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probability/09_random_variables.py
CHANGED
@@ -10,7 +10,7 @@
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import marimo
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__generated_with = "0.11.
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app = marimo.App(width="medium", app_title="Random Variables")
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@@ -72,7 +72,7 @@ def _(mo):
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r"""
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## Properties of Random Variables
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Each random variable has several key properties
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| Property | Description | Example |
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|----------|-------------|---------|
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| Support/Range | Possible values | $\{0,1,2,...,n\}$ for binomial |
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| Distribution | PMF or PDF | $p_X(x)$ or $f_X(x)$ |
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| Expectation | Weighted average | $E[X]$ |
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| Variance | Measure of spread |
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| Standard Deviation | Square root of variance | $\sigma_X$ |
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| Mode | Most likely value | argmax$_x$ $p_X(x)$ |
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def _(np, plt):
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def die_pmf(x):
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if x in [1, 2, 3, 4, 5, 6]:
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return 1/6
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return 0
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# Plot the PMF
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_x = np.arange(1, 7)
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probabilities = [die_pmf(i) for i in _x]
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plt.figure(figsize=(8,
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plt.bar(_x, probabilities)
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plt.title("PMF of Rolling a Fair Die")
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plt.xlabel("Outcome")
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_pdf = stats.norm.pdf(_x, loc=0, scale=1)
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plt.figure(figsize=(8, 4))
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plt.plot(_x, _pdf,
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plt.fill_between(_x, _pdf, where=(_x >= -1) & (_x <= 1), alpha=0.3)
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plt.title("Standard Normal Distribution")
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plt.xlabel("x")
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@@ -214,7 +214,7 @@ def _(np):
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@app.cell
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def _(E_X):
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return
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r"""
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## Variance
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The variance
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This can be computed as:
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Properties:
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1.
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2.
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"""
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)
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return
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@app.cell
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def _(E_X, die_probs, die_values, np):
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def variance_discrete(x_values, probabilities, expected_value):
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squared_diff = [(x - expected_value)**2 for x in x_values]
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return sum(d * p for d, p in zip(squared_diff, probabilities))
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# Example: Variance of a fair die roll
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@app.cell(hide_code=True)
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def _(coin_var, mo, normal_var, uniform_var):
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mo.md(
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-
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Let's look at some calculated variances:
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- Fair coin (X = 0 or 1): Var(X) = {coin_var:.4f}
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- Standard normal distribution (discretized): Var(X) ≈ {normal_var:.4f}
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- Uniform distribution on [0,1] (discretized): Var(X) ≈ {uniform_var:.4f}
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"""
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)
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return
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@@ -320,7 +347,7 @@ def _(coin_var, mo, normal_var, uniform_var):
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def _(mo):
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mo.md(
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r"""
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-
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This example shows the relationship between a Binomial distribution (discrete) and its Normal approximation (continuous).
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The parameters control both distributions:
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@app.cell
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def _(mo, n_trials, p_success):
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mo.hstack([n_trials, p_success], justify=
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return
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@app.cell(hide_code=True)
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def _(n_trials, np, p_success, plt, stats):
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fig, (ax1, ax2) = plt.subplots(1, 2, figsize=(12,
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# Discrete: Binomial PMF
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k = np.arange(0, n_trials.value + 1)
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pmf = stats.binom.pmf(k, n_trials.value, p_success.value)
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ax1.bar(k, pmf, alpha=0.8, color=
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ax1.set_title(f
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ax1.set_xlabel(
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ax1.set_ylabel(
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ax1.grid(True, alpha=0.3)
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# Continuous: Normal PDF approx.
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mu = n_trials.value * p_success.value
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sigma = np.sqrt(n_trials.value * p_success.value * (1-p_success.value))
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x = np.linspace(max(0, mu - 4*sigma), min(n_trials.value, mu + 4*sigma), 100)
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pdf = stats.norm.pdf(x, mu, sigma)
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ax2.plot(x, pdf,
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ax2.fill_between(x, pdf, alpha=0.3, color=
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ax2.set_title(f
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ax2.set_xlabel(
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ax2.set_ylabel(
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ax2.grid(True, alpha=0.3)
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# Set consistent x-axis limits for better comparison
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**Current Distribution Properties:**
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- Mean (μ) = {n_trials.value * p_success.value:.2f}
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- Standard Deviation (σ) = {np.sqrt(n_trials.value * p_success.value * (1-p_success.value)):.2f}
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Notice how the Normal distribution (right) approximates the Binomial distribution (left) better when:
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return
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@app.cell(hide_code=True)
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def _(mo):
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mo.md(
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r"""
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## Common Distributions
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1. Bernoulli Distribution
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- Models a single success/failure experiment
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- $P(X = 1) = p$, $P(X = 0) = 1-p$
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- $E[X] = p$, $Var(X) = p(1-p)$
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2. Binomial Distribution
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- Models number of successes in $n$ independent trials
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- $P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}$
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- $E[X] = np$, $Var(X) = np(1-p)$
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3. Normal Distribution
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- Bell-shaped curve defined by mean $\mu$ and variance $\sigma^2$
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- PDF: $f_X(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$
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- $E[X] = \mu$, $Var(X) = \sigma^2$
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"""
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)
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return
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@app.cell(hide_code=True)
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def _(mo):
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mo.md(
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1. The support of $X$
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2. The PMF $p_X(x)$
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3. $E[X]$ and
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<details>
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<summary>Solution</summary>
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1. The PDF integrates to 1
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2. $E[X] = 1/2$
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3.
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Try solving this yourself first, then check the solution below.
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"""
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$E[X] = \int_0^1 x \cdot 1 \, dx = [\frac{x^2}{2}]_0^1 = \frac{1}{2} - 0 = \frac{1}{2}$
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3. **Variance**:
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First calculate $E[X^2]$:
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$E[X^2] = \int_0^1 x^2 \cdot 1 \, dx = [\frac{x^3}{3}]_0^1 = \frac{1}{3}$
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Then:
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-
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"""
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)
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return (mktext,)
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import marimo
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__generated_with = "0.11.10"
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app = marimo.App(width="medium", app_title="Random Variables")
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r"""
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## Properties of Random Variables
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+
Each random variable has several key properties:
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| Property | Description | Example |
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|----------|-------------|---------|
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| Support/Range | Possible values | $\{0,1,2,...,n\}$ for binomial |
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| Distribution | PMF or PDF | $p_X(x)$ or $f_X(x)$ |
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| Expectation | Weighted average | $E[X]$ |
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| Variance | Measure of spread | $\text{Var}(X)$ |
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| Standard Deviation | Square root of variance | $\sigma_X$ |
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| Mode | Most likely value | argmax$_x$ $p_X(x)$ |
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def _(np, plt):
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def die_pmf(x):
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if x in [1, 2, 3, 4, 5, 6]:
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return 1 / 6
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return 0
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# Plot the PMF
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_x = np.arange(1, 7)
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probabilities = [die_pmf(i) for i in _x]
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plt.figure(figsize=(8, 2))
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plt.bar(_x, probabilities)
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plt.title("PMF of Rolling a Fair Die")
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plt.xlabel("Outcome")
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_pdf = stats.norm.pdf(_x, loc=0, scale=1)
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plt.figure(figsize=(8, 4))
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plt.plot(_x, _pdf, "b-", label="PDF")
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plt.fill_between(_x, _pdf, where=(_x >= -1) & (_x <= 1), alpha=0.3)
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plt.title("Standard Normal Distribution")
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plt.xlabel("x")
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@app.cell
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def _(E_X):
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E_X
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return
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r"""
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## Variance
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The variance $\text{Var}(X)$ measures the spread of a random variable around its mean:
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$\text{Var}(X) = E[(X - E[X])^2]$
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This can be computed as:
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$\text{Var}(X) = E[X^2] - (E[X])^2$
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Properties:
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1. $\text{Var}(aX) = a^2Var(X)$
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2. $\text{Var}(X + b) = Var(X)$
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"""
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)
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return
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@app.cell
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def _(E_X, die_probs, die_values, np):
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def variance_discrete(x_values, probabilities, expected_value):
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squared_diff = [(x - expected_value) ** 2 for x in x_values]
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return sum(d * p for d, p in zip(squared_diff, probabilities))
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# Example: Variance of a fair die roll
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@app.cell(hide_code=True)
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def _(coin_var, mo, normal_var, uniform_var):
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mo.md(
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rf"""
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Let's look at some calculated variances:
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- Fair coin (X = 0 or 1): $\text{{Var}}(X) = {coin_var:.4f}$
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- Standard normal distribution (discretized): $\text{{Var(X)}} ≈ {normal_var:.4f}$
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- Uniform distribution on $[0,1]$ (discretized): $\text{{Var(X)}} ≈ {uniform_var:.4f}$
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"""
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)
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return
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@app.cell(hide_code=True)
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def _(mo):
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mo.md(
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r"""
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## Common Distributions
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+
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1. Bernoulli Distribution
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- Models a single success/failure experiment
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- $P(X = 1) = p$, $P(X = 0) = 1-p$
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- $E[X] = p$, $\text{Var}(X) = p(1-p)$
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+
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2. Binomial Distribution
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- Models number of successes in $n$ independent trials
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- $P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}$
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- $E[X] = np$, $\text{Var}(X) = np(1-p)$
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+
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3. Normal Distribution
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- Bell-shaped curve defined by mean $\mu$ and variance $\sigma^2$
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- PDF: $f_X(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$
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- $E[X] = \mu$, $\text{Var}(X) = \sigma^2$
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"""
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)
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return
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def _(mo):
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mo.md(
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r"""
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### Example: Comparing Discrete and Continuous Distributions
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This example shows the relationship between a Binomial distribution (discrete) and its Normal approximation (continuous).
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The parameters control both distributions:
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@app.cell
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def _(mo, n_trials, p_success):
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mo.hstack([n_trials, p_success], justify="space-around")
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return
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@app.cell(hide_code=True)
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def _(n_trials, np, p_success, plt, stats):
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fig, (ax1, ax2) = plt.subplots(1, 2, figsize=(12, 3))
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# Discrete: Binomial PMF
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k = np.arange(0, n_trials.value + 1)
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pmf = stats.binom.pmf(k, n_trials.value, p_success.value)
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ax1.bar(k, pmf, alpha=0.8, color="#1f77b4", label="PMF")
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ax1.set_title(f"Binomial PMF (n={n_trials.value}, p={p_success.value})")
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ax1.set_xlabel("Number of Successes")
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ax1.set_ylabel("Probability")
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ax1.grid(True, alpha=0.3)
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# Continuous: Normal PDF approx.
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mu = n_trials.value * p_success.value
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sigma = np.sqrt(n_trials.value * p_success.value * (1 - p_success.value))
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x = np.linspace(max(0, mu - 4 * sigma), min(n_trials.value, mu + 4 * sigma), 100)
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pdf = stats.norm.pdf(x, mu, sigma)
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ax2.plot(x, pdf, "r-", linewidth=2, label="PDF")
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ax2.fill_between(x, pdf, alpha=0.3, color="red")
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ax2.set_title(f"Normal PDF (μ={mu:.1f}, σ={sigma:.1f})")
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ax2.set_xlabel("Continuous Approximation")
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ax2.set_ylabel("Density")
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ax2.grid(True, alpha=0.3)
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# Set consistent x-axis limits for better comparison
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**Current Distribution Properties:**
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- Mean (μ) = {n_trials.value * p_success.value:.2f}
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- Standard Deviation (σ) = {np.sqrt(n_trials.value * p_success.value * (1 - p_success.value)):.2f}
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Notice how the Normal distribution (right) approximates the Binomial distribution (left) better when:
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return
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@app.cell(hide_code=True)
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def _(mo):
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mo.md(
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1. The support of $X$
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2. The PMF $p_X(x)$
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3. $E[X]$ and $\text{Var}(X)$
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<details>
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<summary>Solution</summary>
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1. The PDF integrates to 1
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2. $E[X] = 1/2$
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3. $\text{Var}(X) = 1/12$
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Try solving this yourself first, then check the solution below.
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"""
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$E[X] = \int_0^1 x \cdot 1 \, dx = [\frac{x^2}{2}]_0^1 = \frac{1}{2} - 0 = \frac{1}{2}$
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3. **Variance**:
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$\text{Var}(X) = E[X^2] - (E[X])^2$
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First calculate $E[X^2]$:
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$E[X^2] = \int_0^1 x^2 \cdot 1 \, dx = [\frac{x^3}{3}]_0^1 = \frac{1}{3}$
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Then:
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$\text{Var}(X) = \frac{1}{3} - (\frac{1}{2})^2 = \frac{1}{3} - \frac{1}{4} = \frac{1}{12}$
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"""
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)
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return (mktext,)
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