Continuous Random Variables & PDFs

Worked Example 1: Determining Normalization Constant & Interval Probability

A continuous random variable $X$ has probability density function defined by:

$$f(x) = \begin{cases} k x(2 - x) & \text{for } 0 \le x \le 2 \\ 0 & \text{otherwise} \end{cases}$$

(a) Show that $k = \frac{3}{4}$. [3 marks]
(b) Calculate $P(0.5 \le X \le 1.5)$. [3 marks]

Full Markscheme Solution:

(a) Normalization Condition:

For any valid probability density function, the total integral over its support must equal $1$:

$$\int_0^2 f(x)\,dx = 1 \implies \int_0^2 k(2x - x^2)\,dx = 1 \quad [M1]$$

Integrate term-by-term:

$$k \left[ x^2 - \frac{x^3}{3} \right]_0^2 = 1 \quad [A1]$$

Substitute the limits:

$$k \left[ \left(2^2 - \frac{2^3}{3}\right) - 0 \right] = 1 \implies k \left( 4 - \frac{8}{3} \right) = 1 \implies k \left(\frac{4}{3}\right) = 1$$ $$k = \frac{3}{4} \quad [AG]$$

(b) Interval Probability:

$$P(0.5 \le X \le 1.5) = \int_{0.5}^{1.5} \frac{3}{4}(2x - x^2)\,dx \quad [M1]$$ $$= \frac{3}{4}\left[ x^2 - \frac{x^3}{3} \right]_{0.5}^{1.5} \quad [A1]$$

Evaluating at upper limit $x = 1.5 = \frac{3}{2}$:

$$\left(\frac{3}{2}\right)^2 - \frac{(3/2)^3}{3} = \frac{9}{4} - \frac{27/8}{3} = \frac{9}{4} - \frac{9}{8} = \frac{9}{8}$$

Evaluating at lower limit $x = 0.5 = \frac{1}{2}$:

$$\left(\frac{1}{2}\right)^2 - \frac{(1/2)^3}{3} = \frac{1}{4} - \frac{1/8}{3} = \frac{1}{4} - \frac{1}{24} = \frac{5}{24}$$

Subtracting:

$$P(0.5 \le X \le 1.5) = \frac{3}{4} \left( \frac{9}{8} - \frac{5}{24} \right) = \frac{3}{4} \left( \frac{27 - 5}{24} \right) = \frac{3}{4} \left( \frac{22}{24} \right) = \frac{3}{4} \cdot \frac{11}{12} = \frac{11}{16} = 0.6875 \quad [A1]$$

Examiner Commentary: In "Show that" questions, the substitution of limits $\left(4 - \frac{8}{3}\right)$ must be shown explicitly before stating $k = \frac{3}{4}$ to earn the $[AG]$ mark. Note that because $f(x)$ is symmetric about $x = 1$, students could also compute $2 \int_{0.5}^1 f(x)\,dx$ for an equally valid alternative method.

Worked Example 2: Piecewise Cumulative Distribution Function & Exact Median

For the continuous random variable $X$ with PDF $f(x) = \frac{3}{4}(2x - x^2)$ for $0 \le x \le 2$:
(a) Find the cumulative distribution function $F(x)$, defining it for all $x \in \mathbb{R}$. [4 marks]
(b) Determine the exact median of $X$. [3 marks]

Full Markscheme Solution:

(a) Deriving $F(x)$:

• For $x < 0$, no probability mass has accumulated: $F(x) = 0$ $[A1]$.
• For $0 \le x \le 2$, use dummy variable of integration $t$:

$$F(x) = \int_0^x f(t)\,dt = \int_0^x \frac{3}{4}(2t - t^2)\,dt \quad [M1]$$ $$= \frac{3}{4}\left[ t^2 - \frac{t^3}{3} \right]_0^x = \frac{3}{4}\left( x^2 - \frac{x^3}{3} \right) = \frac{3}{4}x^2 - \frac{1}{4}x^3 = \frac{x^2(3 - x)}{4} \quad [A1]$$

• For $x > 2$, all probability mass has accumulated: $F(x) = 1$ $[A1]$.

Complete piecewise definition:

$$F(x) = \begin{cases} 0 & x < 0 \\ \frac{3x^2 - x^3}{4} & 0 \le x \le 2 \\ 1 & x > 2 \end{cases}$$

(b) Finding the Median $m$:

By definition, the median satisfies $F(m) = 0.5$ $[M1]$:

$$\frac{3m^2 - m^3}{4} = \frac{1}{2} \implies 3m^2 - m^3 = 2 \implies m^3 - 3m^2 + 2 = 0 \quad [A1]$$

Test integer factors of $2$ ($\pm 1, \pm 2$):

$$1^3 - 3(1)^2 + 2 = 1 - 3 + 2 = 0 \implies (m - 1) \text{ is a factor.}$$

Polynomial division: $m^3 - 3m^2 + 2 = (m - 1)(m^2 - 2m - 2) = 0$.
Roots of the quadratic: $m = \frac{2 \pm \sqrt{4 - 4(1)(-2)}}{2} = 1 \pm \sqrt{3}$.
Since $1 + \sqrt{3} \approx 2.732 > 2$ and $1 - \sqrt{3} \approx -0.732 < 0$, both quadratic roots lie outside the support $[0, 2]$.
Therefore, the unique median is $m = 1$ $[A1]$.

Alternative Pathway: $f(x) = \frac{3}{4}x(2-x)$ is a downward parabola with line of symmetry at $x = \frac{0 + 2}{2} = 1$. Since $f(x)$ is continuous and perfectly symmetric about $x = 1$, the median and mean must both equal the axis of symmetry: $m = 1$ $[R1][A1]$.

Worked Example 3: Conditional Probability with Continuous Distributions

The continuous random variable $T$ represents the waiting time (in hours) at a hospital clinic, with PDF:

$$f(t) = \begin{cases} 2e^{-2t} & t \ge 0 \\ 0 & t < 0 \end{cases}$$

(a) Find an expression for $P(T > k)$ where $k > 0$. [2 marks]
(b) A patient has already been waiting for $1$ hour. Find the probability that their total wait exceeds $3$ hours. [3 marks]
(c) Interpret this result in relation to the memoryless property. [1 mark]

Full Markscheme Solution:

(a) Survival / Exceedance Probability:

$$P(T > k) = \int_k^\infty 2e^{-2t}\,dt = \lim_{b \to \infty} \left[ -e^{-2t} \right]_k^b \quad [M1]$$ $$= \lim_{b \to \infty} (-e^{-2b}) - (-e^{-2k}) = 0 + e^{-2k} = e^{-2k} \quad [A1]$$

(b) Conditional Probability:

We require $P(T > 3 \mid T > 1)$:

$$P(T > 3 \mid T > 1) = \frac{P(T > 3 \cap T > 1)}{P(T > 1)} \quad [M1]$$

Since the event $\{T > 3\}$ is a subset of $\{T > 1\}$, their intersection is simply $\{T > 3\}$:

$$= \frac{P(T > 3)}{P(T > 1)} = \frac{e^{-2(3)}}{e^{-2(1)}} = \frac{e^{-6}}{e^{-2}} \quad [A1]$$ $$= e^{-6 - (-2)} = e^{-4} \approx 0.0183 \text{ (to 3 s.f.)} \quad [A1]$$

(c) Interpretation:

Notice that $P(T > 1 + 2 \mid T > 1) = e^{-2(2)} = P(T > 2)$. The exponential distribution is memoryless: the probability of waiting an additional $2$ hours is completely independent of how long the patient has already waited $[R1]$.

Worked Example 4: Calculus Determination of Mode & Quartiles

A continuous random variable $X$ has probability density function:

$$f(x) = \begin{cases} 12x^2(1 - x) & 0 \le x \le 1 \\ 0 & \text{otherwise} \end{cases}$$

(a) Find the mode of $X$, fully justifying your answer. [3 marks]
(b) Determine the maximum value of the probability density function $f(x)$. [1 mark]
(c) Show that the lower quartile $Q_1$ satisfies the equation $4Q_1^3 - 3Q_1^4 = 0.25$, and hence find $Q_1$ to 3 significant figures. [3 marks]
(d) Calculate $P(X \ge 0.5 \mid X \le 0.8)$. [3 marks]

Full Markscheme Solution:

(a) Mode Determination:

The mode is the value of $x$ where $f(x)$ attains its global maximum. Expand $f(x) = 12x^2 - 12x^3$ and differentiate:

$$f'(x) = 24x - 36x^2 = 12x(2 - 3x) \quad [M1]$$

Setting $f'(x) = 0 \implies x = 0$ or $x = \frac{2}{3}$.
Second derivative test:

$$f''(x) = 24 - 72x$$ $$f''\left(\frac{2}{3}\right) = 24 - 72\left(\frac{2}{3}\right) = 24 - 48 = -24 < 0 \implies \text{local and global maximum} \quad [R1]$$

At the boundaries: $f(0) = 0$ and $f(1) = 0$.
Therefore, the mode is $x = \frac{2}{3}$ $[A1]$.

(b) Maximum Probability Density:

$$f_{\max} = f\left(\frac{2}{3}\right) = 12\left(\frac{2}{3}\right)^2\left(1 - \frac{2}{3}\right) = 12\left(\frac{4}{9}\right)\left(\frac{1}{3}\right) = \frac{48}{27} = \frac{16}{9} \approx 1.78 \quad [A1]$$

(c) Lower Quartile Equation & Value:

The CDF for $0 \le x \le 1$ is:

$$F(x) = \int_0^x 12(t^2 - t^3)\,dt = 12\left[ \frac{t^3}{3} - \frac{t^4}{4} \right]_0^x = 4x^3 - 3x^4 \quad [M1]$$

By definition of lower quartile, $F(Q_1) = 0.25$:

$$4Q_1^3 - 3Q_1^4 = 0.25 \quad [AG]$$

Using a GDC numerical polynomial solver on $3Q_1^4 - 4Q_1^3 + 0.25 = 0$ in the interval $[0, 1]$:

$$Q_1 \approx 0.456 \quad [A1]$$

(d) Conditional Probability:

$$P(X \ge 0.5 \mid X \le 0.8) = \frac{P(0.5 \le X \le 0.8)}{P(X \le 0.8)} = \frac{F(0.8) - F(0.5)}{F(0.8)} \quad [M1]$$ $$F(0.8) = 4(0.8)^3 - 3(0.8)^4 = 4(0.512) - 3(0.4096) = 2.048 - 1.2288 = 0.8192$$ $$F(0.5) = 4(0.5)^3 - 3(0.5)^4 = 4(0.125) - 3(0.0625) = 0.5 - 0.1875 = 0.3125$$ $$\text{Probability} = \frac{0.8192 - 0.3125}{0.8192} = \frac{0.5067}{0.8192} \approx 0.619 \text{ (or } \frac{5067}{8192}) \quad [A1]$$

Examiner Commentary: Candidates often confuse the mode (the value on the $x$-axis where density peaks, $x = \frac{2}{3}$) with the maximum density value ($f(2/3) = \frac{16}{9}$). Always state the $x$-value for the mode!

Worked Example 5: Piecewise Linear / Triangular Distribution & Upper Quartile

A continuous random variable $X$ has a triangular probability density function defined by:

$$f(x) = \begin{cases} \frac{1}{4}x & 0 \le x \le 2 \\ 1 - \frac{1}{4}x & 2 < x \le 4 \\ 0 & \text{otherwise} \end{cases}$$

(a) Verify that $\int_0^4 f(x)\,dx = 1$. [2 marks]
(b) Derive the cumulative distribution function $F(x)$ for the branch $2 < x \le 4$. [3 marks]
(c) Find the exact value of the upper quartile $Q_3$. [3 marks]

Full Markscheme Solution:

(a) Verifying Total Probability:

Split the integral at the transition boundary $x = 2$:

$$\int_0^4 f(x)\,dx = \int_0^2 \frac{1}{4}x\,dx + \int_2^4 \left(1 - \frac{1}{4}x\right)\,dx \quad [M1]$$ $$= \left[ \frac{x^2}{8} \right]_0^2 + \left[ x - \frac{x^2}{8} \right]_2^4 = \left( \frac{4}{8} - 0 \right) + \left[ (4 - 2) - \left(2 - \frac{4}{8}\right) \right] = \frac{1}{2} + \left( 2 - 1.5 \right) = \frac{1}{2} + \frac{1}{2} = 1 \quad [A1]$$

(b) Cumulative Distribution Function for $2 < x \le 4$:

The accumulated probability up to $x$ equals the area accumulated up to $x = 2$ plus the integral from $2$ to $x$:

$$F(x) = F(2) + \int_2^x \left(1 - \frac{1}{4}t\right)\,dt \quad [M1]$$

Since $F(2) = \frac{1}{2}$:

$$F(x) = \frac{1}{2} + \left[ t - \frac{t^2}{8} \right]_2^x = \frac{1}{2} + \left( x - \frac{x^2}{8} \right) - \left( 2 - \frac{4}{8} \right) \quad [A1]$$ $$= \frac{1}{2} + x - \frac{x^2}{8} - \frac{3}{2} = x - \frac{x^2}{8} - 1 \quad [A1]$$

Check: When $x = 2$, $F(2) = 2 - \frac{4}{8} - 1 = \frac{1}{2}$. When $x = 4$, $F(4) = 4 - \frac{16}{8} - 1 = 1$. The CDF is continuous!

(c) Exact Upper Quartile $Q_3$:

Since $F(2) = 0.5$, the upper quartile ($F(Q_3) = 0.75$) must lie in the second branch $2 < x \le 4$:

$$Q_3 - \frac{Q_3^2}{8} - 1 = \frac{3}{4} \implies 8Q_3 - Q_3^2 - 8 = 6 \implies Q_3^2 - 8Q_3 + 14 = 0 \quad [M1]$$

Apply the quadratic formula:

$$Q_3 = \frac{8 \pm \sqrt{(-8)^2 - 4(1)(14)}}{2} = \frac{8 \pm \sqrt{64 - 56}}{2} = \frac{8 \pm \sqrt{8}}{2} = 4 \pm \sqrt{2} \quad [A1]$$

Since $Q_3$ must lie in $(2, 4]$, we reject the positive root $4 + \sqrt{2} \approx 5.414$:

$$Q_3 = 4 - \sqrt{2} \approx 2.586 \quad [A1]$$

Alternative Geometric Pathway: By symmetry, the area in the upper tail above $Q_3$ must equal $1 - 0.75 = 0.25$. This upper tail is a right triangle with base $(4 - Q_3)$ and height $f(Q_3) = 1 - \frac{1}{4}Q_3 = \frac{4 - Q_3}{4}$.
$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}(4 - Q_3)\left(\frac{4 - Q_3}{4}\right) = \frac{(4 - Q_3)^2}{8} = 0.25 \implies (4 - Q_3)^2 = 2 \implies 4 - Q_3 = \sqrt{2} \implies Q_3 = 4 - \sqrt{2}$ $[R1][A1]$.