> For the complete documentation index, see [llms.txt](https://zavalab.gitbook.io/statschemebook/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://zavalab.gitbook.io/statschemebook/chapter-2-univariate-random-variables/07-uniform-continuous.md).
# Uniform (Continuous)
* A continuous uniform RV has a pdf of the form:
$$
f(x)=\frac{1}{b-a}, ; x\in \mathcal{D}
$$
where domain $$\mathcal{D}=\[a,b]$$ is continuous.
* RV is denoted as $$X\sim \mathcal{U}\_c(a,b)$$.
* Cdf is:
$$
F(x)=\frac{x-a}{b-a}, ; x\in \mathcal{D}
$$
* Recall that pdf is derivative of cdf (it is the slope of the cdf); consequently, cdf has constant slope $$f(x)=1/(b-a)$$.
* Below we show an example of a pdf/cdf of a uniform RV along with realizations.
Python: reproduce this figure (ch2_compare_uniformc.ipynb)
Pdf and cdf of a continuous uniform RV $$X\sim\mathcal{U}\_c(a,b)$$ with $$a=-1,b=1$$ (Victor Z and David Cole, UW-Madison, 2024). Python names: `a`, `b`, pdf `f`, cdf `F`, realizations `x_omega`. In `scipy.stats.uniform`, `loc` $$=a$$ and `scale` $$=b-a$$.
**Note:** This notebook reproduces the figure of the continuous uniform RV in Section 2.2.1 of *Statistics for Chemical Engineers*, with a = -1 and b = 1.
```python
import numpy as np
import matplotlib.pyplot as plt
from scipy.stats import uniform
a, b = -1, 1
x = np.linspace(-2, 2, 1000)
f = uniform.pdf(x, loc=a, scale=b - a) # 1/(b-a) on [a, b]
F = uniform.cdf(x, loc=a, scale=b - a) # (x-a)/(b-a) on [a, b]
np.random.seed(0)
S = 1000
x_omega = np.random.uniform(a, b, S)
plt.figure(figsize=(10, 8))
plt.subplot(2, 2, (1, 2))
plt.plot(x_omega, 'ko', markerfacecolor='w', markersize=4)
plt.xlabel(r'Outcome $\omega$', fontsize=14)
plt.ylabel(r'$x_\omega$', fontsize=14)
plt.grid(True)
plt.axis([0, S, -2, 2])
plt.subplot(2, 2, 3)
plt.plot(x, f, color='black', linewidth=1.5)
plt.grid(True)
plt.xlabel('$x$', fontsize=14)
plt.ylabel('$f(x)$', fontsize=14)
plt.subplot(2, 2, 4)
plt.plot(x, F, color='black', linewidth=1.5)
plt.grid(True)
plt.xlabel('$x$', fontsize=14)
plt.ylabel('$F(x)$', fontsize=14)
plt.tight_layout()
plt.show()
```
**Note:** The pdf is flat on \[a, b] and the cdf grows linearly from 0 to 1.
Open in Colab: [`ch2_compare_uniformc.ipynb`](https://colab.research.google.com/github/vzavala/StatsBook/blob/master/Python/ch2_compare_uniformc.ipynb) · MATLAB live script: [`ch2_compare_uniformc.mlx`](https://github.com/vzavala/StatsBook/blob/master/Matlab-Live/ch2_compare_uniformc.mlx)
* Mean of the RV is:
$$
\begin{aligned}
\mathbb{E}\[X]&=\int\_{a}^b\frac{1}{b-a}xdx\\
&=\frac{1}{2}\frac{1}{b-a}(b^2-a^2)\\
&=\frac{1}{2}(a+b)
\end{aligned}
$$
* From cdf we note that $$\alpha$$-quantiles are values $$x$$ satisfying:
$$
\alpha=F(x)=\frac{x-a}{b-a}
$$
we thus have that $$\mathbb{Q}\_X(\alpha)=a +\alpha,(b-a)$$.
* Mode (point maximizing $$f(x)$$) is not unique because pdf is flat function.
* Variance for this RV is:
$$
\mathbb{V}\[X]=\frac{1}{b-a}\int\_{a}^b\left(x-\frac{1}{2}(a+b)\right)^2dx=\frac{1}{12}(b-a)^2.
$$
When range $$\[a,b]$$ is wide we have that variance (uncertainty) is large.
## Example: Setting up an Alarm
{% hint style="danger" icon="flask" %}
Consider Gibbs reactor and consider situation in which there is no pressure data recorded; all that you know is that, historically, pressure has remained in the range $$63-234$$ bar.
* Is it adequate to model pressure as a uniform RV?
* Thermodynamics tells us that we will have unacceptably low conversions at pressures below 100 bar. What is probability that reactor will exhibit low conversions?
* Reactor vessel is 30-yr-old and you are unsure if it can tolerate high pressures for much longer. To be safe, you want to design an alarm system; you want to set a threshold that triggers alarm only 1% of the time. What is such threshold value?
{% endhint %}
{% hint style="danger" icon="flask" %}
* We only know that pressure is in continuous range $$\[63,234]$$ bar, we thus have no reason to doubt that it is equally likely to find pressure anywhere in the range. Consequently, uniform RV is an adequate model with $$b=234$$ and $$a=63$$.
* Probability that reactor exhibits low conversion is:
$$
\mathbb{P}(X\leq 100)=\frac{100-63}{234-63}= 0.21.
$$
* Want to find value $$x$$ such that $$\mathbb{P}(X>x)=0.01$$. This is equivalent to finding $$x$$ such that $$\mathbb{P}(X\leq x)=0.99$$ (find $$x$$ such that $$F(x)=0.99$$):
$$
0.99=\frac{x-63}{234-63}
$$
Solving for $$x$$ we get that a threshold of 232.3 bar activates alarm 1% of the time.
* Note that threshold value is quantile:
$$
\mathbb{Q}\_X(0.99)=63 + 0.99\cdot (234-63)=232.3.
$$
{% endhint %}
#### Python implementation
{% hint style="success" %}
Notebook [`ch2_example_alarm.ipynb`](https://colab.research.google.com/github/vzavala/StatsBook/blob/master/Python/ch2_example_alarm.ipynb) — open it in Colab and run the cells alongside this section.
{% endhint %}
Setting up a pressure alarm with a uniform model $$X\sim\mathcal{U}\_c(a,b)$$, $$a=63$$, $$b=234$$ bar (written for the book to accompany the slide example). Python names: `a`, `b`, cdf `F(x)`, quantile `Q(alpha)`.
**Note:** This notebook reproduces Example 2.10 (Section 2.2.1) of *Statistics for Chemical Engineers*: setting a pressure alarm when the pressure is modeled as a uniform RV between 63 and 234 bar.
```python
from scipy.stats import uniform
a, b = 63, 234 # bar
X = uniform(loc=a, scale=b - a)
def F(x):
"""Cdf F(x) = (x - a)/(b - a)."""
return X.cdf(x)
def Q(alpha):
"""Quantile Q_X(alpha) = a + alpha (b - a)."""
return X.ppf(alpha)
print(f"P(X <= 100) = F(100) = {F(100):.2f}")
print(f"Alarm threshold Q_X(0.99) = {Q(0.99):.1f} bar (check: a + 0.99(b-a) = {a + 0.99*(b-a):.1f})")
```
```
P(X <= 100) = F(100) = 0.22
Alarm threshold Q_X(0.99) = 232.3 bar (check: a + 0.99(b-a) = 232.3)
```
**Note:** The alarm threshold is the quantile Q(0.99): with a uniform model, the pressure exceeds it only 1% of the time.
