Mixing
Exponential family canonical form
$begingroup$
I cannot figure this myself - suppose I have some distribution $f(y; theta)$ from the exponential family. By Wikipedia (https://en.wikipedia.org/wiki/Exponential_family) let us assume this form:
$
f(y; theta) = h(y)expbig[eta(theta)T(y) - A(theta)big]
$
The wiki defines the canonical form as
If $eta(theta) = theta$, then the exponential family is said to be
in canonical form
Then, it adds the part I do not get:
By defining a transformed parameter $eta(theta) = theta$, it is
always possible to convert an exponential family to canonical form
I do not get what they do with parameters $theta$. For example, let us have the Normal with unknown mean $mu$ AND unknown variance $sigma^2$. There is immediately example in wiki, where they define:
$eta = big(frac{mu}{sigma^2}, -frac{1}{2sigma^2}big)$
and then conclude: then our form is canonical. Why is that? They initially define the normal as $f(y; mu, sigma)$! According to the previous definition, it is clear that $eta(theta) neq (mu, sigma^2)$! Why do they call it a canonical form then?
Actually this can be shown almost for all distributions when written in canonical form, their parameter vector $eta$ is clearly not equal to the parameters in the definition.
Could someone clarify please.
probability-distributions normal-distribution
asked Jan 31 at 18:11
JohnJohn
2531314
$endgroup$
add a comment |
$begingroup$
I cannot figure this myself - suppose I have some distribution $f(y; theta)$ from the exponential family. By Wikipedia (https://en.wikipedia.org/wiki/Exponential_family) let us assume this form:
$
f(y; theta) = h(y)expbig[eta(theta)T(y) - A(theta)big]
$
The wiki defines the canonical form as
If $eta(theta) = theta$, then the exponential family is said to be
in canonical form
Then, it adds the part I do not get:
By defining a transformed parameter $eta(theta) = theta$, it is
always possible to convert an exponential family to canonical form
I do not get what they do with parameters $theta$. For example, let us have the Normal with unknown mean $mu$ AND unknown variance $sigma^2$. There is immediately example in wiki, where they define:
$eta = big(frac{mu}{sigma^2}, -frac{1}{2sigma^2}big)$
and then conclude: then our form is canonical. Why is that? They initially define the normal as $f(y; mu, sigma)$! According to the previous definition, it is clear that $eta(theta) neq (mu, sigma^2)$! Why do they call it a canonical form then?
Actually this can be shown almost for all distributions when written in canonical form, their parameter vector $eta$ is clearly not equal to the parameters in the definition.
Could someone clarify please.
probability-distributions normal-distribution
asked Jan 31 at 18:11
JohnJohn
2531314
$endgroup$
$begingroup$
It's a typo. They define $theta=(cdot,cdot)^{top}$ (look at the next lines where they define $A(eta)$).
$endgroup$
– d.k.o.
Jan 31 at 18:44
$begingroup$
Then does it mean that the real density should be defined as $f(y; frac{mu}{sigma^2}, -frac{1}{2sigma^2})$? They kinda no where mention that parameters can be swapped like that (apart from second quote "By defining a transformed parameter..." - which i did not get - is it what it implies?).
$endgroup$
– John
Jan 31 at 18:56
add a comment |
$begingroup$
I cannot figure this myself - suppose I have some distribution $f(y; theta)$ from the exponential family. By Wikipedia (https://en.wikipedia.org/wiki/Exponential_family) let us assume this form:
$
f(y; theta) = h(y)expbig[eta(theta)T(y) - A(theta)big]
$
The wiki defines the canonical form as
If $eta(theta) = theta$, then the exponential family is said to be
in canonical form
Then, it adds the part I do not get:
By defining a transformed parameter $eta(theta) = theta$, it is
always possible to convert an exponential family to canonical form
I do not get what they do with parameters $theta$. For example, let us have the Normal with unknown mean $mu$ AND unknown variance $sigma^2$. There is immediately example in wiki, where they define:
$eta = big(frac{mu}{sigma^2}, -frac{1}{2sigma^2}big)$
and then conclude: then our form is canonical. Why is that? They initially define the normal as $f(y; mu, sigma)$! According to the previous definition, it is clear that $eta(theta) neq (mu, sigma^2)$! Why do they call it a canonical form then?
Actually this can be shown almost for all distributions when written in canonical form, their parameter vector $eta$ is clearly not equal to the parameters in the definition.
Could someone clarify please.
probability-distributions normal-distribution
asked Jan 31 at 18:11
JohnJohn
2531314
$endgroup$
I cannot figure this myself - suppose I have some distribution $f(y; theta)$ from the exponential family. By Wikipedia (https://en.wikipedia.org/wiki/Exponential_family) let us assume this form:
$
f(y; theta) = h(y)expbig[eta(theta)T(y) - A(theta)big]
$
The wiki defines the canonical form as
If $eta(theta) = theta$, then the exponential family is said to be
in canonical form
Then, it adds the part I do not get:
By defining a transformed parameter $eta(theta) = theta$, it is
always possible to convert an exponential family to canonical form
I do not get what they do with parameters $theta$. For example, let us have the Normal with unknown mean $mu$ AND unknown variance $sigma^2$. There is immediately example in wiki, where they define:
$eta = big(frac{mu}{sigma^2}, -frac{1}{2sigma^2}big)$
and then conclude: then our form is canonical. Why is that? They initially define the normal as $f(y; mu, sigma)$! According to the previous definition, it is clear that $eta(theta) neq (mu, sigma^2)$! Why do they call it a canonical form then?
Actually this can be shown almost for all distributions when written in canonical form, their parameter vector $eta$ is clearly not equal to the parameters in the definition.
Could someone clarify please.
probability-distributions normal-distribution
probability-distributions normal-distribution
asked Jan 31 at 18:11
JohnJohn
2531314
asked Jan 31 at 18:11
JohnJohn
2531314
asked Jan 31 at 18:11
JohnJohn
2531314
asked Jan 31 at 18:11
JohnJohn
2531314
asked Jan 31 at 18:11
JohnJohn
2531314
2531314
$begingroup$
It's a typo. They define $theta=(cdot,cdot)^{top}$ (look at the next lines where they define $A(eta)$).
$endgroup$
– d.k.o.
Jan 31 at 18:44
$begingroup$
Then does it mean that the real density should be defined as $f(y; frac{mu}{sigma^2}, -frac{1}{2sigma^2})$? They kinda no where mention that parameters can be swapped like that (apart from second quote "By defining a transformed parameter..." - which i did not get - is it what it implies?).
$endgroup$
– John
Jan 31 at 18:56
add a comment |
$begingroup$
It's a typo. They define $theta=(cdot,cdot)^{top}$ (look at the next lines where they define $A(eta)$).
$endgroup$
– d.k.o.
Jan 31 at 18:44
$begingroup$
Then does it mean that the real density should be defined as $f(y; frac{mu}{sigma^2}, -frac{1}{2sigma^2})$? They kinda no where mention that parameters can be swapped like that (apart from second quote "By defining a transformed parameter..." - which i did not get - is it what it implies?).
$endgroup$
– John
Jan 31 at 18:56
$begingroup$
It's a typo. They define $theta=(cdot,cdot)^{top}$ (look at the next lines where they define $A(eta)$).
$endgroup$
– d.k.o.
Jan 31 at 18:44
$begingroup$
It's a typo. They define $theta=(cdot,cdot)^{top}$ (look at the next lines where they define $A(eta)$).
$endgroup$
– d.k.o.
Jan 31 at 18:44
$begingroup$
Then does it mean that the real density should be defined as $f(y; frac{mu}{sigma^2}, -frac{1}{2sigma^2})$? They kinda no where mention that parameters can be swapped like that (apart from second quote "By defining a transformed parameter..." - which i did not get - is it what it implies?).
$endgroup$
– John
Jan 31 at 18:56
$begingroup$
Then does it mean that the real density should be defined as $f(y; frac{mu}{sigma^2}, -frac{1}{2sigma^2})$? They kinda no where mention that parameters can be swapped like that (apart from second quote "By defining a transformed parameter..." - which i did not get - is it what it implies?).
$endgroup$
– John
Jan 31 at 18:56
add a comment |
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$begingroup$
It's a typo. They define $theta=(cdot,cdot)^{top}$ (look at the next lines where they define $A(eta)$).
$endgroup$
– d.k.o.
Jan 31 at 18:44
$begingroup$
Then does it mean that the real density should be defined as $f(y; frac{mu}{sigma^2}, -frac{1}{2sigma^2})$? They kinda no where mention that parameters can be swapped like that (apart from second quote "By defining a transformed parameter..." - which i did not get - is it what it implies?).
$endgroup$
– John
Jan 31 at 18:56