Merge pull request #76 from marimo-team/haleshot/015_poisson

probability/15_poisson_distribution.py

@@ -0,0 +1,805 @@
+# /// script
+# requires-python = ">=3.10"
+# dependencies = [
+#     "marimo",
+#     "matplotlib==3.10.0",
+#     "numpy==2.2.4",
+#     "scipy==1.15.2",
+#     "altair==5.2.0",
+#     "wigglystuff==0.1.10",
+#     "pandas==2.2.3",
+# ]
+# ///
+
+import marimo
+
+__generated_with = "0.11.25"
+app = marimo.App(width="medium", app_title="Poisson Distribution")
+
+
+@app.cell(hide_code=True)
+def _(mo):
+    mo.md(
+        r"""
+        # Poisson Distribution
+
+        _This notebook is a computational companion to ["Probability for Computer Scientists"](https://chrispiech.github.io/probabilityForComputerScientists/en/part2/poisson/), by Stanford professor Chris Piech._
+
+        A Poisson random variable gives the probability of a given number of events in a fixed interval of time (or space). It makes the Poisson assumption that events occur with a known constant mean rate and independently of the time since the last event.
+        """
+    )
+    return
+
+
+@app.cell(hide_code=True)
+def _(mo):
+    mo.md(
+        r"""
+        ## Poisson Random Variable Definition
+
+        $X \sim \text{Poisson}(\lambda)$ represents a Poisson random variable where:
+
+        - $X$ is our random variable (number of events)
+        - $\text{Poisson}$ indicates it follows a Poisson distribution
+        - $\lambda$ is the rate parameter (average number of events per time interval)
+
+        ```
+        X ~ Poisson(λ)
+         ↑     ↑    ↑
+         |     |    +-- Rate parameter:
+         |     |        average number of
+         |     |        events per interval
+         |     +-- Indicates Poisson
+         |         distribution
+         |
+        Our random variable
+          counting number of events
+        ```
+
+        The Poisson distribution is particularly useful when:
+
+        1. Events occur independently of each other
+        2. The average rate of occurrence is constant
+        3. Two events cannot occur at exactly the same instant
+        4. The probability of an event is proportional to the length of the time interval
+        """
+    )
+    return
+
+
+@app.cell(hide_code=True)
+def _(mo):
+    mo.md(
+        r"""
+        ## Properties of Poisson Distribution
+
+        | Property | Formula |
+        |----------|---------|
+        | Notation | $X \sim \text{Poisson}(\lambda)$ |
+        | Description | Number of events in a fixed time frame if (a) events occur with a constant mean rate and (b) they occur independently of time since last event |
+        | Parameters | $\lambda \in \mathbb{R}^{+}$, the constant average rate |
+        | Support | $x \in \{0, 1, \dots\}$ |
+        | PMF equation | $P(X=x) = \frac{\lambda^x e^{-\lambda}}{x!}$ |
+        | Expectation | $E[X] = \lambda$ |
+        | Variance | $\text{Var}(X) = \lambda$ |
+
+        Note that unlike many other distributions, the Poisson distribution's mean and variance are equal, both being $\lambda$.
+
+        Let's explore how the Poisson distribution changes with different rate parameters.
+        """
+    )
+    return
+
+
+@app.cell(hide_code=True)
+def _(TangleSlider, mo):
+    # interactive elements using TangleSlider
+    lambda_slider = mo.ui.anywidget(TangleSlider(
+        amount=5, 
+        min_value=0.1, 
+        max_value=20, 
+        step=0.1,
+        digits=1,
+        suffix=" events"
+    ))
+
+    # interactive controls
+    _controls = mo.vstack([
+        mo.md("### Adjust the Rate Parameter to See How Poisson Distribution Changes"),
+        mo.hstack([
+            mo.md("**Rate parameter (λ):** "),
+            lambda_slider,
+            mo.md("**events per interval.** Higher values shift the distribution rightward and make it more spread out.")
+        ], justify="start"),
+    ])
+    _controls
+    return (lambda_slider,)
+
+
+@app.cell(hide_code=True)
+def _(lambda_slider, np, plt, stats):
+    def create_poisson_pmf_plot(lambda_value):
+        """Create a visualization of Poisson PMF with annotations for mean and variance."""
+        # PMF for values
+        max_x = max(20, int(lambda_value * 3))  # Show at least up to 3*lambda
+        x = np.arange(0, max_x + 1)
+        pmf = stats.poisson.pmf(x, lambda_value)
+
+        # Relevant key statistics
+        mean = lambda_value  # For Poisson, mean = lambda
+        variance = lambda_value  # For Poisson, variance = lambda
+        std_dev = np.sqrt(variance)
+
+        # plot
+        fig, ax = plt.subplots(figsize=(10, 6))
+
+        # PMF as bars
+        ax.bar(x, pmf, color='royalblue', alpha=0.7, label=f'PMF: P(X=k)')
+
+        # for the PMF values
+        ax.plot(x, pmf, 'ro-', alpha=0.6, label='PMF line')
+
+        # Vertical lines - mean and key values
+        ax.axvline(x=mean, color='green', linestyle='--', linewidth=2, 
+                label=f'Mean: {mean:.2f}')
+
+        # Stdev region
+        ax.axvspan(mean - std_dev, mean + std_dev, alpha=0.2, color='green',
+                label=f'±1 Std Dev: {std_dev:.2f}')
+
+        ax.set_xlabel('Number of Events (k)')
+        ax.set_ylabel('Probability: P(X=k)')
+        ax.set_title(f'Poisson Distribution with λ={lambda_value:.1f}')
+
+        # annotations
+        ax.annotate(f'E[X] = {mean:.2f}', 
+                    xy=(mean, stats.poisson.pmf(int(mean), lambda_value)), 
+                    xytext=(mean + 1, max(pmf) * 0.8),
+                    arrowprops=dict(facecolor='black', shrink=0.05, width=1))
+
+        ax.annotate(f'Var(X) = {variance:.2f}', 
+                    xy=(mean, stats.poisson.pmf(int(mean), lambda_value) / 2), 
+                    xytext=(mean + 1, max(pmf) * 0.6),
+                    arrowprops=dict(facecolor='black', shrink=0.05, width=1))
+
+        ax.grid(alpha=0.3)
+        ax.legend()
+
+        plt.tight_layout()
+        return plt.gca()
+
+    # Get parameter from slider and create plot
+    _lambda = lambda_slider.amount
+    create_poisson_pmf_plot(_lambda)
+    return (create_poisson_pmf_plot,)
+
+
+@app.cell(hide_code=True)
+def _(mo):
+    mo.md(
+        r"""
+        ## Poisson Intuition: Relation to Binomial Distribution
+
+        The Poisson distribution can be derived as a limiting case of the [binomial distribution](http://marimo.app/https://github.com/marimo-team/learn/blob/main/probability/14_binomial_distribution.py). 
+
+        Let's work on a practical example: predicting the number of ride-sharing requests in a specific area over a one-minute interval. From historical data, we know that the average number of requests per minute is $\lambda = 5$.
+
+        We could approximate this using a binomial distribution by dividing our minute into smaller intervals. For example, we can divide a minute into 60 seconds and treat each second as a [Bernoulli trial](http://marimo.app/https://github.com/marimo-team/learn/blob/main/probability/13_bernoulli_distribution.py) - either there's a request (success) or there isn't (failure).
+
+        Let's visualize this concept:
+        """
+    )
+    return
+
+
+@app.cell(hide_code=True)
+def _(fig_to_image, mo, plt):
+    def create_time_division_visualization():
+        # visualization of dividing a minute into 60 seconds
+        fig, ax = plt.subplots(figsize=(12, 2))
+
+        # Example events hardcoded at 2.75s and 7.12s
+        events = [2.75, 7.12]
+
+        # array of 60 rectangles
+        for i in range(60):
+            color = 'royalblue' if any(i <= e < i+1 for e in events) else 'lightgray'
+            ax.add_patch(plt.Rectangle((i, 0), 0.9, 1, color=color))
+
+        # markers for events
+        for e in events:
+            ax.plot(e, 0.5, 'ro', markersize=10)
+
+        # labels
+        ax.set_xlim(0, 60)
+        ax.set_ylim(0, 1)
+        ax.set_yticks([])
+        ax.set_xticks([0, 15, 30, 45, 60])
+        ax.set_xticklabels(['0s', '15s', '30s', '45s', '60s'])
+        ax.set_xlabel('Time (seconds)')
+        ax.set_title('One Minute Divided into 60 Second Intervals')
+
+        plt.tight_layout()
+        plt.gca()
+        return fig, events, i
+
+    # Create visualization and convert to image
+    _fig, _events, i = create_time_division_visualization()
+    _img = mo.image(fig_to_image(_fig), width="100%")
+
+    # explanation
+    _explanation = mo.md(
+        r"""
+        In this visualization:
+    
+        - Each rectangle represents a 1-second interval
+        - Blue rectangles indicate intervals where an event occurred
+        - Red dots show the actual event times (2.75s and 7.12s)
+
+        If we treat this as a binomial experiment with 60 trials (seconds), we can calculate probabilities using the binomial PMF. But there's a problem: what if multiple events occur within the same second? To address this, we can divide our minute into smaller intervals.
+        """
+    )
+    mo.vstack([_fig, _explanation])
+    return create_time_division_visualization, i
+
+
+@app.cell(hide_code=True)
+def _(mo):
+    mo.md(
+        r"""
+        The total number of requests received over the minute can be approximated as the sum of the sixty indicator variables, which conveniently matches the description of a binomial — a sum of Bernoullis. 
+
+        Specifically, if we define $X$ to be the number of requests in a minute, $X$ is a binomial with $n=60$ trials. What is the probability, $p$, of a success on a single trial? To make the expectation of $X$ equal the observed historical average $\lambda$, we should choose $p$ so that:
+
+        \begin{align}
+        \lambda &= E[X] && \text{Expectation matches historical average} \\
+        \lambda &= n \cdot p && \text{Expectation of a Binomial is } n \cdot p \\
+        p &= \frac{\lambda}{n} && \text{Solving for $p$}
+        \end{align}
+
+        In this case, since $\lambda=5$ and $n=60$, we should choose $p=\frac{5}{60}=\frac{1}{12}$ and state that $X \sim \text{Bin}(n=60, p=\frac{5}{60})$. Now we can calculate the probability of different numbers of requests using the binomial PMF:
+
+        $P(X = x) = {n \choose x} p^x (1-p)^{n-x}$
+
+        For example:
+
+        \begin{align}
+        P(X=1) &= {60 \choose 1} (5/60)^1 (55/60)^{60-1} \approx 0.0295 \\
+        P(X=2) &= {60 \choose 2} (5/60)^2 (55/60)^{60-2} \approx 0.0790 \\
+        P(X=3) &= {60 \choose 3} (5/60)^3 (55/60)^{60-3} \approx 0.1389
+        \end{align}
+
+        This is a good approximation, but it doesn't account for the possibility of multiple events in a single second. One solution is to divide our minute into even more fine-grained intervals. Let's try 600 deciseconds (tenths of a second):
+        """
+    )
+    return
+
+
+@app.cell(hide_code=True)
+def _(fig_to_image, mo, plt):
+    def create_decisecond_visualization(e_value):
+        # (Just showing the first 100 for clarity)
+        fig, ax = plt.subplots(figsize=(12, 2))
+
+        # Example events at 2.75s and 7.12s (convert to deciseconds)
+        events = [27.5, 71.2]
+    
+        for i in range(100):
+            color = 'royalblue' if any(i <= event_val < i + 1 for event_val in events) else 'lightgray'
+            ax.add_patch(plt.Rectangle((i, 0), 0.9, 1, color=color))
+
+        # Markers for events
+        for event in events:
+            if event < 100:  # Only show events in our visible range
+                ax.plot(event/10, 0.5, 'ro', markersize=10)  # Divide by 10 to convert to deciseconds
+
+        # Add labels
+        ax.set_xlim(0, 100)
+        ax.set_ylim(0, 1)
+        ax.set_yticks([])
+        ax.set_xticks([0, 20, 40, 60, 80, 100])
+        ax.set_xticklabels(['0s', '2s', '4s', '6s', '8s', '10s'])
+        ax.set_xlabel('Time (first 10 seconds shown)')
+        ax.set_title('One Minute Divided into 600 Decisecond Intervals (first 100 shown)')
+
+        plt.tight_layout()
+        plt.gca()
+        return fig
+
+    # Create viz and convert to image
+    _fig = create_decisecond_visualization(e_value=5)
+    _img = mo.image(fig_to_image(_fig), width="100%")
+
+    # Explanation
+    _explanation = mo.md(
+        r"""
+        With $n=600$ and $p=\frac{5}{600}=\frac{1}{120}$, we can recalculate our probabilities:
+
+        \begin{align}
+        P(X=1) &= {600 \choose 1} (5/600)^1 (595/600)^{600-1} \approx 0.0333 \\
+        P(X=2) &= {600 \choose 2} (5/600)^2 (595/600)^{600-2} \approx 0.0837 \\
+        P(X=3) &= {600 \choose 3} (5/600)^3 (595/600)^{600-3} \approx 0.1402
+        \end{align}
+
+        As we make our intervals smaller (increasing $n$), our approximation becomes more accurate.
+        """
+    )
+    mo.vstack([_fig, _explanation])
+    return (create_decisecond_visualization,)
+
+
+@app.cell(hide_code=True)
+def _(mo):
+    mo.md(
+        r"""
+        ## The Binomial Distribution in the Limit
+
+        What happens if we continue dividing our time interval into smaller and smaller pieces? Let's explore how the probabilities change as we increase the number of intervals:
+        """
+    )
+    return
+
+
+@app.cell(hide_code=True)
+def _(mo):
+    intervals_slider = mo.ui.slider(
+        start = 60, 
+        stop = 10000,
+        step=100,
+        value=600,
+        label="Number of intervals to divide a minute")
+    return (intervals_slider,)
+
+
+@app.cell(hide_code=True)
+def _(intervals_slider):
+    intervals_slider
+    return
+
+
+@app.cell(hide_code=True)
+def _(intervals_slider, np, pd, plt, stats):
+    def create_comparison_plot(n, lambda_value):
+        # Calculate probability
+        p = lambda_value / n
+
+        # Binomial probabilities
+        x_values = np.arange(0, 15)
+        binom_pmf = stats.binom.pmf(x_values, n, p)
+
+        # True Poisson probabilities
+        poisson_pmf = stats.poisson.pmf(x_values, lambda_value)
+
+        # DF for comparison
+        df = pd.DataFrame({
+            'Events': x_values,
+            f'Binomial(n={n}, p={p:.6f})': binom_pmf,
+            f'Poisson(λ=5)': poisson_pmf,
+            'Difference': np.abs(binom_pmf - poisson_pmf)
+        })
+
+        # Plot both PMFs
+        fig, ax = plt.subplots(figsize=(10, 6))
+
+        # Bar plot for the binomial
+        ax.bar(x_values - 0.2, binom_pmf, width=0.4, alpha=0.7, 
+            color='royalblue', label=f'Binomial(n={n}, p={p:.6f})')
+
+        # Bar plot for the Poisson
+        ax.bar(x_values + 0.2, poisson_pmf, width=0.4, alpha=0.7,
+            color='crimson', label='Poisson(λ=5)')
+
+        # Labels and title
+        ax.set_xlabel('Number of Events (k)')
+        ax.set_ylabel('Probability')
+        ax.set_title(f'Comparison of Binomial and Poisson PMFs with n={n}')
+        ax.legend()
+        ax.set_xticks(x_values)
+        ax.grid(alpha=0.3)
+
+        plt.tight_layout()
+        return df, fig, n, p
+
+    # Number of intervals from the slider
+    n = intervals_slider.value
+    _lambda = 5  # Fixed lambda for our example
+
+    # Cromparison plot
+    df, fig, n, p = create_comparison_plot(n, _lambda)
+    return create_comparison_plot, df, fig, n, p
+
+
+@app.cell(hide_code=True)
+def _(df, fig, fig_to_image, mo, n, p):
+    # table of values
+    _styled_df = df.style.format({
+        f'Binomial(n={n}, p={p:.6f})': '{:.6f}',
+        f'Poisson(λ=5)': '{:.6f}',
+        'Difference': '{:.6f}'
+    })
+
+    # Calculate the max absolute difference
+    _max_diff = df['Difference'].max()
+
+    # output
+    _chart = mo.image(fig_to_image(fig), width="100%")
+    _explanation = mo.md(f"**Maximum absolute difference between distributions: {_max_diff:.6f}**")
+    _table = mo.ui.table(df)
+
+    mo.vstack([_chart, _explanation, _table])
+    return
+
+
+@app.cell(hide_code=True)
+def _(mo):
+    mo.md(
+        r"""
+        As you can see from the interactive comparison above, as the number of intervals increases, the binomial distribution approaches the Poisson distribution! This is not a coincidence - the Poisson distribution is actually the limiting case of the binomial distribution when:
+
+        - The number of trials $n$ approaches infinity
+        - The probability of success $p$ approaches zero
+        - The product $np = \lambda$ remains constant
+
+        This relationship is why the Poisson distribution is so useful - it's easier to work with than a binomial with a very large number of trials and a very small probability of success.
+
+        ## Derivation of the Poisson PMF
+
+        Let's derive the Poisson PMF by taking the limit of the binomial PMF as $n \to \infty$. We start with:
+
+        $P(X=x) = \lim_{n \rightarrow \infty} {n \choose x} (\lambda / n)^x(1-\lambda/n)^{n-x}$
+
+        While this expression looks intimidating, it simplifies nicely:
+
+        \begin{align}
+        P(X=x) 
+        &= \lim_{n \rightarrow \infty} {n \choose x} (\lambda / n)^x(1-\lambda/n)^{n-x}
+            && \text{Start: binomial in the limit}\\
+        &= \lim_{n \rightarrow \infty}
+            {n \choose x} \cdot
+            \frac{\lambda^x}{n^x} \cdot
+            \frac{(1-\lambda/n)^{n}}{(1-\lambda/n)^{x}} 
+            && \text{Expanding the power terms} \\
+        &= \lim_{n \rightarrow \infty}
+            \frac{n!}{(n-x)!x!} \cdot
+            \frac{\lambda^x}{n^x} \cdot
+            \frac{(1-\lambda/n)^{n}}{(1-\lambda/n)^{x}} 
+            && \text{Expanding the binomial term} \\
+        &= \lim_{n \rightarrow \infty}
+            \frac{n!}{(n-x)!x!} \cdot
+            \frac{\lambda^x}{n^x} \cdot
+            \frac{e^{-\lambda}}{(1-\lambda/n)^{x}} 
+            && \text{Using limit rule } \lim_{n \rightarrow \infty}(1-\lambda/n)^{n} = e^{-\lambda}\\
+        &= \lim_{n \rightarrow \infty}
+            \frac{n!}{(n-x)!x!} \cdot
+            \frac{\lambda^x}{n^x} \cdot
+            \frac{e^{-\lambda}}{1} 
+            && \text{As } n \to \infty \text{, } \lambda/n \to 0\\
+        &= \lim_{n \rightarrow \infty}
+            \frac{n!}{(n-x)!} \cdot
+            \frac{1}{x!} \cdot
+            \frac{\lambda^x}{n^x} \cdot
+            e^{-\lambda}
+            && \text{Rearranging terms}\\
+        &= \lim_{n \rightarrow \infty}
+            \frac{n^x}{1} \cdot
+            \frac{1}{x!} \cdot
+            \frac{\lambda^x}{n^x} \cdot
+            e^{-\lambda}
+            && \text{As } n \to \infty \text{, } \frac{n!}{(n-x)!} \approx n^x\\
+        &= \lim_{n \rightarrow \infty}
+            \frac{\lambda^x}{x!} \cdot
+            e^{-\lambda}
+            && \text{Canceling } n^x\\
+        &= 
+            \frac{\lambda^x \cdot e^{-\lambda}}{x!} 
+            && \text{Simplifying}\\
+        \end{align}
+
+        This gives us our elegant Poisson PMF formula: $P(X=x) = \frac{\lambda^x \cdot e^{-\lambda}}{x!}$
+        """
+    )
+    return
+
+
+@app.cell(hide_code=True)
+def _(mo):
+    mo.md(
+        r"""
+        ## Poisson Distribution in Python
+
+        Python's `scipy.stats` module provides functions to work with the Poisson distribution. Let's see how to calculate probabilities and generate random samples.
+
+        First, let's calculate some probabilities for our ride-sharing example with $\lambda = 5$:
+        """
+    )
+    return
+
+
+@app.cell
+def _(stats):
+    _lambda = 5
+
+    # Calculate probabilities for X = 1, 2, 3
+    p_1 = stats.poisson.pmf(1, _lambda)
+    p_2 = stats.poisson.pmf(2, _lambda)
+    p_3 = stats.poisson.pmf(3, _lambda)
+
+    print(f"P(X=1) = {p_1:.5f}")
+    print(f"P(X=2) = {p_2:.5f}")
+    print(f"P(X=3) = {p_3:.5f}")
+
+    # Calculate cumulative probability P(X ≤ 3)
+    p_leq_3 = stats.poisson.cdf(3, _lambda)
+    print(f"P(X≤3) = {p_leq_3:.5f}")
+
+    # Calculate probability P(X > 10)
+    p_gt_10 = 1 - stats.poisson.cdf(10, _lambda)
+    print(f"P(X>10) = {p_gt_10:.5f}")
+    return p_1, p_2, p_3, p_gt_10, p_leq_3
+
+
+@app.cell(hide_code=True)
+def _(mo):
+    mo.md(r"""We can also generate random samples from a Poisson distribution and visualize their distribution:""")
+    return
+
+
+@app.cell(hide_code=True)
+def _(np, plt, stats):
+    def create_samples_plot(lambda_value, sample_size=1000):
+        # Random samples
+        samples = stats.poisson.rvs(lambda_value, size=sample_size)
+
+        # theoretical PMF
+        x_values = np.arange(0, max(samples) + 1)
+        pmf_values = stats.poisson.pmf(x_values, lambda_value)
+
+        # histograms to compare
+        fig, ax = plt.subplots(figsize=(10, 6))
+
+        # samples as a histogram
+        ax.hist(samples, bins=np.arange(-0.5, max(samples) + 1.5, 1), 
+                alpha=0.7, density=True, label='Random Samples')
+
+        # theoretical PMF
+        ax.plot(x_values, pmf_values, 'ro-', label='Theoretical PMF')
+
+        # labels and title
+        ax.set_xlabel('Number of Events')
+        ax.set_ylabel('Relative Frequency / Probability')
+        ax.set_title(f'1000 Random Samples from Poisson(λ={lambda_value})')
+        ax.legend()
+        ax.grid(alpha=0.3)
+
+        # annotations
+        ax.annotate(f'Sample Mean: {np.mean(samples):.2f}', 
+                    xy=(0.7, 0.9), xycoords='axes fraction',
+                    bbox=dict(boxstyle='round,pad=0.5', fc='yellow', alpha=0.3))
+        ax.annotate(f'Theoretical Mean: {lambda_value:.2f}', 
+                    xy=(0.7, 0.8), xycoords='axes fraction',
+                    bbox=dict(boxstyle='round,pad=0.5', fc='lightgreen', alpha=0.3))
+
+        plt.tight_layout()
+        return plt.gca()
+
+    # Use a lambda value of 5 for this example
+    _lambda = 5
+    create_samples_plot(_lambda)
+    return (create_samples_plot,)
+
+
+@app.cell(hide_code=True)
+def _(mo):
+    mo.md(
+        r"""
+        ## Changing Time Frames
+
+        One important property of the Poisson distribution is that the rate parameter $\lambda$ scales linearly with the time interval. If events occur at a rate of $\lambda$ per unit time, then over a period of $t$ units, the rate parameter becomes $\lambda \cdot t$.
+
+        For example, if a website receives an average of 5 requests per minute, what is the distribution of requests over a 20-minute period?
+
+        The rate parameter for the 20-minute period would be $\lambda = 5 \cdot 20 = 100$ requests.
+        """
+    )
+    return
+
+
+@app.cell(hide_code=True)
+def _(mo):
+    rate_slider = mo.ui.slider(
+        start = 0.1,
+        stop = 10,
+        step=0.1,
+        value=5,
+        label="Rate per unit time (λ)"
+    )
+
+    time_slider = mo.ui.slider(
+        start = 1,
+        stop = 60,
+        step=1,
+        value=20,
+        label="Time period (t units)"
+    )
+
+    controls = mo.vstack([
+        mo.md("### Adjust Parameters to See How Time Scaling Works"),
+        mo.hstack([rate_slider, time_slider], justify="space-between")
+    ])
+    return controls, rate_slider, time_slider
+
+
+@app.cell
+def _(controls):
+    controls.center()
+    return
+
+
+@app.cell(hide_code=True)
+def _(mo, np, plt, rate_slider, stats, time_slider):
+    def create_time_scaling_plot(rate, time_period):
+        # scaled rate parameter
+        lambda_value = rate * time_period
+
+        # PMF for values
+        max_x = max(30, int(lambda_value * 1.5))
+        x = np.arange(0, max_x + 1)
+        pmf = stats.poisson.pmf(x, lambda_value)
+
+        # plot
+        fig, ax = plt.subplots(figsize=(10, 6))
+
+        # PMF as bars
+        ax.bar(x, pmf, color='royalblue', alpha=0.7, 
+            label=f'PMF: Poisson(λ={lambda_value:.1f})')
+
+        # vertical line for mean
+        ax.axvline(x=lambda_value, color='red', linestyle='--', linewidth=2,
+                label=f'Mean = {lambda_value:.1f}')
+
+        # labels and title
+        ax.set_xlabel('Number of Events')
+        ax.set_ylabel('Probability')
+        ax.set_title(f'Poisson Distribution Over {time_period} Units (Rate = {rate}/unit)')
+
+        # better visualization if lambda is large
+        if lambda_value > 10:
+            ax.set_xlim(lambda_value - 4*np.sqrt(lambda_value), 
+                         lambda_value + 4*np.sqrt(lambda_value))
+
+        ax.legend()
+        ax.grid(alpha=0.3)
+
+        plt.tight_layout()
+
+        # Create relevant info markdown
+        info_text = f"""
+        When the rate is **{rate}** events per unit time and we observe for **{time_period}** units:
+
+        - The expected number of events is **{lambda_value:.1f}**
+        - The variance is also **{lambda_value:.1f}**
+        - The standard deviation is **{np.sqrt(lambda_value):.2f}**
+        - P(X=0) = {stats.poisson.pmf(0, lambda_value):.4f} (probability of no events)
+        - P(X≥10) = {1 - stats.poisson.cdf(9, lambda_value):.4f} (probability of 10 or more events)
+        """
+
+        return plt.gca(), info_text
+
+    # parameters from sliders
+    _rate = rate_slider.value
+    _time = time_slider.value
+
+    # store
+    _plot, _info_text = create_time_scaling_plot(_rate, _time)
+
+    # Display info as markdown
+    info = mo.md(_info_text)
+
+    mo.vstack([_plot, info], justify="center")
+    return create_time_scaling_plot, info
+
+
+@app.cell(hide_code=True)
+def _(mo):
+    mo.md(
+        r"""
+        ## 🤔 Test Your Understanding
+        Pick which of these statements about Poisson distributions you think are correct:
+
+        /// details | The variance of a Poisson distribution is always equal to its mean
+        ✅ Correct! For a Poisson distribution with parameter $\lambda$, both the mean and variance equal $\lambda$.
+        ///
+
+        /// details | The Poisson distribution can be used to model the number of successes in a fixed number of trials
+        ❌ Incorrect! That's the binomial distribution. The Poisson distribution models the number of events in a fixed interval of time or space, not a fixed number of trials.
+        ///
+
+        /// details | If $X \sim \text{Poisson}(\lambda_1)$ and $Y \sim \text{Poisson}(\lambda_2)$ are independent, then $X + Y \sim \text{Poisson}(\lambda_1 + \lambda_2)$
+        ✅ Correct! The sum of independent Poisson random variables is also a Poisson random variable with parameter equal to the sum of the individual parameters.
+        ///
+
+        /// details | As $\lambda$ increases, the Poisson distribution approaches a normal distribution
+        ✅ Correct! For large values of $\lambda$ (generally $\lambda > 10$), the Poisson distribution is approximately normal with mean $\lambda$ and variance $\lambda$.
+        ///
+
+        /// details | The probability of zero events in a Poisson process is always less than the probability of one event
+        ❌ Incorrect! For $\lambda < 1$, the probability of zero events ($e^{-\lambda}$) is actually greater than the probability of one event ($\lambda e^{-\lambda}$).
+        ///
+
+        /// details | The Poisson distribution has a single parameter $\lambda$, which always equals the average number of events per time period
+        ✅ Correct! The parameter $\lambda$ represents the average rate of events, and it uniquely defines the distribution.
+        ///
+        """
+    )
+    return
+
+
+@app.cell(hide_code=True)
+def _(mo):
+    mo.md(
+        r"""
+        ## Summary
+
+        The Poisson distribution is one of those incredibly useful tools that shows up all over the place. I've always found it fascinating how such a simple formula can model so many real-world phenomena - from website traffic to radioactive decay.
+
+        What makes the Poisson really cool is that it emerges naturally as we try to model rare events occurring over a continuous interval. Remember that visualization where we kept dividing time into smaller and smaller chunks? As we showed, when you take a binomial distribution and let the number of trials approach infinity while keeping the expected value constant, you end up with the elegant Poisson formula.
+
+        The key things to remember about the Poisson distribution:
+
+        - It models the number of events occurring in a fixed interval of time or space, assuming events happen at a constant average rate and independently of each other
+
+        - Its PMF is given by the elegantly simple formula $P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!}$
+
+        - Both the mean and variance equal the parameter $\lambda$, which represents the average number of events per interval
+
+        - It's related to the binomial distribution as a limiting case when $n \to \infty$, $p \to 0$, and $np = \lambda$ remains constant
+
+        - The rate parameter scales linearly with the length of the interval - if events occur at rate $\lambda$ per unit time, then over $t$ units, the parameter becomes $\lambda t$
+
+        From modeling website traffic and customer arrivals to defects in manufacturing and radioactive decay, the Poisson distribution provides a powerful and mathematically elegant way to understand random occurrences in our world.
+        """
+    )
+    return
+
+
+@app.cell(hide_code=True)
+def _(mo):
+    mo.md(r"""Appendix code (helper functions, variables, etc.):""")
+    return
+
+
+@app.cell
+def _():
+    import marimo as mo
+    return (mo,)
+
+
+@app.cell(hide_code=True)
+def _():
+    import numpy as np
+    import matplotlib.pyplot as plt
+    import scipy.stats as stats
+    import pandas as pd
+    import altair as alt
+    from wigglystuff import TangleSlider
+    return TangleSlider, alt, np, pd, plt, stats
+
+
+@app.cell(hide_code=True)
+def _():
+    import io
+    import base64
+    from matplotlib.figure import Figure
+
+    # Helper function to convert mpl figure to an image format mo.image can hopefully handle
+    def fig_to_image(fig):
+        buf = io.BytesIO()
+        fig.savefig(buf, format='png')
+        buf.seek(0)
+        data = f"data:image/png;base64,{base64.b64encode(buf.read()).decode('utf-8')}"
+        return data
+    return Figure, base64, fig_to_image, io
+
+
+if __name__ == "__main__":
+    app.run()