Skip to content
GitLab
Explore
Sign in
Primary navigation
Search or go to…
Project
P
pit
Manage
Activity
Members
Labels
Plan
Issues
Issue boards
Milestones
Wiki
Code
Merge requests
Repository
Branches
Commits
Tags
Repository graph
Compare revisions
Snippets
Build
Pipelines
Jobs
Pipeline schedules
Artifacts
Deploy
Releases
Model registry
Operate
Environments
Monitor
Incidents
Analyze
Value stream analytics
Contributor analytics
CI/CD analytics
Repository analytics
Model experiments
Help
Help
Support
GitLab documentation
Compare GitLab plans
Community forum
Contribute to GitLab
Provide feedback
Keyboard shortcuts
?
Snippets
Groups
Projects
Show more breadcrumbs
Erik Strand
pit
Commits
cf03eb17
Commit
cf03eb17
authored
6 years ago
by
Erik Strand
Browse files
Options
Downloads
Patches
Plain Diff
Editing for clarity
parent
755f048b
Branches
pset3
Branches containing commit
No related tags found
No related merge requests found
Changes
1
Show whitespace changes
Inline
Side-by-side
Showing
1 changed file
_psets/3.md
+16
-15
16 additions, 15 deletions
_psets/3.md
with
16 additions
and
15 deletions
_psets/3.md
+
16
−
15
View file @
cf03eb17
...
...
@@ -35,9 +35,10 @@ continuity of entropy with respect to the topologies of $$\mathbb{R}_{\geq 0}^n$
$$
\m
athbb{R}$$.
First let's show that $$x
\l
og x$$ is continuous. I take as given that $$
\l
og(x)$$ is a continuous
function on its domain. Then $$x
\l
og(x)$$ is also continuous, since finite products of continuous
functions are continuous. This suffices for $$x > 0$$. At zero, $$x
\l
og x$$ is continuous because
we have defined it to be equal to the limit we found above.
function on its domain (after all it's the inverse of $$e^x$$, which is strictly monotonic and
$$C^
\i
nfty$$). Then $$x
\l
og(x)$$ is also continuous, since finite products of continuous functions
are continuous. This suffices for $$x > 0$$. At zero, $$x
\l
og x$$ is continuous because we have
defined it to be equal to the limit we found above.
Thus each term of the entropy function is a continuous function from $$
\m
athbb{R}_{
\g
eq 0}$$ to
$$
\m
athbb{R}$$. But we can also view each term as a function from $$
\m
athbb{R}_{
\g
eq 0}^n$$ to
...
...
@@ -431,9 +432,9 @@ $$
\b
egin{align
*
}
\l
angle
\h
at{x_0}
\r
angle
&=
\l
eft
\l
angle
\f
rac{1}{n}
\s
um_{i = 1}^n x_i
\r
ight
\r
angle
\\
&=
\i
nt_{
-
\i
nfty}^
\i
nfty
\c
dots
\i
nt_{-
\i
nfty}^
\i
nfty
\l
eft(
\p
rod_{i = 1}^n
\f
rac{1}{
\s
qrt{2
\p
i
\s
igma^2}}
e^{-
\f
rac{(x_i - x_0)^2}{2
\s
igma^2}}
\r
ight)
\l
eft(
\f
rac{1}{n}
\s
um_{i = 1}^n x_i
\r
ight)
\m
athrm{d} x_1
\l
dots
\m
athrm{d} x_n
\\
&=
\i
nt_{
\m
athbb{R}^n}
\l
eft(
\p
rod_{i = 1}^n
\f
rac{1}{
\s
qrt{2
\p
i
\s
igma^2}}
e^{-
\f
rac{(x_i - x_0)^2}{2
\s
igma^2}}
\r
ight)
\l
eft(
\f
rac{1}{n}
\s
um_{i = 1}^n x_i
\r
ight)
\m
athrm{d} x_1
\l
dots
\m
athrm{d} x_n
\\
&=
\f
rac{1}{n}
\s
um_{i = 1}^n
\p
rod_{j = 1}^n
\i
nt_{-
\i
nfty}^
\i
nfty
\f
rac{1}{
\s
qrt{2
\p
i
\s
igma^2}} e^{-
\f
rac{(x_j - x_0)^2}{2
\s
igma^2}}
...
...
@@ -488,15 +489,15 @@ $$
\b
egin{align
*
}
\l
eft
\l
angle
\h
at{x_0}^2
\r
ight
\r
angle
&=
\l
eft
\l
angle
\l
eft(
\f
rac{1}{n}
\s
um_{i = 1}^n x_i
\r
ight)^2
\r
ight
\r
angle
\\
&=
\i
nt_{
-
\i
nfty}^
\i
nfty
\c
dots
\i
nt_{-
\i
nfty}^
\i
nfty
\l
eft(
\p
rod_{i = 1}^n
\f
rac{1}{
\s
qrt{2
\p
i
\s
igma^2}}
e^{-
\f
rac{(x_i - x_0)^2}{2
\s
igma^2}}
\r
ight)
\l
eft(
\f
rac{1}{n}
\s
um_{i = 1}^n x_i
\r
ight)^2
\m
athrm{d} x_1
\l
dots
\m
athrm{d} x_n
\\
&=
\f
rac{1}{n^2}
\i
nt_{
-
\i
nfty}^
\i
nfty
\c
dots
\i
nt_{-
\i
nfty}^
\i
nfty
\l
eft(
\p
rod_{i = 1}^n
\f
rac{1}{
\s
qrt{2
\p
i
\s
igma^2}} e^{-
\f
rac{(x_i - x_0)^2}{2
\s
igma^2}}
\r
ight)
\l
eft(
\s
um_{i, j} x_i x_j
\r
ight)
\m
athrm{d} x_1
\l
dots
\m
athrm{d} x_n
\\
&=
\f
rac{1}{n^2}
\s
um_{i, j}
\i
nt_{
-
\i
nfty}^
\i
nfty
\c
dots
\i
nt_{-
\i
nfty}^
\i
nfty
\p
rod_{k = 1}^n
\f
rac{1}{
\s
qrt{2
\p
i
\s
igma^2}}
e^{-
\f
rac{(x_k - x_0)^2}{2
\s
igma^2}} x_i x_j
\m
athrm{d} x_1
\l
dots
\m
athrm{d} x_n
\\
&=
\i
nt_{
\m
athbb{R}^n}
\l
eft(
\p
rod_{i = 1}^n
\f
rac{1}{
\s
qrt{2
\p
i
\s
igma^2}}
e^{-
\f
rac{(x_i - x_0)^2}{2
\s
igma^2}}
\r
ight)
\l
eft(
\f
rac{1}{n}
\s
um_{i = 1}^n x_i
\r
ight)^2
\m
athrm{d} x_1
\l
dots
\m
athrm{d} x_n
\\
&=
\f
rac{1}{n^2}
\i
nt_{
\m
athbb{R}^n}
\l
eft(
\p
rod_{i = 1}^n
\f
rac{1}{
\s
qrt{2
\p
i
\s
igma^2}} e^{-
\f
rac{(x_i - x_0)^2}{2
\s
igma^2}}
\r
ight)
\l
eft(
\s
um_{i, j} x_i x_j
\r
ight)
\m
athrm{d} x_1
\l
dots
\m
athrm{d} x_n
\\
&=
\f
rac{1}{n^2}
\s
um_{i, j}
\i
nt_{
\m
athbb{R}^n}
\l
eft(
\p
rod_{k = 1}^n
\f
rac{1}{
\s
qrt{2
\p
i
\s
igma^2}}
e^{-
\f
rac{(x_k - x_0)^2}{2
\s
igma^2}}
\r
ight) x_i x_j
\m
athrm{d} x_1
\l
dots
\m
athrm{d} x_n
\\
&=
\f
rac{1}{n^2}
\l
eft(
\s
um_{i=j}
\i
nt_{-
\i
nfty}^
\i
nfty
\f
rac{1}{
\s
qrt{2
\p
i
\s
igma^2}}
e^{-
\f
rac{(x_i - x_0)^2}{2
\s
igma^2}} x_i^2
\m
athrm{d} x_i
\r
ight.
\\
...
...
This diff is collapsed.
Click to expand it.
Preview
0%
Loading
Try again
or
attach a new file
.
Cancel
You are about to add
0
people
to the discussion. Proceed with caution.
Finish editing this message first!
Save comment
Cancel
Please
register
or
sign in
to comment