Make the total area one
Let $f(x)=kx$ for $0\le x\le2$, and $f(x)=0$ otherwise.
$$\int_0^2kx\,dx=1\Rightarrow 2k=1\Rightarrow k=\frac12.$$
So $f(x)=x/2$ on the support.
Continuous random variables
Interpret a probability density as area and use integration to find constants, probabilities and summary values.
By the end of this lesson
- normalize a probability density function
- calculate interval probabilities and expectation
- distinguish mean, median and mode for a density
Probability is area
A continuous value can fall anywhere in an interval. To find its probability, count the area above that interval under the density curve. A single point has no width, so its area is zero; the curve height alone is not a probability.
Check the logic
For f(x)=x/2 on [0,2], what is P(0≤X≤1)?
Explore the result
The triangle has base 1 and height 1/2, so its area is 1/4.
Move the upper bound to 1, then 2. Explain why doubling the bound quadruples the area, not doubles it.
For f(x) = x/2 on 0 ≤ x ≤ 2, the whole triangle has area 1. The area from 0 to u is u²/4. At u = √2 it is 1/2, so √2 is the median. The area from 1 to 2 is 3/4.
The whole possible range must carry probability 1. A density is nonnegative, and integration adds the small areas across the chosen interval:
$$P(a\le X\le b)=\int_a^b f(x)\,dx,\qquad \int_{-\infty}^{\infty}f(x)\,dx=1.$$
Why is the total area 1?
The random value must lie somewhere in its support. That certain event has probability 1. Normalizing a density chooses its scale so that the total area represents this whole probability.
Checkpoint
Correct. A single point has zero width and therefore zero area.
Not quite. Density height is not point probability.
Show explanation
A single point has zero width and therefore zero area.
Example 1
Find the normalizing constant
Find interval probabilities
For the density above,
$$P(1\lt X\lt 2)=\int_1^2\frac{x}{2}\,dx=\left[\frac{x^2}{4}\right]_1^2=\frac34.$$
Endpoint choices do not change a continuous probability.
Checkpoint
Correct. The rectangle has area $5c=1$.
Not quite. Normalize using total area, not a point value.
Show explanation
The rectangle has area $5c=1$.
Example 2
Calculate expectation
Weight values by density
For $f(x)=x/2$ on $0\le x\le2$:
$$E(X)=\int_0^2x f(x)\,dx=\int_0^2\frac{x^2}{2}\,dx=\frac43.$$
The mean lies to the right of the midpoint because the density increases with $x$.
Why multiply by $x$?
Expectation is a weighted average. The density supplies the weight assigned near each value $x$.
Separate mean, median and mode
- The mean is $E(X)$.
- A median $m$ satisfies $P(X\le m)=1/2$.
- A mode occurs where the density is greatest.
Checkpoint
Correct. The median divides total probability area in half.
Not quite. The height at the median need not be $1/2$.
Show explanation
The median divides total probability area in half.
Example 3
Find a median and mode
An increasing density
For $f(x)=x/2$ on $[0,2]$, solve
$$\int_0^m\frac{x}{2}\,dx=\frac12\Rightarrow \frac{m^2}{4}=\frac12\Rightarrow m=\sqrt2.$$
The negative root is outside the support. Since $f$ increases throughout $[0,2]$, the mode is $2$.
Here mean $4/3$, median $\sqrt2$, and mode $2$ are different.
In the exam
Quick check
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A density distributes one unit of total probability.
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A density distributes one unit of total probability.
Interval probability is accumulated density.
Show explanation
Interval probability is accumulated density.
The single endpoint has probability zero.
Show explanation
The single endpoint has probability zero.
Expectation weights each value by its density.
Show explanation
Expectation weights each value by its density.
$P(X\le m)=1/2$.
Show explanation
$P(X\le m)=1/2$.
Lesson Rewards
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