Deriving eq 9.15 in PRML by CM Bishop
$begingroup$
In PRML by C.M. Bishop, chapter 9, (9.15) is as follows:
$$
mathcal{N}(mathbf{x}_n|mathbf{x}_n, sigma_j^2mathbf{I}) = frac{1}{(2pi)^{1/2}}frac{1}{sigma_j}
$$
I've tried to derive this myself:
$$
mathcal{N}(mathbf{x}_n|mathbf{x}_n, sigma_j^2mathbf{I}) = frac{1}{(2pi)^{d/2}|sigma_j^2mathbf{I}|^{1/2}}e^{(mathbf{x}_n-mathbf{x}_n)frac{1}{sigma_j^2}mathbf{I}(mathbf{x}_n - mathbf{x}_n)^T}
$$
$$
= frac{1}{(2pi)^{d/2}|sigma_j^2mathbf{I}|^{1/2}}e^{0}
$$
$$
= frac{1}{(2pi)^{d/2}(prod_{i=1}^{d}sigma_j^2)^{1/2}}
$$
$$
= frac{1}{(2pi)^{d/2}(sigma_j^{2d})^{1/2}}
$$
$$
= frac{1}{(2pi)^{d/2}sigma_j^{d}} = frac{1}{(2pi)^{d/2}}frac{1}{sigma_j^{d}}
$$
which gives a clearly different result. What does Bishop mean? It seems to me like the input dimensionality $d$ is not 1, in his equation, otherwise, why have a multivariate distribution? And if it's not d, how can (9.15) be true?
probability-theory probability-distributions
asked Dec 13 '18 at 14:31
SandiSandi
255112
$endgroup$
add a comment |
$begingroup$
In PRML by C.M. Bishop, chapter 9, (9.15) is as follows:
$$
mathcal{N}(mathbf{x}_n|mathbf{x}_n, sigma_j^2mathbf{I}) = frac{1}{(2pi)^{1/2}}frac{1}{sigma_j}
$$
I've tried to derive this myself:
$$
mathcal{N}(mathbf{x}_n|mathbf{x}_n, sigma_j^2mathbf{I}) = frac{1}{(2pi)^{d/2}|sigma_j^2mathbf{I}|^{1/2}}e^{(mathbf{x}_n-mathbf{x}_n)frac{1}{sigma_j^2}mathbf{I}(mathbf{x}_n - mathbf{x}_n)^T}
$$
$$
= frac{1}{(2pi)^{d/2}|sigma_j^2mathbf{I}|^{1/2}}e^{0}
$$
$$
= frac{1}{(2pi)^{d/2}(prod_{i=1}^{d}sigma_j^2)^{1/2}}
$$
$$
= frac{1}{(2pi)^{d/2}(sigma_j^{2d})^{1/2}}
$$
$$
= frac{1}{(2pi)^{d/2}sigma_j^{d}} = frac{1}{(2pi)^{d/2}}frac{1}{sigma_j^{d}}
$$
which gives a clearly different result. What does Bishop mean? It seems to me like the input dimensionality $d$ is not 1, in his equation, otherwise, why have a multivariate distribution? And if it's not d, how can (9.15) be true?
probability-theory probability-distributions
asked Dec 13 '18 at 14:31
SandiSandi
255112
$endgroup$
add a comment |
$begingroup$
In PRML by C.M. Bishop, chapter 9, (9.15) is as follows:
$$
mathcal{N}(mathbf{x}_n|mathbf{x}_n, sigma_j^2mathbf{I}) = frac{1}{(2pi)^{1/2}}frac{1}{sigma_j}
$$
I've tried to derive this myself:
$$
mathcal{N}(mathbf{x}_n|mathbf{x}_n, sigma_j^2mathbf{I}) = frac{1}{(2pi)^{d/2}|sigma_j^2mathbf{I}|^{1/2}}e^{(mathbf{x}_n-mathbf{x}_n)frac{1}{sigma_j^2}mathbf{I}(mathbf{x}_n - mathbf{x}_n)^T}
$$
$$
= frac{1}{(2pi)^{d/2}|sigma_j^2mathbf{I}|^{1/2}}e^{0}
$$
$$
= frac{1}{(2pi)^{d/2}(prod_{i=1}^{d}sigma_j^2)^{1/2}}
$$
$$
= frac{1}{(2pi)^{d/2}(sigma_j^{2d})^{1/2}}
$$
$$
= frac{1}{(2pi)^{d/2}sigma_j^{d}} = frac{1}{(2pi)^{d/2}}frac{1}{sigma_j^{d}}
$$
which gives a clearly different result. What does Bishop mean? It seems to me like the input dimensionality $d$ is not 1, in his equation, otherwise, why have a multivariate distribution? And if it's not d, how can (9.15) be true?
probability-theory probability-distributions
asked Dec 13 '18 at 14:31
SandiSandi
255112
$endgroup$
In PRML by C.M. Bishop, chapter 9, (9.15) is as follows:
$$
mathcal{N}(mathbf{x}_n|mathbf{x}_n, sigma_j^2mathbf{I}) = frac{1}{(2pi)^{1/2}}frac{1}{sigma_j}
$$
I've tried to derive this myself:
$$
mathcal{N}(mathbf{x}_n|mathbf{x}_n, sigma_j^2mathbf{I}) = frac{1}{(2pi)^{d/2}|sigma_j^2mathbf{I}|^{1/2}}e^{(mathbf{x}_n-mathbf{x}_n)frac{1}{sigma_j^2}mathbf{I}(mathbf{x}_n - mathbf{x}_n)^T}
$$
$$
= frac{1}{(2pi)^{d/2}|sigma_j^2mathbf{I}|^{1/2}}e^{0}
$$
$$
= frac{1}{(2pi)^{d/2}(prod_{i=1}^{d}sigma_j^2)^{1/2}}
$$
$$
= frac{1}{(2pi)^{d/2}(sigma_j^{2d})^{1/2}}
$$
$$
= frac{1}{(2pi)^{d/2}sigma_j^{d}} = frac{1}{(2pi)^{d/2}}frac{1}{sigma_j^{d}}
$$
which gives a clearly different result. What does Bishop mean? It seems to me like the input dimensionality $d$ is not 1, in his equation, otherwise, why have a multivariate distribution? And if it's not d, how can (9.15) be true?
probability-theory probability-distributions
probability-theory probability-distributions
asked Dec 13 '18 at 14:31
SandiSandi
255112
asked Dec 13 '18 at 14:31
SandiSandi
255112
asked Dec 13 '18 at 14:31
SandiSandi
255112
asked Dec 13 '18 at 14:31
SandiSandi
255112
asked Dec 13 '18 at 14:31
SandiSandi
255112
255112
add a comment |
add a comment |
1 Answer
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I agree with you.$$det(2pi sigma_j^2 I )^{-frac12}=(2pisigma_j^2)^{-frac{d}{2}}$$
Here is an errata of PRML, equation $(9.15)$ is one of them.
answered Dec 13 '18 at 15:04
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
I agree with you.$$det(2pi sigma_j^2 I )^{-frac12}=(2pisigma_j^2)^{-frac{d}{2}}$$
Here is an errata of PRML, equation $(9.15)$ is one of them.
answered Dec 13 '18 at 15:04
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
add a comment |
$begingroup$
I agree with you.$$det(2pi sigma_j^2 I )^{-frac12}=(2pisigma_j^2)^{-frac{d}{2}}$$
Here is an errata of PRML, equation $(9.15)$ is one of them.
answered Dec 13 '18 at 15:04
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
add a comment |
$begingroup$
I agree with you.$$det(2pi sigma_j^2 I )^{-frac12}=(2pisigma_j^2)^{-frac{d}{2}}$$
Here is an errata of PRML, equation $(9.15)$ is one of them.
answered Dec 13 '18 at 15:04
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
I agree with you.$$det(2pi sigma_j^2 I )^{-frac12}=(2pisigma_j^2)^{-frac{d}{2}}$$
Here is an errata of PRML, equation $(9.15)$ is one of them.
answered Dec 13 '18 at 15:04
Siong Thye GohSiong Thye Goh
101k1466118
answered Dec 13 '18 at 15:04
Siong Thye GohSiong Thye Goh
101k1466118
answered Dec 13 '18 at 15:04
Siong Thye GohSiong Thye Goh
101k1466118
answered Dec 13 '18 at 15:04
Siong Thye GohSiong Thye Goh
101k1466118
101k1466118
add a comment |
add a comment |
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