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Erik Strand
pit
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9e9fd2ef
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9e9fd2ef
authored
6 years ago
by
Erik Strand
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Answer through 3.2
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_posts/2019-02-14-pset2.md
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...
@@ -6,7 +6,8 @@ title: Problem Set 2
...
@@ -6,7 +6,8 @@ title: Problem Set 2
### (a)
### (a)
Derive equation (3.16) from the binomial distribution and Stirling’s approximation.
Derive equation (3.16) (i.e. the Poisson probability density function) from the binomial
distribution and Stirling’s approximation.
Fix any $$
\l
ambda$$ between 0 and 1. For any $$n$$, the binomial probability mass function for $$n$$
Fix any $$
\l
ambda$$ between 0 and 1. For any $$n$$, the binomial probability mass function for $$n$$
trials with probability of success $$p =
\l
ambda / n$$ is
trials with probability of success $$p =
\l
ambda / n$$ is
...
@@ -31,11 +32,56 @@ term approaches $$\lambda^k e^{-\lambda}/k!$$ which is the Poisson distribution.
...
@@ -31,11 +32,56 @@ term approaches $$\lambda^k e^{-\lambda}/k!$$ which is the Poisson distribution.
### (b)
### (b)
Use it to derive equation (3.18).
Use it to derive equation (3.18): $$
\l
angle k (k - 1)
\c
dots (k - m + 1)
\r
angle =
\l
ambda^m$$ when
k follows the Poisson distribution with average number of events $$
\l
ambda$$.
First, let's verify that $$
\l
angle k
\r
angle =
\l
ambda$$. This will serve as the base case for an
inductive argument.
$$
\b
egin{align
*
}
\l
angle k
\r
angle &=
\s
um_{k = 0}^
\i
nfty k
\f
rac{
\l
ambda^k e^{-
\l
ambda}}{k!}
\\
&=
\l
ambda
\s
um_{k = 1}^
\i
nfty
\f
rac{
\l
ambda^{k - 1} e^{-
\l
ambda}}{(k - 1)!}
\\
&=
\l
ambda
\s
um_{k = 0}^
\i
nfty
\f
rac{
\l
ambda^k e^{-
\l
ambda}}{k!}
\\
&=
\l
ambda
\e
nd{align
*
}
$$
Now assume that $$
\l
angle k (k - 1)
\c
dots (k - m + 1)
\r
angle =
\l
ambda^m$$.
$$
\b
egin{align
*
}
\l
angle k (k - 1)
\c
dots (k - m + 1) (k - m)
\r
angle
&=
\s
um_{k = 0}^
\i
nfty k
\c
dots (k - m)
\f
rac{
\l
ambda^k e^{-
\l
ambda}}{k!}
\\
&=
\l
ambda
\s
um_{k = 1}^
\i
nfty (k - 1)
\c
dots (k - m)
\f
rac{
\l
ambda^{k - 1} e^{-
\l
ambda}}{(k - 1)!}
\\
&=
\l
ambda
\s
um_{k = 0}^
\i
nfty k
\c
dots (k - m + 1)
\f
rac{
\l
ambda^k e^{-
\l
ambda}}{k!}
\\
&=
\l
ambda
\l
ambda^m
\\
&=
\l
ambda^{m + 1}
\e
nd{align
*
}
$$
### (c)
### (c)
Use (b) to derive equation (3.19).
Use (b) to derive equation (3.19): $$
\s
igma /
\l
angle k
\r
angle = 1 /
\s
qrt{
\l
ambda}$$.
To compute $$
\s
igma$$, we need to know what $$
\l
angle k^2
\r
angle$$ is. It can be found using the
same trick we relied on for the last problem.
$$
\b
egin{align
*
}
\l
angle k^2
\r
angle
&=
\s
um_{k = 0}^
\i
nfty k^2
\f
rac{
\l
ambda^k e^{-
\l
ambda}}{k!}
\\
&=
\l
ambda
\s
um_{k = 1}^
\i
nfty k
\f
rac{
\l
ambda^{k - 1} e^{-
\l
ambda}}{(k - 1)!}
\\
&=
\l
ambda
\s
um_{k = 0}^
\i
nfty (k + 1)
\f
rac{
\l
ambda^k e^{-
\l
ambda}}{k!}
\\
&=
\l
ambda
\l
eft(
\s
um_{k = 0}^
\i
nfty k
\f
rac{
\l
ambda^k e^{-
\l
ambda}}{k!} +
\s
um_{k = 0}^
\i
nfty
\f
rac{
\l
ambda^k e^{-
\l
ambda}}{k!}
\r
ight)
\\
&=
\l
ambda (
\l
ambda + 1)
\e
nd{align
*
}
$$
Thus $$
\s
igma^2 =
\l
angle k^2
\r
angle -
\l
angle k
\r
angle^2 =
\l
ambda (
\l
ambda + 1) -
\l
ambda^2 =
\l
ambda$$, and $$
\s
igma /
\l
angle k
\r
angle =
\s
qrt{
\l
ambda} /
\l
ambda = 1 /
\s
qrt{
\l
ambda}$$.
## (3.2)
## (3.2)
...
@@ -44,6 +90,43 @@ N per second. How many must be counted by a photodetector per second to be able
...
@@ -44,6 +90,43 @@ N per second. How many must be counted by a photodetector per second to be able
rate to within 1%? To within 1 part per million? How many watts do these cases correspond to for
rate to within 1%? To within 1 part per million? How many watts do these cases correspond to for
visible light?
visible light?
We have already found that $$
\s
igma =
\s
qrt{
\l
ambda}$$. For large $$
\l
ambda$$, the Poisson
distribution is very close to the normal distribution. So about two thirds of the probability mass
lies between $$
\l
ambda -
\s
igma$$ and $$
\l
ambda +
\s
igma$$. Thus if $$
\s
igma
\l
eq 0.01
\l
ambda$$, in
any given second it's more likely than not that the number of photons emitted is within one percent
of the true mean. Thus we'd need $$
\s
qrt{
\l
ambda}
\l
eq 0.01
\l
ambda$$, i.e. $$
\l
ambda
\g
eq 10^4$$.
To have the same probability that the number of observed photons is within $$10^{-6}
\l
ambda$$ of
the true value, we need $$
\l
ambda
\g
eq 10^{12}$$.
The wavelength of visible light is about $$
\n
um{500e-9}
\s
i{m}$$, so the energy of each photon will
be
$$
E =
\f
rac{h c}{
\l
ambda} =
\f
rac{
\n
um{6.626e-34}
\s
i{J.s}
\c
dot
\n
um{3e8}
\s
i{m/s}}
{
\n
um{500e-9}
\s
i{m}}
=
\n
um{3.8e-19}
\s
i{J}
$$
Thus $$10^4$$ photons per second is $$
\n
um{3.8e-15}
\s
i{W}$$, and $$10^{12}$$ photons per second is
$$
\n
um{3.8e-7}
\s
i{W}$$.
Some caveats about this answer: the Poisson distribution is strictly greater than zero on its whole
domain, so one can never be
*certain*
that they have found the correct value within any error
bounds. That is, even if the real value of $$
\l
ambda$$ is one, there's still a (vanishingly) slight
chance you'll see a million. The best we can do is find an answer that's
[
probably approximately
correct
](
https://en.wikipedia.org/wiki/Probably_approximately_correct_learning
)
.
Second, I only found limits on the probability that the number of emitted photons is near the true
(known) average. It's a more subtle question to ask for the probability that a given estimate is
within one percent of the true (latent) mean. That could be tackled with confidence intervals,
p-values, Bayes factors, or direct posterior estimation, though none would be quite so simple. As an
unapologetic Bayesian I'm required to point out that my confidence I've determined $$
\l
ambda$$ with
some precision depends on my prior for $$
\l
ambda$$. This would be very real if we were, for
instance, counting photons in a lab module for this course. I would be more confident in my answer
if I measured something close to $$
\n
um{1e12}$$ photons, than if I measured something like
$$
\n
um{3.64e12}$$, since I think it's more likely that the instructor would choose a nice round
number with lots of zeros.
## (3.3)
## (3.3)
Consider an audio amplifier with a 20 kHz bandwidth.
Consider an audio amplifier with a 20 kHz bandwidth.
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