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Deriving the canonical link for a binomial distribution
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I define an exponential dispersion family as any distribution whose PMF/PDF is
$$f(y mid boldsymboltheta) = expleftphi[ytheta - b(theta)] + c(y, phi) righttext, y in Omega$$
where $Omega$ is in the support of a random variable $Y$ in the family.
Suppose $Y_1, dots, Y_m$ are independent and binomially distributed ($n$ trials, success probability $p_i$). I've already shown that the Binomial distribution satisfies the above, with
$$beginalign
phi &= 1 \
theta_i &= logleft(dfracp_i1-p_i right) \
b(theta_i) &= nlogleft(dfrac11-p_iright) \
c(phi, y_i) &= logbinomny_itext.
endalign$$
After some work, I showed that, as a function of $theta_i$,
$$b(theta_i) = nlog(e^theta_i + 1)$$
(this is consistent with what I found at http://sfb649.wiwi.hu-berlin.de/fedc_homepage/xplore/tutorials/xlghtmlnode38.html)
and I understand that
$$mu_i = b^prime(theta_i) = n cdot dfrace^theta_ie^theta_i+1text.$$
I also understand that what we need to do is solve for $theta_i$ in the above, and the canonical link function would be $g(mu_i) = theta_i$ according to above. But one thing bothers me: when I run the above through WolframAlpha, I obtain
$$g(mu_i) = theta_i = logleft( dfracmu_in-mu_iright)text.$$
Every source I've seen says that the $n$ above should be a $1$ for the binomial canonical link function. Did I do something wrong?
self-study mathematical-statistics generalized-linear-model exponential-family
asked 1 hour ago
Clarinetist
1,0581142
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up vote
3
down vote
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I define an exponential dispersion family as any distribution whose PMF/PDF is
$$f(y mid boldsymboltheta) = expleftphi[ytheta - b(theta)] + c(y, phi) righttext, y in Omega$$
where $Omega$ is in the support of a random variable $Y$ in the family.
Suppose $Y_1, dots, Y_m$ are independent and binomially distributed ($n$ trials, success probability $p_i$). I've already shown that the Binomial distribution satisfies the above, with
$$beginalign
phi &= 1 \
theta_i &= logleft(dfracp_i1-p_i right) \
b(theta_i) &= nlogleft(dfrac11-p_iright) \
c(phi, y_i) &= logbinomny_itext.
endalign$$
After some work, I showed that, as a function of $theta_i$,
$$b(theta_i) = nlog(e^theta_i + 1)$$
(this is consistent with what I found at http://sfb649.wiwi.hu-berlin.de/fedc_homepage/xplore/tutorials/xlghtmlnode38.html)
and I understand that
$$mu_i = b^prime(theta_i) = n cdot dfrace^theta_ie^theta_i+1text.$$
I also understand that what we need to do is solve for $theta_i$ in the above, and the canonical link function would be $g(mu_i) = theta_i$ according to above. But one thing bothers me: when I run the above through WolframAlpha, I obtain
$$g(mu_i) = theta_i = logleft( dfracmu_in-mu_iright)text.$$
Every source I've seen says that the $n$ above should be a $1$ for the binomial canonical link function. Did I do something wrong?
self-study mathematical-statistics generalized-linear-model exponential-family
asked 1 hour ago
Clarinetist
1,0581142
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I define an exponential dispersion family as any distribution whose PMF/PDF is
$$f(y mid boldsymboltheta) = expleftphi[ytheta - b(theta)] + c(y, phi) righttext, y in Omega$$
where $Omega$ is in the support of a random variable $Y$ in the family.
Suppose $Y_1, dots, Y_m$ are independent and binomially distributed ($n$ trials, success probability $p_i$). I've already shown that the Binomial distribution satisfies the above, with
$$beginalign
phi &= 1 \
theta_i &= logleft(dfracp_i1-p_i right) \
b(theta_i) &= nlogleft(dfrac11-p_iright) \
c(phi, y_i) &= logbinomny_itext.
endalign$$
After some work, I showed that, as a function of $theta_i$,
$$b(theta_i) = nlog(e^theta_i + 1)$$
(this is consistent with what I found at http://sfb649.wiwi.hu-berlin.de/fedc_homepage/xplore/tutorials/xlghtmlnode38.html)
and I understand that
$$mu_i = b^prime(theta_i) = n cdot dfrace^theta_ie^theta_i+1text.$$
I also understand that what we need to do is solve for $theta_i$ in the above, and the canonical link function would be $g(mu_i) = theta_i$ according to above. But one thing bothers me: when I run the above through WolframAlpha, I obtain
$$g(mu_i) = theta_i = logleft( dfracmu_in-mu_iright)text.$$
Every source I've seen says that the $n$ above should be a $1$ for the binomial canonical link function. Did I do something wrong?
self-study mathematical-statistics generalized-linear-model exponential-family
asked 1 hour ago
Clarinetist
1,0581142
I define an exponential dispersion family as any distribution whose PMF/PDF is
$$f(y mid boldsymboltheta) = expleftphi[ytheta - b(theta)] + c(y, phi) righttext, y in Omega$$
where $Omega$ is in the support of a random variable $Y$ in the family.
Suppose $Y_1, dots, Y_m$ are independent and binomially distributed ($n$ trials, success probability $p_i$). I've already shown that the Binomial distribution satisfies the above, with
$$beginalign
phi &= 1 \
theta_i &= logleft(dfracp_i1-p_i right) \
b(theta_i) &= nlogleft(dfrac11-p_iright) \
c(phi, y_i) &= logbinomny_itext.
endalign$$
After some work, I showed that, as a function of $theta_i$,
$$b(theta_i) = nlog(e^theta_i + 1)$$
(this is consistent with what I found at http://sfb649.wiwi.hu-berlin.de/fedc_homepage/xplore/tutorials/xlghtmlnode38.html)
and I understand that
$$mu_i = b^prime(theta_i) = n cdot dfrace^theta_ie^theta_i+1text.$$
I also understand that what we need to do is solve for $theta_i$ in the above, and the canonical link function would be $g(mu_i) = theta_i$ according to above. But one thing bothers me: when I run the above through WolframAlpha, I obtain
$$g(mu_i) = theta_i = logleft( dfracmu_in-mu_iright)text.$$
Every source I've seen says that the $n$ above should be a $1$ for the binomial canonical link function. Did I do something wrong?
self-study mathematical-statistics generalized-linear-model exponential-family
self-study mathematical-statistics generalized-linear-model exponential-family
asked 1 hour ago
Clarinetist
1,0581142
asked 1 hour ago
Clarinetist
1,0581142
edited 59 mins ago
asked 1 hour ago
Clarinetist
1,0581142
asked 1 hour ago
Clarinetist
1,0581142
asked 1 hour ago
Clarinetist
1,0581142
1,0581142
add a comment |Â
add a comment |Â
1 Answer
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up vote
3
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You're almost right, and it's such an easy fix:
$$mu_i = p_i n$$ so $$log(fracmu_in - mu_i) = log(fracnp_in - np_i) = ...$$
So, the $n$ can be a 1, as long as you swap out $mu_i$ for $p_i$.
answered 1 hour ago
eric_kernfeld
2,340421
So if I'm understanding what you're saying correctly, it's this: if I wanted to write the canonical link in terms of $p_i$, it would be the logit, but if I wanted to write it in terms of $mu_i$, it would be what I have above (and in both cases, they are identical for when $n = 1$ , i.e., the Bernoulli)?
– Clarinetist
1 hour ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
You're almost right, and it's such an easy fix:
$$mu_i = p_i n$$ so $$log(fracmu_in - mu_i) = log(fracnp_in - np_i) = ...$$
So, the $n$ can be a 1, as long as you swap out $mu_i$ for $p_i$.
answered 1 hour ago
eric_kernfeld
2,340421
So if I'm understanding what you're saying correctly, it's this: if I wanted to write the canonical link in terms of $p_i$, it would be the logit, but if I wanted to write it in terms of $mu_i$, it would be what I have above (and in both cases, they are identical for when $n = 1$ , i.e., the Bernoulli)?
– Clarinetist
1 hour ago
add a comment |Â
up vote
3
down vote
You're almost right, and it's such an easy fix:
$$mu_i = p_i n$$ so $$log(fracmu_in - mu_i) = log(fracnp_in - np_i) = ...$$
So, the $n$ can be a 1, as long as you swap out $mu_i$ for $p_i$.
answered 1 hour ago
eric_kernfeld
2,340421
So if I'm understanding what you're saying correctly, it's this: if I wanted to write the canonical link in terms of $p_i$, it would be the logit, but if I wanted to write it in terms of $mu_i$, it would be what I have above (and in both cases, they are identical for when $n = 1$ , i.e., the Bernoulli)?
– Clarinetist
1 hour ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
You're almost right, and it's such an easy fix:
$$mu_i = p_i n$$ so $$log(fracmu_in - mu_i) = log(fracnp_in - np_i) = ...$$
So, the $n$ can be a 1, as long as you swap out $mu_i$ for $p_i$.
answered 1 hour ago
eric_kernfeld
2,340421
You're almost right, and it's such an easy fix:
$$mu_i = p_i n$$ so $$log(fracmu_in - mu_i) = log(fracnp_in - np_i) = ...$$
So, the $n$ can be a 1, as long as you swap out $mu_i$ for $p_i$.
answered 1 hour ago
eric_kernfeld
2,340421
answered 1 hour ago
eric_kernfeld
2,340421
answered 1 hour ago
eric_kernfeld
2,340421
answered 1 hour ago
eric_kernfeld
2,340421
2,340421
So if I'm understanding what you're saying correctly, it's this: if I wanted to write the canonical link in terms of $p_i$, it would be the logit, but if I wanted to write it in terms of $mu_i$, it would be what I have above (and in both cases, they are identical for when $n = 1$ , i.e., the Bernoulli)?
– Clarinetist
1 hour ago
add a comment |Â
So if I'm understanding what you're saying correctly, it's this: if I wanted to write the canonical link in terms of $p_i$, it would be the logit, but if I wanted to write it in terms of $mu_i$, it would be what I have above (and in both cases, they are identical for when $n = 1$ , i.e., the Bernoulli)?
– Clarinetist
1 hour ago
So if I'm understanding what you're saying correctly, it's this: if I wanted to write the canonical link in terms of $p_i$, it would be the logit, but if I wanted to write it in terms of $mu_i$, it would be what I have above (and in both cases, they are identical for when $n = 1$ , i.e., the Bernoulli)?
– Clarinetist
1 hour ago
So if I'm understanding what you're saying correctly, it's this: if I wanted to write the canonical link in terms of $p_i$, it would be the logit, but if I wanted to write it in terms of $mu_i$, it would be what I have above (and in both cases, they are identical for when $n = 1$ , i.e., the Bernoulli)?
– Clarinetist
1 hour ago
add a comment |Â
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