Probability Density Functions

For a continuous random variable, like a waiting time or a length, probability is area under a curve. With calculus you can now work with any curve, not just rectangles and the normal distribution: the curve is a probability density function, and probabilities, the median, the mean and the variance all come from integrals. This page uses calculus, so any trigonometric functions are in radians.

Key ideas

Probability density functions

A continuous random variable $X$ has a probability density function (pdf) $f(x)$ with two properties:

• $f(x) \ge 0$ for all $x$ (the curve never goes below the axis);
• the total area under the curve is $1$: $\displaystyle\int_{-\infty}^{\infty} f(x)\,dx = 1$.

Most pdfs are zero outside some interval $a \le x \le b$, so the condition becomes $\displaystyle\int_a^b f(x)\,dx = 1$. If a pdf contains an unknown constant $k$, this condition is how you find it.

Probabilities are integrals

$$P(c \le X \le d) = \int_c^d f(x)\,dx$$

This is the area under the curve between $c$ and $d$ (see evaluating definite integrals). A single value has no area, so $P(X = c) = 0$, and it doesn’t matter whether you write $\lt$ or $\le$.

Note that $f(x)$ itself is not a probability: it can be bigger than $1$. Only areas are probabilities.

Mode and median

• The mode is the value of $x$ where $f(x)$ is greatest. Find it with $f'(x) = 0$ (checking it’s a maximum), but remember the maximum can also be at an endpoint of the interval.
• The median $m$ splits the area in half:

$$\int_{-\infty}^{m} f(x)\,dx = \frac{1}{2}$$

Solve this equation for $m$, by hand if you can, or with your GDC’s solver.

Mean and variance

These are the continuous versions of the formulas for discrete random variables, with sums replaced by integrals:

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

The linear transformation rules still hold: $E(aX + b) = aE(X) + b$ and $\text{Var}(aX + b) = a^2\,\text{Var}(X)$.

Piecewise pdfs

A pdf can be defined by different formulas on different intervals. Integrate each piece over its own interval and add the results. For the median, first find which piece it’s in by checking the area of the first piece.

Shape tells you about mode, median and mean

For a symmetric pdf, the mode (if there’s a single peak), median and mean are all in the middle. If the pdf has a long tail to the right (positively skewed), the mean is pulled right: usually mode $\lt$ median $\lt$ mean. Example 3 shows this.

With or without technology

On calculator papers you can find integrals, solve for the median and find the maximum with your GDC. Without a calculator, you need to integrate by hand, so practise both. Exact answers are expected when the integrals are straightforward.

Worked examples

Example 1: Finding k and a probability

$X$ has pdf $f(x) = kx^2$ for $0 \le x \le 3$, and $f(x) = 0$ otherwise.

• (a) Find $k$.
• (b) Find $P(1 \le X \le 2)$.

Solution.

(a) The total area must be $1$:

$$\int_0^3 kx^2\,dx = k\left[\frac{x^3}{3}\right]_0^3 = 9k = 1 \quad\Rightarrow\quad k = \frac{1}{9}$$

(b)

$$P(1 \le X \le 2) = \int_1^2 \frac{x^2}{9}\,dx = \left[\frac{x^3}{27}\right]_1^2 = \frac{8}{27} - \frac{1}{27} = \frac{7}{27}$$

Example 2: Mode, median, mean and variance by hand

$X$ has pdf $f(x) = \dfrac{x}{8}$ for $0 \le x \le 4$, and $0$ otherwise. Find the mode, median, mean and standard deviation of $X$.

Solution.

Mode. $f(x) = \dfrac{x}{8}$ is increasing, so its largest value is at the endpoint: the mode is $4$.

Median.

$$\int_0^m \frac{x}{8}\,dx = \frac{m^2}{16} = \frac{1}{2} \quad\Rightarrow\quad m^2 = 8 \quad\Rightarrow\quad m = 2\sqrt{2} \approx 2.83$$

(Reject $m = -2\sqrt{2}$, which is outside $0 \le x \le 4$.)

Mean.

$$E(X) = \int_0^4 x \cdot \frac{x}{8}\,dx = \left[\frac{x^3}{24}\right]_0^4 = \frac{64}{24} = \frac{8}{3}$$

Variance.

$$E(X^2) = \int_0^4 \frac{x^3}{8}\,dx = \left[\frac{x^4}{32}\right]_0^4 = 8$$ $$\text{Var}(X) = 8 - \left(\frac{8}{3}\right)^2 = \frac{72 - 64}{9} = \frac{8}{9}$$

The standard deviation is $\sqrt{8/9} = \dfrac{2\sqrt{2}}{3} \approx 0.943$ (3 s.f.).

Example 3: Mode by calculus, median by GDC

$X$ has pdf $f(x) = \dfrac{3}{64}x(4 - x)^2$ for $0 \le x \le 4$, and $0$ otherwise.

• (a) Find the mode.
• (b) Find the median.
• (c) Show that $E(X) = 1.6$ and find $\text{Var}(X)$.

Solution.

(a) Differentiate $x(4 - x)^2$ with the product rule:

$$\frac{d}{dx}\Big[x(4 - x)^2\Big] = (4 - x)^2 - 2x(4 - x) = (4 - x)(4 - 3x)$$

This is $0$ at $x = 4$ (where $f(x) = 0$, a minimum) and at $x = \dfrac{4}{3}$, where $f$ has its maximum. The mode is $\dfrac{4}{3} \approx 1.33$.

(b) Solve $\displaystyle\int_0^m \frac{3}{64}x(4 - x)^2\,dx = 0.5$ with the GDC (enter the integral as a function of $m$ and use the solver, or graph it and intersect with $y = 0.5$):

$$m = 1.54 \text{ (3 s.f.)}$$

(c) Expand: $x(4 - x)^2 = 16x - 8x^2 + x^3$.

$$E(X) = \frac{3}{64}\int_0^4 (16x^2 - 8x^3 + x^4)\,dx = \frac{3}{64}\left[\frac{16x^3}{3} - 2x^4 + \frac{x^5}{5}\right]_0^4 = \frac{3}{64}\left(\frac{1024}{3} - 512 + \frac{1024}{5}\right) = \frac{3}{64}\cdot\frac{512}{15} = 1.6$$ $$E(X^2) = \frac{3}{64}\int_0^4 (16x^3 - 8x^4 + x^5)\,dx = \frac{3}{64}\left(1024 - \frac{8192}{5} + \frac{2048}{3}\right) = \frac{3}{64}\cdot\frac{1024}{15} = 3.2$$ $$\text{Var}(X) = 3.2 - 1.6^2 = 0.64$$

So the standard deviation is $0.8$.

A pdf with a long tail to the right: mode $\approx 1.33$, median $\approx 1.54$, mean $= 1.6$.

Example 4: A piecewise pdf

$X$ has pdf

$$f(x) = \begin{cases} kx, & 0 \le x \le 2 \\ \dfrac{k(6 - x)}{2}, & 2 \lt x \le 6 \\ 0, & \text{otherwise} \end{cases}$$

• (a) Find $k$.
• (b) Find $P(X \gt 4)$.
• (c) Find the median.
• (d) Find $E(X)$.

Solution.

(a) Add the areas of the two pieces:

$$\int_0^2 kx\,dx + \int_2^6 \frac{k(6 - x)}{2}\,dx = 2k + \frac{k}{2}\left[6x - \frac{x^2}{2}\right]_2^6 = 2k + \frac{k}{2}(18 - 10) = 6k = 1$$

So $k = \dfrac{1}{6}$. (The graph is a triangle with base $6$ and height $f(2) = 2k$, so the area is $\tfrac{1}{2}(6)(2k) = 6k$. ✓)

The pdf is $f(x) = \dfrac{x}{6}$ on $0 \le x \le 2$ and $f(x) = \dfrac{6 - x}{12}$ on $2 \lt x \le 6$.

(b)

$$P(X \gt 4) = \int_4^6 \frac{6 - x}{12}\,dx = \frac{1}{12}\left[6x - \frac{x^2}{2}\right]_4^6 = \frac{1}{12}(18 - 16) = \frac{1}{6}$$

(c) The first piece has area $\displaystyle\int_0^2 \frac{x}{6}\,dx = \frac{1}{3} \lt \frac{1}{2}$, so the median is in the second piece. It’s easiest to use the area to the right of $m$, which must also be $\tfrac{1}{2}$:

$$\int_m^6 \frac{6 - x}{12}\,dx = \frac{(6 - m)^2}{24} = \frac{1}{2} \quad\Rightarrow\quad (6 - m)^2 = 12 \quad\Rightarrow\quad m = 6 - 2\sqrt{3} \approx 2.54$$

(Take $6 - m = +\sqrt{12}$, since $m = 6 + 2\sqrt{3}$ is outside the interval.)

(d)

$$E(X) = \int_0^2 \frac{x^2}{6}\,dx + \int_2^6 \frac{x(6 - x)}{12}\,dx = \frac{8}{18} + \frac{1}{12}\left[3x^2 - \frac{x^3}{3}\right]_2^6 = \frac{4}{9} + \frac{1}{12}\cdot\frac{80}{3} = \frac{4}{9} + \frac{20}{9} = \frac{8}{3}$$

The piecewise pdf of Example 4. Half the area lies left of $m = 6 - 2\sqrt{3} \approx 2.54$, and $P(X \gt 4) = \tfrac{1}{6}$.

Common mistakes

Treating f(x) as a probability. $f(2) = \tfrac{1}{3}$ in Example 4 does not mean $P(X = 2) = \tfrac{1}{3}$. For a continuous variable, $P(X = 2) = 0$; probabilities come from areas.

Forgetting the x in the mean. $E(X) = \int x f(x)\,dx$, not $\int f(x)\,dx$ (that’s always $1$). For $E(X^2)$, multiply by $x^2$.

Assuming the mode is where f′(x) = 0. The maximum might be at an endpoint, as in Example 2, where $f$ is increasing all the way to $x = 4$. Always compare with the endpoints.

Integrating over the wrong interval for the median. For a piecewise pdf, check the area of the first piece first. If it’s less than $\tfrac{1}{2}$, the median is in a later piece, and you must include the whole first piece’s area.

Keeping a root outside the domain. Equations for the median often have two solutions. Keep only the one inside the interval where the pdf is defined.

Writing Var(X) = E(X²). Subtract $[E(X)]^2$. If your variance is bigger than a quarter of the square of the interval’s width, something has gone wrong.

Practice

1. (Warm-up) $X$ has pdf $f(x) = kx$ for $0 \le x \le 2$, and $0$ otherwise. Find $k$ and $P(X \lt 1)$.

Solution

$$\int_0^2 kx\,dx = k\left[\frac{x^2}{2}\right]_0^2 = 2k = 1 \quad\Rightarrow\quad k = \frac{1}{2}$$$$P(X \lt 1) = \int_0^1 \frac{x}{2}\,dx = \left[\frac{x^2}{4}\right]_0^1 = \frac{1}{4}$$

2. (Warm-up) $X$ has pdf $f(x) = \dfrac{1}{5}$ for $2 \le x \le 7$, and $0$ otherwise. Find $P(X \gt 6)$, the median and $E(X)$.

Solution

$P(X \gt 6) = \dfrac{1}{5}(7 - 6) = \dfrac{1}{5}$ (a rectangle of width $1$ and height $\tfrac{1}{5}$).

The pdf is symmetric about the centre of the interval, so the median is $\dfrac{2 + 7}{2} = 4.5$.

$$E(X) = \int_2^7 \frac{x}{5}\,dx = \left[\frac{x^2}{10}\right]_2^7 = \frac{49 - 4}{10} = 4.5$$

3. (Core) $X$ has pdf $f(x) = k(4 - x^2)$ for $0 \le x \le 2$, and $0$ otherwise.

• (a) Find $k$.
• (b) Find $P(X \lt 1)$.
• (c) State the mode and find $E(X)$.

Solution

(a)

$$\int_0^2 k(4 - x^2)\,dx = k\left[4x - \frac{x^3}{3}\right]_0^2 = k\left(8 - \frac{8}{3}\right) = \frac{16k}{3} = 1 \quad\Rightarrow\quad k = \frac{3}{16}$$

(b)

$$P(X \lt 1) = \frac{3}{16}\left[4x - \frac{x^3}{3}\right]_0^1 = \frac{3}{16}\cdot\frac{11}{3} = \frac{11}{16}$$

(c) $f$ is decreasing on $0 \le x \le 2$, so the mode is $0$.

$$E(X) = \frac{3}{16}\int_0^2 (4x - x^3)\,dx = \frac{3}{16}\left[2x^2 - \frac{x^4}{4}\right]_0^2 = \frac{3}{16}(8 - 4) = \frac{3}{4}$$

4. (Core) $X$ has pdf $f(x) = \dfrac{3}{8}x^2$ for $0 \le x \le 2$, and $0$ otherwise. Find the median, the mean and the variance of $X$.

Solution

Median:

$$\int_0^m \frac{3}{8}x^2\,dx = \frac{m^3}{8} = \frac{1}{2} \quad\Rightarrow\quad m^3 = 4 \quad\Rightarrow\quad m = \sqrt[3]{4} \approx 1.59$$

Mean and variance:

$$E(X) = \frac{3}{8}\int_0^2 x^3\,dx = \frac{3}{8}\cdot 4 = 1.5 \qquad E(X^2) = \frac{3}{8}\int_0^2 x^4\,dx = \frac{3}{8}\cdot\frac{32}{5} = 2.4$$$$\text{Var}(X) = 2.4 - 1.5^2 = 0.15$$

5. (Core) $X$ has pdf $f(x) = k\sin x$ for $0 \le x \le \pi$ (radians), and $0$ otherwise.

• (a) Find $k$.
• (b) Find $P\left(X \lt \dfrac{\pi}{3}\right)$.
• (c) Write down the mode and the median.

Solution

(a)

$$\int_0^{\pi} k\sin x\,dx = k\Big[-\cos x\Big]_0^{\pi} = k(1 + 1) = 2k = 1 \quad\Rightarrow\quad k = \frac{1}{2}$$

(b)

$$P\left(X \lt \frac{\pi}{3}\right) = \frac{1}{2}\Big[-\cos x\Big]_0^{\pi/3} = \frac{1}{2}\left(-\frac{1}{2} + 1\right) = \frac{1}{4}$$

(c) $\sin x$ has its maximum at $x = \dfrac{\pi}{2}$, so the mode is $\dfrac{\pi}{2}$. The graph is symmetric about $x = \dfrac{\pi}{2}$, so the median is also $\dfrac{\pi}{2}$.

6. (Core) $X$ has pdf

$$f(x) = \begin{cases} \dfrac{x}{10}, & 0 \le x \le 2 \\ k, & 2 \lt x \le 6 \\ 0, & \text{otherwise} \end{cases}$$

Find $k$, the median and $E(X)$.

Solution

Area of the first piece: $\displaystyle\int_0^2 \frac{x}{10}\,dx = \frac{4}{20} = 0.2$. Area of the second piece: $4k$. So $0.2 + 4k = 1$ and $k = 0.2$.

The first piece has area $0.2 \lt 0.5$, so the median is in the second piece:

$$0.2 + 0.2(m - 2) = 0.5 \quad\Rightarrow\quad m - 2 = 1.5 \quad\Rightarrow\quad m = 3.5$$$$E(X) = \int_0^2 \frac{x^2}{10}\,dx + \int_2^6 0.2x\,dx = \frac{8}{30} + 0.1\Big[x^2\Big]_2^6 = \frac{4}{15} + 3.2 = \frac{52}{15} \approx 3.47$$

7. (Core) $X$ has pdf $f(x) = k\,e^{-x/2}$ for $0 \le x \le 4$, and $0$ otherwise.

• (a) Show that $k = \dfrac{1}{2(1 - e^{-2})}$.
• (b) Use your GDC to find the median and the mean of $X$.

Solution

(a)

$$\int_0^4 k\,e^{-x/2}\,dx = k\Big[-2e^{-x/2}\Big]_0^4 = k(2 - 2e^{-2}) = 1 \quad\Rightarrow\quad k = \frac{1}{2(1 - e^{-2})}$$

($k \approx 0.578$.)

(b) Solving $\displaystyle\int_0^m k\,e^{-x/2}\,dx = 0.5$ with the GDC gives $m = 1.13$ (3 s.f.). (By hand: $e^{-m/2} = \dfrac{1 + e^{-2}}{2}$, so $m = -2\ln\left(\dfrac{1 + e^{-2}}{2}\right)$.)

$$E(X) = \int_0^4 x \cdot k\,e^{-x/2}\,dx = 1.37 \text{ (3 s.f., GDC)}$$

8. (Challenge) $X$ has pdf $f(x) = ax + b$ for $0 \le x \le 3$, and $0$ otherwise. Given that $E(X) = 1.75$, find $a$ and $b$.

Solution

Total area:

$$\int_0^3 (ax + b)\,dx = \frac{9a}{2} + 3b = 1$$

Mean:

$$\int_0^3 (ax^2 + bx)\,dx = 9a + \frac{9b}{2} = 1.75$$

From the first equation, $b = \dfrac{1 - 4.5a}{3}$. Substitute into the second:

$$9a + 1.5(1 - 4.5a) = 1.75 \quad\Rightarrow\quad 2.25a = 0.25 \quad\Rightarrow\quad a = \frac{1}{9}$$

Then $b = \dfrac{1 - 0.5}{3} = \dfrac{1}{6}$.

Check: $f(x) = \dfrac{x}{9} + \dfrac{1}{6} \ge 0$ on $0 \le x \le 3$, so it’s a valid pdf. ✓

9. (Challenge) $X$ has the pdf from Example 2, $f(x) = \dfrac{x}{8}$ for $0 \le x \le 4$. Let $Y = 2X + 3$.

• (a) Find $E(Y)$ and $\text{Var}(Y)$.
• (b) Find $P(Y \gt 9)$.

Solution

(a) From Example 2, $E(X) = \dfrac{8}{3}$ and $\text{Var}(X) = \dfrac{8}{9}$.

$$E(Y) = 2\cdot\frac{8}{3} + 3 = \frac{25}{3} \qquad \text{Var}(Y) = 2^2\cdot\frac{8}{9} = \frac{32}{9}$$

(b) $Y \gt 9$ means $2X + 3 \gt 9$, so $X \gt 3$:

$$P(X \gt 3) = \int_3^4 \frac{x}{8}\,dx = \left[\frac{x^2}{16}\right]_3^4 = 1 - \frac{9}{16} = \frac{7}{16}$$