Lesson 8: Continuous Random Variables

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Understand the purpose of a probability density function for a continuous random variable

Remember this from previous classes?

\[ P(X=x) = \begin{cases} \dfrac{1}{6}, & x=1, \\[6pt] \dfrac{1}{6}, & x=2, \\[6pt] \dfrac{1}{6}, & x=3, \\[6pt] \dfrac{1}{6}, & x=4, \\[6pt] \dfrac{1}{6}, & x=5, \\[6pt] \dfrac{1}{6}, & x=6, \\[6pt] 0, & \text{otherwise}. \end{cases} \]

This is the probability mass function of a fair die.

What do we know about it?

  • Probabilities are non-negative: \(P(X=x) \geq 0\).
  • The total probability is 1:
    \[ \sum_x P(X=x) = 1 \]
  • We can compute probabilities of events by adding up the masses.
    • Example: \(P(X \leq 3) = \tfrac{1}{6} + \tfrac{1}{6} + \tfrac{1}{6} = \tfrac{1}{2}\).

So what about continuous random variables?

This probability distribution function isn’t all that different from the discrete version.

Here are the rules that all PDFs must follow:

  • \(f(x) \geq 0\) for all \(x\).
  • The total area under the curve is 1:
    \[ \int_{-\infty}^{\infty} f(x)\,dx = 1 \]
  • Probabilities are found by areas, not points:
    \[ P(a \leq X \leq b) = \int_a^b f(x)\,dx \]
  • For any single point,
    \[ P(X=c) = 0 \]

Example of a PDF

Consider the function

\[ f(x) = \begin{cases} 2x, & 0 \leq x \leq 1, \\ 0, & \text{otherwise}. \end{cases} \]

Check the properties:

  • Nonnegative?
    \(f(x) = 2x \geq 0\) on \([0,1]\). ✅

  • Total area?
    \[ \int_{-\infty}^\infty f(x)\,dx = \int_0^1 2x\,dx = \left[x^2\right]_0^1 = 1 \] So it integrates to 1. ✅

  • Probability of an interval?
    \[ P(0.2 \leq X \leq 0.5) = \int_{0.2}^{0.5} 2x \, dx = \left[x^2\right]_{0.2}^{0.5} = (0.25 - 0.04) = 0.21 \]

So this is a valid PDF! It’s a “triangle-shaped” distribution on \([0,1]\) that places more weight near 1 than near 0.

  • Probability of a single point is zero: \(P(X=c)=0\).
    For a continuous random variable with PDF \(f(x)\), probability comes from area, so any single point has zero width: \[ P(X=c) \;=\; \int_{c}^{c} f(x)\,dx \;=\; 0. \]

For example: \(P(X=.6)=0\)

\[ P(X=.6) \;=\; \int_{.6}^{.6} f(x)\,dx \;=\; 0. \]

Understand the purpose of a cumulative density function for a continuous random variable

Remember this from previous classes?

\[ F(x) = \begin{cases} 0, & x < 1, \\[6pt] \dfrac{1}{6}, & 1 \leq x < 2, \\[6pt] \dfrac{2}{6}, & 2 \leq x < 3, \\[6pt] \dfrac{3}{6}, & 3 \leq x < 4, \\[6pt] \dfrac{4}{6}, & 4 \leq x < 5, \\[6pt] \dfrac{5}{6}, & 5 \leq x < 6, \\[6pt] 1, & x \geq 6. \end{cases} \]

This step function is the CDF of the discrete die. It accumulates probability as we move to the right.

What is a Cumulative Distribution Function (in general)?

For any random variable \(X\) (discrete or continuous), the cumulative distribution function is
\[ F(x) \;=\; P(X \le x). \]

Things that are true of all CDFs:
- Nondecreasing: if \(a < b\) then \(F(a) \le F(b)\).
- Right-continuous: \(\lim_{x \downarrow c} F(x) = F(c)\).
- Event probabilities from \(F\): for any \(a < b\),
\[ P(a \le X \le b) \;=\; F(b) - F(a). \]

Deriving a CDF from our PDF

Recall our PDF: \[ f(x) = \begin{cases} 2x, & 0 \le x \le 1,\\ 0, & \text{otherwise}. \end{cases} \]

Integrate to get \(F\):

\[ F(x) \;=\; \int_{0}^{x} 2t\,dt \;=\; x^{2}. \]

So for any \(x\) in \([0,1]\),
\[ F(x) = x^2. \]

Quick checks:
- \(F\) is nondecreasing and continuous (no jumps).
- \(F(0)=0\), \(F(1)=1\).
- Example reads: \(F(0.5)=0.5^2=0.25\), \(F(0.7)=0.49\).
- Interval probability via CDF:
\[ P(0.2 \le X \le 0.5) = F(0.5) - F(0.2) = 0.25 - 0.04 = 0.21. \]

Reading probabilities from the CDF:

  • \(P(X \le 0.5) = F(0.5) = 0.25\)
    \[ P(X \le 0.5) = \int_{0}^{0.5} 2t\,dt = 0.25 \]

  • \(P(X > 0.7) = 1 - F(0.7) = 1 - 0.49 = 0.51\)
    \[ P(X > 0.7) = \int_{0.7}^{1} 2t\,dt = 0.51 \]

  • \(P(0.2 \le X \le 0.5) = F(0.5) - F(0.2) = 0.21\)
    \[ P(0.2 \le X \le 0.5) = \int_{0.2}^{0.5} 2t\,dt = 0.21 \]

Calculate and interpret the expected value of a continuous random variable

Remember this from previous classes?

For a discrete random variable (like a fair die), the expected value is the weighted average of the possible outcomes:

\[ E[X] = \sum_x x \cdot P(X=x). \]

For the fair die: \[ E[X] = 1\cdot \tfrac{1}{6} + 2\cdot \tfrac{1}{6} + 3\cdot \tfrac{1}{6} + 4\cdot \tfrac{1}{6} + 5\cdot \tfrac{1}{6} + 6\cdot \tfrac{1}{6} = 3.5 \]

Continuous random variables

For a continuous random variable with PDF \(f(x)\), the expected value is defined by an integral:

\[ E[X] = \int_{-\infty}^{\infty} x \, f(x)\,dx. \]

This is the continuous version of the same weighted average idea — instead of summing over points, we integrate over the real line.

Example: our PDF

Recall our PDF:
\[ f(x) = \begin{cases} 2x, & 0 \leq x \leq 1,\\ 0, & \text{otherwise}. \end{cases} \]

Compute the expected value:
\[ E[X] = \int_{0}^{1} x \cdot (2x)\,dx = \int_{0}^{1} 2x^{2}\,dx. \]

\[ E[X] = \left[\tfrac{2}{3}x^{3}\right]_{0}^{1} = \tfrac{2}{3}. \]

So the mean of this distribution is \(\tfrac{2}{3} \approx 0.667\).

Calculate and interpret the variance and standard deviation of a continuous random variable

Remember this from previous classes?

For a discrete random variable, variance measures how spread out the values are around the mean:

\[ \text{Var}(X) = \sum_x (x - \mu)^2 \, P(X=x), \]

where \(\mu = E[X]\).

The standard deviation is just the square root of the variance:

\[ \sigma = \sqrt{\text{Var}(X)}. \]

For a fair six-sided die:

  • We already know the mean is
    \[ E[X] = \frac{1+2+3+4+5+6}{6} = 3.5. \]

  • Compute the variance:
    \[ \text{Var}(X) = \sum_{x=1}^{6} (x - 3.5)^2 \cdot \tfrac{1}{6}. \]

\[ = \frac{(1-3.5)^2 + (2-3.5)^2 + (3-3.5)^2 + (4-3.5)^2 + (5-3.5)^2 + (6-3.5)^2}{6} = \frac{35}{12} \approx 2.92. \]

  • Then the standard deviation is
    \[ \sigma = \sqrt{\tfrac{35}{12}} \approx 1.71. \]

So the die’s outcomes are typically about 1.7 away from the mean value of 3.5.

Continuous random variables

For a continuous random variable with PDF \(f(x)\), we replace the sum with an integral:

\[ \text{Var}(X) = \int_{-\infty}^{\infty} (x - \mu)^2 f(x)\,dx, \]

where \(\mu = E[X]\).

But this integral is often hard to compute directly, so there is a shortcut formula we can use:

\[ \text{Var}(X) = E[X^2] - \big(E[X]\big)^2, \]

where

\[ E[X^2] = \int_{-\infty}^{\infty} x^2 f(x)\,dx. \]

Example: our PDF

\[ f(x) = \begin{cases} 2x, & 0 \leq x \leq 1,\\ 0, & \text{otherwise}. \end{cases} \]

  • From before, the mean is \(\mu = E[X] = \tfrac{2}{3}\).

  • Compute \(E[X^2]\):
    \[ E[X^2] = \int_{0}^{1} x^2 \cdot (2x)\,dx = \int_{0}^{1} 2x^3\,dx = \left[\tfrac{1}{2}x^4\right]_{0}^{1} = \tfrac{1}{2}. \]

  • Now variance:
    \[ \text{Var}(X) = \tfrac{1}{2} - \left(\tfrac{2}{3}\right)^2 = \tfrac{1}{2} - \tfrac{4}{9} = \tfrac{1}{18}. \]

  • Standard deviation:
    \[ \sigma = \sqrt{\tfrac{1}{18}} \approx 0.236. \]

Practice Problem

Consider the random variable \(X\) with PDF

\[ f(x) = \begin{cases} 2(1-x), & 0 \le x \le 1, \\ 0, & \text{otherwise}. \end{cases} \]

1. Plot the PDF

2. Find \(P(X = 0.4)\)

For continuous random variables:

\[ P(X=0.4) = \int_{0.4}^{0.4} f(x)\,dx = 0. \]

3. Find \(P(0.2 \le X \le 0.6)\)

\[ P(0.2 \le X \le 0.6) = \int_{0.2}^{0.6} 2(1-x)\,dx = \Big[2x - x^2\Big]_{0.2}^{0.6}. \]

At \(0.6\): \(2(0.6)-0.6^2=0.84\).
At \(0.2\): \(0.36\).

Difference: \(0.48\).

4. Find the CDF \(F(x)\)

For \(0 \le x \le 1\):

\[ F(x) = \int_{0}^{x} 2(1-t)\,dt = 2x - x^2. \]

So

\[ F(x) = \begin{cases} 0, & x < 0, \\ 2x - x^2, & 0 \le x \le 1, \\ 1, & x \ge 1. \end{cases} \]

5. Find \(P(0.2 \le X \le 0.6)\) using the CDF

\[ P(0.2 \le X \le 0.6) = F(0.6) - F(0.2). \]

\(F(0.6)=0.84\), \(F(0.2)=0.36\), difference = \(0.48\).
✅ Same as before.

6. Find the expected value \(E[X]\)

\[ E[X] = \int_{0}^{1} x \cdot 2(1-x)\,dx = \int_{0}^{1} (2x - 2x^2)\,dx. \]

\[ = \Big[x^2 - \tfrac{2}{3}x^3\Big]_0^1 = \tfrac{1}{3}. \]

7. Find the variance \(\text{Var}(X)\)

Compute \(E[X^2]\):

\[ E[X^2] = \int_{0}^{1} x^2 \cdot 2(1-x)\,dx = \tfrac{1}{6}. \]

So
\[ \text{Var}(X) = E[X^2] - (E[X])^2 = \tfrac{1}{6} - \left(\tfrac{1}{3}\right)^2 = \tfrac{1}{18}. \]

8. Find the standard deviation

\[ \sigma = \sqrt{\tfrac{1}{18}} \approx 0.236. \]

9. Find \(P(\mu - 2\sigma \le X \le \mu + 2\sigma)\)

\(\mu=\tfrac{1}{3}\), \(\sigma \approx 0.236\), so \(2\sigma \approx 0.472\).

Range: \([\tfrac{1}{3} - 0.472,\;\tfrac{1}{3} + 0.472] = [-0.139,\;0.805]\).
Since the support is \([0,1]\), we use \([0,0.805]\).

\[ P(0 \le X \le 0.805) = F(0.805) - F(0) = F(0.805). \]

\(F(0.805) = 2(0.805) - (0.805)^2 \approx 0.962\).

10. Find \(P(X < \mu - \sigma \;\text{ or }\; X > \mu + \sigma)\)

\(\mu=\tfrac{1}{3}\), \(\sigma \approx 0.236\).

Interval within 1 SD: \([0.097,\,0.569]\).

So the probability outside is
\[ P(X < 0.097) + P(X > 0.569). \]

Using the CDF:
\[ P(X < 0.097) = F(0.097) \approx 0.185, \] \[ P(X > 0.569) = 1 - F(0.569) \approx 1 - 0.814 = 0.186. \]

Total = \(0.185 + 0.186 = 0.371\).

Board Problem

Consider the random variable \(X\) with PDF

\[ f(x) = \begin{cases} \dfrac{1}{4}, & 2 \le x \le 6,\\ 0, & \text{otherwise}. \end{cases} \]

  1. Plot the PDF
  2. Find \(P(X = 3)\)
  3. Find \(P(2.5 \le X \le 4.5)\)
  4. Find the CDF \(F(x)\)
  5. Find \(P(2.5 \le X \le 4.5)\) using the CDF. Is it the same as before?
  6. Find the expected value \(E[X]\)
  7. Find the variance \(\mathrm{Var}(X)\)
  8. Find the standard deviation
  9. Find \(P(\mu - 2\sigma \le X \le \mu + 2\sigma)\)
  10. Find \(P(X < \mu - \sigma \;\text{ or }\; X > \mu + \sigma)\)

1. Plot the PDF

2. \(P(X=3)\)

For a continuous RV:

\[ P(X=3) = \int_{3}^{3} f(x)\,dx = 0. \]

3. \(P(2.5 \le X \le 4.5)\)

\[ P(2.5 \le X \le 4.5) = \int_{2.5}^{4.5} \tfrac{1}{4}\,dx = \tfrac{1}{4}\,(4.5-2.5) = \tfrac{1}{2}. \]

4. CDF \(F(x)\)

\[ F(x) = \begin{cases} 0, & x < 2,\\[4pt] \tfrac{x-2}{4}, & 2 \le x \le 6,\\[8pt] 1, & x \ge 6. \end{cases} \]

5. \(P(2.5 \le X \le 4.5)\) using \(F\)

\[ P(2.5 \le X \le 4.5) = F(4.5) - F(2.5). \]

\(F(4.5) = (4.5-2)/4 = 0.625\),
\(F(2.5) = (2.5-2)/4 = 0.125\).

Difference = \(0.5\). ✅ Same as before.

6. Expected value \(E[X]\)

\[ E[X] = \int_{2}^{6} x \cdot \tfrac{1}{4}\,dx = \tfrac{1}{4}\,\left[\tfrac{x^2}{2}\right]_{2}^{6} = \tfrac{1}{4}\,\Big(\tfrac{36}{2}-\tfrac{4}{2}\Big) = \tfrac{1}{4}\,(18-2) = 4. \]

7. Variance \(\mathrm{Var}(X)\)

First compute

\[ E[X^2] = \int_{2}^{6} x^2 \cdot \tfrac{1}{4}\,dx = \tfrac{1}{4}\,\left[\tfrac{x^3}{3}\right]_{2}^{6} = \tfrac{1}{4}\,\Big(\tfrac{216}{3}-\tfrac{8}{3}\Big) = \tfrac{1}{4}\cdot \tfrac{208}{3} = \tfrac{52}{3}. \]

So

\[ \mathrm{Var}(X) = E[X^2] - (E[X])^2 = \tfrac{52}{3} - 4^2 = \tfrac{52}{3} - 16 = \tfrac{4}{3}. \]

8. Standard deviation

\[ \sigma = \sqrt{\tfrac{4}{3}} = \tfrac{2}{\sqrt{3}} \approx 1.155. \]

9. Probability within \(2\sigma\) of the mean

Here \(\mu=4\), \(2\sigma \approx 2.309\).

Interval \([4-2.309,\,4+2.309] = [1.691,\,6.309]\).
Intersect with support \([2,6]\) gives \([2,6]\).

So

\[ P(\mu-2\sigma \le X \le \mu+2\sigma) = 1. \]

10. Probability more than \(1\sigma\) from the mean

Interval within \(1\sigma\):

\[ [\mu-\sigma,\;\mu+\sigma] = [4-1.155,\,4+1.155] = [2.845,\,5.155]. \]

Length of this interval = \(2.31\).
Since the PDF is flat on \([2,6]\) (length 4):

\[ P(|X-\mu|\le\sigma) = \tfrac{2.31}{4} \approx 0.577. \]

Therefore

\[ P(|X-\mu| > \sigma) = 1 - 0.577 = 0.423. \]

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