Probability Density Function

15 min · 2 drills

Why do we need density?

Imagine you're waiting for a bus that comes "roughly every 10 minutes." How long will you wait? 3 minutes? 7.5 minutes? 9.234 minutes? Your waiting time could be any number from 0 to 10.

This is different from rolling a die, where the result is only 1, 2, 3, 4, 5, or 6. No 3.7 or 5.123 possible.

Discrete variables	Continuous variables
Only specific values	Any value in an interval
Die: 1, 2, 3, 4, 5, 6	Waiting time: 0...10 min
Coin: heads or tails	Height: 150...200 cm

For continuous variables, we can't ask "what's the probability of waiting exactly 3 minutes?" The answer is always zero. Why? Because between 2.999 and 3.001 there are infinitely many numbers, so the chance of hitting one exact point is zero.

Instead, we ask: "what's the probability of waiting between 2 and 5 minutes?" That's where density comes in.

The main idea: probability = area

The density function $f (x)$ is a curve showing where values are "concentrated." The higher the curve at some point, the more often values fall in that region.

Probability of landing in an interval = area under the curve on that interval.

P (a \leq X \leq b) = \int_{a}^{b} f (x) d x

a = 2b = 9

P(2 ≤ X ≤ 9) = 0.700

Move a and b so that P(a ≤ X ≤ b) = 0.5.

The integral is just the area. That's why $\int_{a}^{b} f (x) d x$ gives us the probability.

Why does the total area equal 1?

The probability that something happens is 100%, or 1. So the total area under the entire density curve must equal 1:

\int_{- \infty}^{+ \infty} f (x) d x = 1

This is called the normalization condition. If the function is only defined on interval $[a, b]$ , then:

\int_{a}^{b} f (x) d x = = 1

This condition is our main tool for finding unknown constants in problems.

How to find the constant

Problem: Given $f (x) = c \cdot x$ on $[0, 2]$ , and $f (x) = 0$ outside this interval. Find $c$ .

Logic: Since this is a density, the area under it must equal 1.

Solution:

Calculate the area (integral) and set it equal to 1:

\int_{0}^{2} c \cdot \cdot x d x = = 1

The integral of $x$ is $\frac{x ^{2}}{2}$ . Substitute the limits:

c \cdot \cdot \frac{x ^{2}}{2}_{0}^{2} = = c \cdot \cdot (\frac{4}{2} - \frac{0}{2}) = = c \cdot \cdot 2 = = 1

Therefore:

c = = \frac{1}{2}

Check: The area of a triangle with base 2 and height $f (2) = \frac{1}{2} \cdot 2 = 1$ is $\frac{1}{2} \cdot 2 \cdot 1 = 1$ . It works!

How to calculate probabilities

Once the constant is found, we calculate probability as the area over the required interval.

Example: Density $f (x) = \frac{1}{2} x$ on $[0, 2]$ . Find $P (1 \leq X \leq 2)$ .

a = 1b = 2

P(1 ≤ X ≤ 2) = 0.750

We need the shaded area (a trapezoid). We can calculate it geometrically or via integral:

P (1 \leq \leq X \leq \leq 2) = = \int_{1}^{2} \frac{1}{2} x d x = = \frac{1}{2} \cdot \cdot \frac{x ^{2}}{2}_{1}^{2} = = \frac{1}{4} (4 - 1) = = \frac{3}{4}

So in 75% of cases, the value will be between 1 and 2.

Uniform distribution: the simplest case

When all values are equally likely, the density is a horizontal line.

Example: A bus arrives every 15 minutes. You arrive at the stop at a random time. Waiting time is uniformly distributed from 0 to 15.

a = 0b = 15

P(0 ≤ X ≤ 15) = 1.000

Density: $f (x) = \frac{1}{15}$ on interval $[0, 15]$ .

Why $\frac{1}{15}$ ? Because the rectangle area is $15 \times h = 1$ , so $h = \frac{1}{15}$ .

What's the probability of waiting at most 5 minutes?

P (X \leq \leq 5) = = \frac{5}{15} = = \frac{1}{3} \approx \approx 33%

For uniform distribution, probability is just the ratio of interval lengths!

Formula for uniform distribution on $[a, b]$ :

f (x) = = \frac{1}{b - a}

Expected value: the "average" result

Expected value $E (X)$ is the "center of mass" of the distribution: if you repeated the experiment many times and averaged the results, you'd get approximately $E (X)$ .

E (X) = \int x \cdot f (x) d x

Why this formula? We take each value $x$ , multiply by its "weight" $f (x)$ (how likely it is), and sum everything up (integrate).

Example: $f (x) = \frac{1}{2} x$ on $[0, 2]$ .

Where's the "center" of this distribution? The density is higher on the right (at $x = 2$ density is maximum), so the center is shifted right from the interval's midpoint.

E (X) = = \int_{0}^{2} x \cdot \cdot \frac{1}{2} x d x = = \frac{1}{2} \int_{0}^{2} x^{2} d x = = \frac{1}{2} \cdot \cdot \frac{8}{3} = = \frac{4}{3} \approx \approx 1.33

The midpoint of $[0, 2]$ is 1, but our $E (X) = 1.33$ is further right, because values are "concentrated" closer to 2.

For uniform distribution:

E (X) = = \frac{a + b}{2}

Just the midpoint, which makes sense since all values are equally likely.

Cumulative distribution function F(x)

Sometimes instead of density $f (x)$ you're given the distribution function $F (x)$ .

$F (x)$ is the probability that the value is less than $x$ :

F (x) = = P (X < < x)

Connection:

$F (x)$ is the integral of $f (x)$
$f (x)$ is the derivative of $F (x)$

F (x) = = \int_{- \infty}^{x} f (t) d t and f (x) = = F^{'} (x)

Properties of $F (x)$ :

$F (- \infty) = 0$ (probability that $X$ is less than minus infinity is zero)
$F (+ \infty) = 1$ (probability that $X$ is less than plus infinity is 100%)
$F (x)$ always increases (or stays flat)

Useful formula:

P (a \leq \leq X \leq \leq b) = = F (b) - - F (a)

Complete worked example

Problem: A random variable $X$ has a density whose graph is a triangle on $[0, 4]$ with its peak at $x = 2$ .

a = 1b = 3

P(1 ≤ X ≤ 3) = 0.750

a) Find the height $h$ .

b) Write $f (x)$ as a piecewise function.

c) Find $P (1 \leq X \leq 3)$ .

Solution:

a) Find $h$ from normalization:

Area of the triangle = 1:

\frac{1}{2} \cdot \cdot 4 \cdot \cdot h = = 1 ⟹ ⟹ h = = \frac{1}{2}

b) Write $f (x)$ piecewise:

Left side of the triangle: a line from $(0, 0)$ to $(2, \frac{1}{2})$ :

f (x) = = \frac{x}{4}, 0 \leq \leq x \leq \leq 2

Right side: a line from $(2, \frac{1}{2})$ to $(4, 0)$ :

f (x) = = \frac{4 - x}{4}, 2 < < x \leq \leq 4

Outside $[0, 4]$ : $f (x) = 0$ .

c) Find $P (1 \leq X \leq 3)$ :

The interval $[1, 3]$ crosses both pieces, so we split at $x = 2$ :

P (1 \leq \leq X \leq \leq 3) = = \int_{1}^{2} \frac{x}{4} d x + + \int_{2}^{3} \frac{4 - x}{4} d x

First integral:

\int_{1}^{2} \frac{x}{4} d x = = \frac{1}{4} \cdot \cdot \frac{x ^{2}}{2}_{1}^{2} = = \frac{1}{8} (4 - 1) = = \frac{3}{8}

Second integral:

\int_{2}^{3} \frac{4 - x}{4} d x = = \frac{1}{4} [4 x - \frac{x ^{2}}{2}]_{2}^{3} = = \frac{1}{4} [(12 - \frac{9}{2}) - (8 - 2)] = = \frac{1}{4} \cdot \cdot \frac{3}{2} = = \frac{3}{8}

Total:

P (1 \leq \leq X \leq \leq 3) = = \frac{3}{8} + + \frac{3}{8} = = \frac{3}{4}

This makes sense: a symmetric interval around the peak contains 75% of the probability.

Common mistakes

Mistake 1: Forgetting normalization

If you're given $f (x)$ and not told it's a density, check for yourself: the integral should equal 1.

Mistake 2: Integrating where $f (x) = 0$

If $f (x) = 0$ outside $[0, 2]$ , then when calculating $P (X \leq 5)$ , integrate only up to 2, not to 5.

Mistake 3: Confusing $f (x)$ and $F (x)$

$f (x)$ : density (can be greater than 1!)
$F (x)$ : probability (always between 0 and 1)

Mistake 4: Thinking $P (X = a) \neq = 0$

For a continuous variable, the probability of hitting a point is always zero. It's not an error in the problem, it's a mathematical fact.

Normal distribution: the most important example

The normal distribution (bell curve) is the most famous continuous distribution in statistics. Heights, measurement errors, exam scores: all of them follow it approximately.

Its density function:

f (x) = = \frac{1}{σ 2 π} e^{- \frac{1}{2} (\frac{x - μ}{σ})^{2}}

Here $μ = 0$ (mean) and $σ = 1$ (standard deviation) stay fixed. Drag the red and green points to compute $P (a < X < b)$ .

Move a and b until P(a ≤ X ≤ b) ≈ 90 %.

Key properties:

Symmetric around $μ$
~68% of probability falls within $[μ - σ, μ + σ]$
~95% falls within $[μ - 2 σ, μ + 2 σ]$

Check Your Understanding

Your turn1 of 2

The function $f (x) = c \cdot (3 - x)$ for $0 \leq x \leq 3$ and $f (x) = 0$ otherwise is a density. What is $P (X \leq 1)$ ?

\frac{10}{18}

\frac{15}{18}

\frac{8}{18}

\frac{12}{18}

Your turn2 of 2

$X$ is uniformly distributed on the interval $[2, 10]$ . What are the values of $P (X < 4)$ and $P (X \leq 4)$ ?

P (X < 4) = \frac{1}{2}

and

P (X \leq 4) = \frac{1}{2}

P (X < 4) = \frac{1}{4}

and

P (X \leq 4) = \frac{3}{8}

P (X < 4) = \frac{1}{4}

and

P (X \leq 4) = \frac{1}{4}

P (X < 4) = \frac{1}{2}

and

P (X \leq 4) = \frac{5}{8}

Formulas to remember

Probability = area:

P (a \leq \leq X \leq \leq b) = = \int_{a}^{b} f (x) d x

Normalization condition:

\int f (x) d x = = 1

Expected value:

E (X) = = \int x \cdot \cdot f (x) d x

Uniform distribution on $[a, b]$ :

Density: $f (x) = \frac{1}{b - a}$
Mean: $E (X) = \frac{a + b}{2}$

Connection between $f$ and $F$ :

f (x) = = F^{'} (x), F (x) = = \int f (x) d x

Now practise it

Probability Density Function

Trainer · Probability & Combinatorics

Practise

Next: Combinatorics (Kombinatorik)

0/2

drills solved

Probability Density Function

Why do we need density?

The main idea: probability = area

Why does the total area equal 1?

How to find the constant

How to calculate probabilities

Uniform distribution: the simplest case

Expected value: the "average" result

Cumulative distribution function F(x)

Complete worked example

Common mistakes

Mistake 1: Forgetting normalization

Mistake 2: Integrating where f(x)=0

Mistake 3: Confusing f(x) and F(x)

Mistake 4: Thinking P(X=a)=0

Normal distribution: the most important example

Check Your Understanding

Formulas to remember

Mistake 2: Integrating where $f (x) = 0$

Mistake 3: Confusing $f (x)$ and $F (x)$

Mistake 4: Thinking $P (X = a) \neq = 0$