CourseCourseQ BankQuestion BankDocsDocuments

6.3 Continuous random variables

Syllabus
9709–2028–2029
Topic
6.3
Level
A2

Learning objectives

6.3.1Continuous random variables• understand the concept of a continuous random variable, and recall and use properties of a probability density function For density functions defined over a single interval only; the domain may be infinite, e.g. x 3 4 for x 1H.6.3.2Density functions• use a probability density function to solve problems involving probabilities, and to calculate the mean and variance of a distribution. Including location of the median or other percentiles of a distribution by direct consideration of an area using the density function. Explicit knowledge of the cumulative distribution function is not included.

Density height becomes probability only through area

Density property on its support II Meaning
f(x)≥0f(x)\ge0 area cannot create negative probability
∫If(x) dx=1\int_I f(x)\,dx=1 all possible values have total probability 1
P(a<X<b)=∫abf(x) dxP(a<X<b)=\int_a^b f(x)\,dx interval probability is area
P(X=x)=0P(X=x)=0 a single point has zero width

State the single support interval and take f(x)=0f(x)=0 outside it. The support may be finite or infinite. Endpoint inclusion makes no difference to a continuous probability.

For f(x)=3/x4f(x)=3/x^4 on x≥1x\ge1, ∫1∞3x−4 dx=1,\int_1^\infty 3x^{-4}\,dx=1, so it is a valid density. Also P(X>2)=∫2∞3x−4 dx=1/8P(X>2)=\int_2^\infty3x^{-4}\,dx=1/8.

A density can exceed 1 over a narrow interval; only its total area must be 1. Density carries inverse units so that integrated probability is unitless.

f(x)f(x) is not P(X=x)P(X=x). Work directly with density areas; explicit cumulative-distribution-function knowledge is not part of this syllabus objective.

One density gives probabilities, moments and percentiles

Goal Direct density calculation
unknown constant set total support area equal to 1
probability integrate f(x)f(x) over the event interval
mean E(X)=∫xf(x) dxE(X)=\int x f(x)\,dx
variance E(X2)=∫x2f(x) dxE(X^2)=\int x^2f(x)\,dx, then Var(X)=E(X2)−[E(X)]2Var(X)=E(X^2)-[E(X)]^2
ppth percentile cc set area to the left of cc equal to pp

Let f(x)=kxf(x)=kx for 0≤x≤20\le x\le2. Normalisation gives 1=∫02kx dx=2k,1=\int_0^2kx\,dx=2k, so k=1/2k=1/2. Therefore P(X>1)=∫12x/2 dx=3/4P(X>1)=\int_1^2x/2\,dx=3/4.

For this density, E(X)=∫02x(x/2) dx=4/3,E(X)=\int_0^2x(x/2)\,dx=4/3, E(X2)=∫02x2(x/2) dx=2,E(X^2)=\int_0^2x^2(x/2)\,dx=2, so Var(X)=2−(4/3)2=2/9.Var(X)=2-(4/3)^2=2/9.

The median mm divides the total area in half: ∫0mx/2 dx=1/2.\int_0^m x/2\,dx=1/2. Hence m2/4=1/2m^2/4=1/2 and, since mm lies in the support, m=2m=\sqrt2. Other percentiles use the same direct-area equation.

Normalise before every later calculation. A median is found from accumulated area, not from f(m)=0.5f(m)=0.5; explicit CDF notation is neither needed nor included.