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Commit 9e9fd2ef authored by Erik Strand's avatar Erik Strand
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Answer through 3.2

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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 $$\lambda$$ between 0 and 1. For any $$n$$, the binomial probability mass function for $$n$$ Fix any $$\lambda$$ between 0 and 1. For any $$n$$, the binomial probability mass function for $$n$$
trials with probability of success $$p = \lambda / n$$ is trials with probability of success $$p = \lambda / n$$ 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): $$\langle k (k - 1) \cdots (k - m + 1) \rangle = \lambda^m$$ when
k follows the Poisson distribution with average number of events $$\lambda$$.
First, let's verify that $$\langle k \rangle = \lambda$$. This will serve as the base case for an
inductive argument.
$$
\begin{align*}
\langle k \rangle &= \sum_{k = 0}^\infty k \frac{\lambda^k e^{-\lambda}}{k!} \\
&= \lambda \sum_{k = 1}^\infty \frac{\lambda^{k - 1} e^{-\lambda}}{(k - 1)!} \\
&= \lambda \sum_{k = 0}^\infty \frac{\lambda^k e^{-\lambda}}{k!} \\
&= \lambda
\end{align*}
$$
Now assume that $$\langle k (k - 1) \cdots (k - m + 1) \rangle = \lambda^m$$.
$$
\begin{align*}
\langle k (k - 1) \cdots (k - m + 1) (k - m) \rangle
&= \sum_{k = 0}^\infty k \cdots (k - m) \frac{\lambda^k e^{-\lambda}}{k!} \\
&= \lambda \sum_{k = 1}^\infty (k - 1) \cdots (k - m)
\frac{\lambda^{k - 1} e^{-\lambda}}{(k - 1)!} \\
&= \lambda \sum_{k = 0}^\infty k \cdots (k - m + 1) \frac{\lambda^k e^{-\lambda}}{k!} \\
&= \lambda \lambda^m \\
&= \lambda^{m + 1}
\end{align*}
$$
### (c) ### (c)
Use (b) to derive equation (3.19). Use (b) to derive equation (3.19): $$\sigma / \langle k \rangle = 1 / \sqrt{\lambda}$$.
To compute $$\sigma$$, we need to know what $$\langle k^2 \rangle$$ is. It can be found using the
same trick we relied on for the last problem.
$$
\begin{align*}
\langle k^2 \rangle
&= \sum_{k = 0}^\infty k^2 \frac{\lambda^k e^{-\lambda}}{k!} \\
&= \lambda \sum_{k = 1}^\infty k \frac{\lambda^{k - 1} e^{-\lambda}}{(k - 1)!} \\
&= \lambda \sum_{k = 0}^\infty (k + 1) \frac{\lambda^k e^{-\lambda}}{k!} \\
&= \lambda \left( \sum_{k = 0}^\infty k \frac{\lambda^k e^{-\lambda}}{k!} +
\sum_{k = 0}^\infty \frac{\lambda^k e^{-\lambda}}{k!} \right) \\
&= \lambda (\lambda + 1)
\end{align*}
$$
Thus $$\sigma^2 = \langle k^2 \rangle - \langle k \rangle^2 = \lambda (\lambda + 1) - \lambda^2 =
\lambda$$, and $$\sigma / \langle k \rangle = \sqrt{\lambda} / \lambda = 1 / \sqrt{\lambda}$$.
## (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 $$\sigma = \sqrt{\lambda}$$. For large $$\lambda$$, the Poisson
distribution is very close to the normal distribution. So about two thirds of the probability mass
lies between $$\lambda - \sigma$$ and $$\lambda + \sigma$$. Thus if $$\sigma \leq 0.01 \lambda$$, 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 $$\sqrt{\lambda} \leq 0.01 \lambda$$, i.e. $$\lambda \geq 10^4$$.
To have the same probability that the number of observed photons is within $$10^{-6} \lambda$$ of
the true value, we need $$\lambda \geq 10^{12}$$.
The wavelength of visible light is about $$\num{500e-9} \si{m}$$, so the energy of each photon will
be
$$
E = \frac{h c}{\lambda} = \frac{\num{6.626e-34} \si{J.s} \cdot \num{3e8} \si{m/s}}
{\num{500e-9} \si{m}}
= \num{3.8e-19} \si{J}
$$
Thus $$10^4$$ photons per second is $$\num{3.8e-15} \si{W}$$, and $$10^{12}$$ photons per second is
$$\num{3.8e-7} \si{W}$$.
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 $$\lambda$$ 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 $$\lambda$$ with
some precision depends on my prior for $$\lambda$$. 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 $$\num{1e12}$$ photons, than if I measured something like
$$\num{3.64e12}$$, 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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