**THE GOAL**$~~$ Compare the differential entropy computed analytically using the probability
density function (PDF) with those estimated using a collection of samples.

# An Analytical Example

Consider a uniform distribution between $[0, 1/2]$

The differential entropy is

There are two ways to compute this numerically. The first is to use analytical calculation from the distribution. We do so by discretizing the probability distribution function into bins. We do so by normalizing the sequence of probabilities. This gives us $H(x) = -\log 2 = -0.69$:

```
def H_d(ps):
ps_norm = ps / ps.sum()
return - np.sum(np.log(ps) * ps_norm)
```

```
xs = np.linspace(0, 1 / 2, 1001)
ps = np.ones(1001) * 2
h_d = H_d(ps)
doc.print(f"analytical: {h_d:0.2f}")
```

`analytical: -0.69`

## Can We Compute This Via `scipy entropy`

?

The `scipy.stats.entropy`

function computes the `Shannon entropy`

of a discrete distribution. The scipy entropy does not assume that the probability is normalized,
so it normalizes the `ps`

internally first. This means that in order to convert this to the
differential entropy, we need to scale the result by the log sum.

```
h_d_analytic = stats.entropy(ps) - np.log(ps.sum())
doc.print(f"analytic entropy w/ scipy: {h_d_analytic:0.3f}")
```

`analytic entropy w/ scipy: -0.693`

This shows that we can get the discrete `scipy.stats.entropy`

to agree with our differential entropy function `H_d:ps -> R`

that is defined analytically.

# Estimating Entropy Using Samples from A Probability Density Function

When we do not have access to the PDF, we can also estimate the entropy from a set of samples. We can do this using the `differential_entropy`

function from the `stats`

package in `scipy`

:

```
from scipy import stats
samples = np.random.uniform(0, 1/2, 10001)
doc.print(f"verify the distribution: min: {samples.min():0.2f}, max: {samples.max():0.2f}")
```

`verify the distribution: min: 0.00, max: 0.50`

```
h_d_sample = stats.differential_entropy(samples)
doc.print(f"sample-based: {h_d_sample:0.3f}")
```

`sample-based: -0.702`

# Effect of Sampling On Differential Entropy

**THE GOAL**$~~$ vary the number of samples from the distribution, and compare
the sample-based estimate against those computed from the analytical solution.
This is done on the scipy documentation page ^{1}.

Again, consider the uniform distribution from above.

```
def H_d(ps):
ps_norm = ps / ps.sum()
return - np.sum(np.log(ps) * ps_norm)
def get_H_d(N):
xs = np.linspace(0, 1 / 2, N)
ps = np.ones(N) * 2
h_d = H_d(ps)
doc.print(f"N={N} => h_d={h_d:0.2f}")
for n in [101, 201, 401, 801, 1601]:
get_H_d(n)
```

```
N=101 => h_d=-0.69
N=201 => h_d=-0.69
N=401 => h_d=-0.69
N=801 => h_d=-0.69
N=1601 => h_d=-0.69
```

The differential entropy does not depend on the number of points in the discretized distribution as long as the discretization is done properly.

# Effect of Change of Variable On Differential Entropy

**THE GOAL**$~~$ Investigate the effect of the change of variable, by varying the
scale of the distribution and looking at the entropy. We can look at both
the differential entropy analytically, and the sample-based estimates ^{1}.

Now consider a more general case, a uniform distribution between $[a, b]$

The differential entropy is $\log \left(\vert b - a \vert\right)$. When
`a=0`

and `b=0.5`

, $H(x) = - \log 2$.

We can verify this numerically. First let's define the entropy function

```
def H_d(ps):
ps_norm = ps / ps.sum()
return - np.sum(np.log(ps) * ps_norm)
```

We can plot the delta against $\log(b - a)$ -- it turned out to be zero across the range variants.

```
def get_H_d(a, b, N=201):
xs = np.linspace(a, b, N)
ps = np.ones(N) / (b - a)
h_d = H_d(ps)
delta = h_d - np.log(b - a)
doc.print(f"a={a}, b={b} => h_d={h_d:0.2f}, δ={delta}")
for b in [1/4, 1/2, 1, 2, 4]:
get_H_d(0, b)
```

```
a=0, b=0.25 => h_d=-1.39, δ=0.0
a=0, b=0.5 => h_d=-0.69, δ=0.0
a=0, b=1 => h_d=-0.00, δ=-0.0
a=0, b=2 => h_d=0.69, δ=0.0
a=0, b=4 => h_d=1.39, δ=0.0
```