t}{c}} f^Y(x)dx $$
define $ u = cx Rightarrow " du = c dx" iff dx "=" frac{1}{c}du.$
This yields
$$ P(T< t) = int_{-infty}^{frac{t}{c}} f^Y(x)dx = int_{-infty}^t frac{1}{c} f^Y(u/c) du .$$
And $c^{-1}f^Y(xc^{-1})$ is a probability density function for $T$.
answered Jan 3 at 16:09
DigitalisDigitalis
528216
$endgroup$
$begingroup$
The originial exercise was about computing the Fisher information based on $T$, i.e. $I_{T}(sigma^2)$. Is this step necessary because we only know the distribution of $cT$ and not $T$? Or is there another reason for having to change the variables?
$endgroup$
– S. Crim
Jan 3 at 16:16
$begingroup$
I'm afraid I'm not able to help you with that, I'm not very gifted when it comes to computing Fischer information. I'm also not familiar with the concept of calculating the fisher information based on a given statistic $T$. :(
$endgroup$
– Digitalis
Jan 3 at 16:55
$begingroup$
Fair enough, thanks for the help anyway!
$endgroup$
– S. Crim
Jan 3 at 17:16
$begingroup$
Hope I can revive this question. Shouldn't there be a $t/c$ as the upper boundary of your last integral in the last line? If not, why did it turn into a $t$?
$endgroup$
– S. Crim
2 days ago
1
$begingroup$
@S.Crim By using the substitution $u = cx$ then $u$ is a function of $x$ and we can write it as $u(x) = cx$. Then when $x$ approches $-infty$ then so does $u$ since $c > 0$. When $x = frac{t}{c}$ then $u = u(frac{t}{c}) = c cdot frac{t}{c} = t.$
$endgroup$
– Digitalis
yesterday
add a comment |
$begingroup$
You can see this as a simple change of variable in an integral:
$$ P(T < t) = P(cY < t) = P(Y < frac{t}{c}) =int_{-infty}^{frac{t}{c}} f^Y(x)dx $$
define $ u = cx Rightarrow " du = c dx" iff dx "=" frac{1}{c}du.$
This yields
$$ P(T< t) = int_{-infty}^{frac{t}{c}} f^Y(x)dx = int_{-infty}^t frac{1}{c} f^Y(u/c) du .$$
And $c^{-1}f^Y(xc^{-1})$ is a probability density function for $T$.
answered Jan 3 at 16:09
DigitalisDigitalis
528216
$endgroup$
$begingroup$
The originial exercise was about computing the Fisher information based on $T$, i.e. $I_{T}(sigma^2)$. Is this step necessary because we only know the distribution of $cT$ and not $T$? Or is there another reason for having to change the variables?
$endgroup$
– S. Crim
Jan 3 at 16:16
$begingroup$
I'm afraid I'm not able to help you with that, I'm not very gifted when it comes to computing Fischer information. I'm also not familiar with the concept of calculating the fisher information based on a given statistic $T$. :(
$endgroup$
– Digitalis
Jan 3 at 16:55
$begingroup$
Fair enough, thanks for the help anyway!
$endgroup$
– S. Crim
Jan 3 at 17:16
$begingroup$
Hope I can revive this question. Shouldn't there be a $t/c$ as the upper boundary of your last integral in the last line? If not, why did it turn into a $t$?
$endgroup$
– S. Crim
2 days ago
1
$begingroup$
@S.Crim By using the substitution $u = cx$ then $u$ is a function of $x$ and we can write it as $u(x) = cx$. Then when $x$ approches $-infty$ then so does $u$ since $c > 0$. When $x = frac{t}{c}$ then $u = u(frac{t}{c}) = c cdot frac{t}{c} = t.$
$endgroup$
– Digitalis
yesterday
add a comment |
$begingroup$
You can see this as a simple change of variable in an integral:
$$ P(T < t) = P(cY < t) = P(Y < frac{t}{c}) =int_{-infty}^{frac{t}{c}} f^Y(x)dx $$
define $ u = cx Rightarrow " du = c dx" iff dx "=" frac{1}{c}du.$
This yields
$$ P(T< t) = int_{-infty}^{frac{t}{c}} f^Y(x)dx = int_{-infty}^t frac{1}{c} f^Y(u/c) du .$$
And $c^{-1}f^Y(xc^{-1})$ is a probability density function for $T$.
answered Jan 3 at 16:09
DigitalisDigitalis
528216
$endgroup$
You can see this as a simple change of variable in an integral:
$$ P(T < t) = P(cY < t) = P(Y < frac{t}{c}) =int_{-infty}^{frac{t}{c}} f^Y(x)dx $$
define $ u = cx Rightarrow " du = c dx" iff dx "=" frac{1}{c}du.$
This yields
$$ P(T< t) = int_{-infty}^{frac{t}{c}} f^Y(x)dx = int_{-infty}^t frac{1}{c} f^Y(u/c) du .$$
And $c^{-1}f^Y(xc^{-1})$ is a probability density function for $T$.
answered Jan 3 at 16:09
DigitalisDigitalis
528216
answered Jan 3 at 16:09
DigitalisDigitalis
528216
answered Jan 3 at 16:09
DigitalisDigitalis
528216
answered Jan 3 at 16:09
DigitalisDigitalis
528216
528216
$begingroup$
The originial exercise was about computing the Fisher information based on $T$, i.e. $I_{T}(sigma^2)$. Is this step necessary because we only know the distribution of $cT$ and not $T$? Or is there another reason for having to change the variables?
$endgroup$
– S. Crim
Jan 3 at 16:16
$begingroup$
I'm afraid I'm not able to help you with that, I'm not very gifted when it comes to computing Fischer information. I'm also not familiar with the concept of calculating the fisher information based on a given statistic $T$. :(
$endgroup$
– Digitalis
Jan 3 at 16:55
$begingroup$
Fair enough, thanks for the help anyway!
$endgroup$
– S. Crim
Jan 3 at 17:16
$begingroup$
Hope I can revive this question. Shouldn't there be a $t/c$ as the upper boundary of your last integral in the last line? If not, why did it turn into a $t$?
$endgroup$
– S. Crim
2 days ago
1
$begingroup$
@S.Crim By using the substitution $u = cx$ then $u$ is a function of $x$ and we can write it as $u(x) = cx$. Then when $x$ approches $-infty$ then so does $u$ since $c > 0$. When $x = frac{t}{c}$ then $u = u(frac{t}{c}) = c cdot frac{t}{c} = t.$
$endgroup$
– Digitalis
yesterday
add a comment |
$begingroup$
The originial exercise was about computing the Fisher information based on $T$, i.e. $I_{T}(sigma^2)$. Is this step necessary because we only know the distribution of $cT$ and not $T$? Or is there another reason for having to change the variables?
$endgroup$
– S. Crim
Jan 3 at 16:16
$begingroup$
I'm afraid I'm not able to help you with that, I'm not very gifted when it comes to computing Fischer information. I'm also not familiar with the concept of calculating the fisher information based on a given statistic $T$. :(
$endgroup$
– Digitalis
Jan 3 at 16:55
$begingroup$
Fair enough, thanks for the help anyway!
$endgroup$
– S. Crim
Jan 3 at 17:16
$begingroup$
Hope I can revive this question. Shouldn't there be a $t/c$ as the upper boundary of your last integral in the last line? If not, why did it turn into a $t$?
$endgroup$
– S. Crim
2 days ago
1
$begingroup$
@S.Crim By using the substitution $u = cx$ then $u$ is a function of $x$ and we can write it as $u(x) = cx$. Then when $x$ approches $-infty$ then so does $u$ since $c > 0$. When $x = frac{t}{c}$ then $u = u(frac{t}{c}) = c cdot frac{t}{c} = t.$
$endgroup$
– Digitalis
yesterday
$begingroup$
The originial exercise was about computing the Fisher information based on $T$, i.e. $I_{T}(sigma^2)$. Is this step necessary because we only know the distribution of $cT$ and not $T$? Or is there another reason for having to change the variables?
$endgroup$
– S. Crim
Jan 3 at 16:16
$begingroup$
The originial exercise was about computing the Fisher information based on $T$, i.e. $I_{T}(sigma^2)$. Is this step necessary because we only know the distribution of $cT$ and not $T$? Or is there another reason for having to change the variables?
$endgroup$
– S. Crim
Jan 3 at 16:16
$begingroup$
I'm afraid I'm not able to help you with that, I'm not very gifted when it comes to computing Fischer information. I'm also not familiar with the concept of calculating the fisher information based on a given statistic $T$. :(
$endgroup$
– Digitalis
Jan 3 at 16:55
$begingroup$
I'm afraid I'm not able to help you with that, I'm not very gifted when it comes to computing Fischer information. I'm also not familiar with the concept of calculating the fisher information based on a given statistic $T$. :(
$endgroup$
– Digitalis
Jan 3 at 16:55
$begingroup$
Fair enough, thanks for the help anyway!
$endgroup$
– S. Crim
Jan 3 at 17:16
$begingroup$
Fair enough, thanks for the help anyway!
$endgroup$
– S. Crim
Jan 3 at 17:16
$begingroup$
Hope I can revive this question. Shouldn't there be a $t/c$ as the upper boundary of your last integral in the last line? If not, why did it turn into a $t$?
$endgroup$
– S. Crim
2 days ago
$begingroup$
Hope I can revive this question. Shouldn't there be a $t/c$ as the upper boundary of your last integral in the last line? If not, why did it turn into a $t$?
$endgroup$
– S. Crim
2 days ago
1
1
$begingroup$
@S.Crim By using the substitution $u = cx$ then $u$ is a function of $x$ and we can write it as $u(x) = cx$. Then when $x$ approches $-infty$ then so does $u$ since $c > 0$. When $x = frac{t}{c}$ then $u = u(frac{t}{c}) = c cdot frac{t}{c} = t.$
$endgroup$
– Digitalis
yesterday
$begingroup$
@S.Crim By using the substitution $u = cx$ then $u$ is a function of $x$ and we can write it as $u(x) = cx$. Then when $x$ approches $-infty$ then so does $u$ since $c > 0$. When $x = frac{t}{c}$ then $u = u(frac{t}{c}) = c cdot frac{t}{c} = t.$
$endgroup$
– Digitalis
yesterday
add a comment |
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