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Continuous random variables — chapter review — 9231 FPS

Integrate on the correct branch and keep a cumulative distribution continuous. Find a transformed distribution through its CDF and support. Normalise a density and compute probabilities and moments by integration.

Notes and key methods

Normalise a piecewise density by adding its separate integrals. The cumulative distribution \(F(x)=P(X\le x)\) must include all area to the left, including completed earlier pieces.

For an absolutely continuous variable, F is continuous, nondecreasing, zero below the support and one above it. At a point where f changes formula, match the two CDF expressions rather than resetting the accumulated probability.

To find the distribution of \(\displaystyle Y=g(X)\), first determine its support and write \(F_Y(y)=P(g(X)\le y)\). On a monotone increasing branch, invert g to obtain \(F_X(g^{-1}(y))\), then differentiate.

A decreasing transformation reverses the inequality. If g is not one-to-one, collect every inverse branch that contributes. The Jacobian formula uses the absolute derivative of the inverse.

A density f is nonnegative and integrates to one over its support. Probabilities are areas: \(P(a<X<b)=\int_a^b f(x)\,dx\). A density may be larger than 1; a probability cannot.

Use \(E(X)=\int xf(x)\,dx\), \(E(X^2)=\int x^2f(x)\,dx\) and \(\operatorname{Var}(X)=E(X^2)-E(X)^2\). A median m solves \(\displaystyle \int_{- \infty }^{m} f(x)\, dx= \frac{1}{2}\).

Check the conditions

Three worked examples

Example 1 · Piecewise densities and cumulative functions

The density is \(f(x)=kx\ (0\le x\le 3),\quad f(x)=k(6-x)\ (3<x\le 6)\), and zero elsewhere.

Diagram for this worked example
Use the labelled scales alongside the numerical method.
  1. Find k.
  2. Find E(X).
  3. Find Var(X).
  4. Find \(\displaystyle P(X\le 1)\).
Show the reasoning

(a)

\(\displaystyle k(\frac{3^{2}}{2}+ \frac{3^{2}}{2})= 1\implies k= \frac{1}{9}\)

Answer: \(\displaystyle \frac{1}{9}\).

(b)

\(E(X)=3\) by symmetry about the centre of the support.

Answer: \(3\).

(c)

\(\displaystyle \int(x-E(X))^2f(x)\,dx=\frac{3}{2}\)

Answer: \(\displaystyle \frac{3}{2}\).

(d)

\(\displaystyle \int_{0}^{1} \frac{x}{9}\, dx= \frac{1}{18}\)

Answer: \(\displaystyle \frac{1}{18}\).

Example 2 · Transformations of continuous variables

X has density \(\displaystyle \frac{2x}{16}\) on \(0\le x\le 4\). Define \(\displaystyle Y=X^{2}\). For \(0\le y\le 16\), \(\displaystyle F_{Y}(y)= P(X\le \sqrt{y})= \frac{y}{16}\).

  1. Find the constant density of Y on its support.
  2. Find E(Y).
  3. Find Var(Y).
  4. Find \(\displaystyle P(Y\le 1)\).
Show the reasoning

(a)

\(\displaystyle f_{Y}(y)= F_{Y}'(y)= \frac{1}{16}\)

Answer: \(\displaystyle \frac{1}{16}\).

(b)

\(\displaystyle E(Y)= \frac{16}{2}= 8\)

Answer: \(8\).

(c)

\(\displaystyle \operatorname{Var}(Y)= \frac{(16)^{2}}{12}= \frac{64}{3}\)

Answer: \(\displaystyle \frac{64}{3}\).

(d)

\(\displaystyle F_{Y}(1)= \frac{1}{16}\)

Answer: \(\displaystyle \frac{1}{16}\).

Example 3 · Continuous densities and moments

X has density \(f(x)=kx^{1}\) for \(0\le x\le 5\) and zero elsewhere.

Diagram for this worked example
Use the labelled scales alongside the numerical method.
  1. Find k.
  2. Find E(X).
  3. Find Var(X).
  4. Find the median of X.
Show the reasoning

(a)

\(\displaystyle k\frac{5^{2}}{2}=1\implies k=\frac{2}{25}\)

Answer: \(\displaystyle \frac{2}{25}\).

(b)

\(\displaystyle \int_0^L xf(x)\,dx=\frac{10}{3}\)

Answer: \(\displaystyle \frac{10}{3}\).

(c)

\(\displaystyle E(X^2)-E(X)^2=\frac{25}{18}\)

Answer: \(\displaystyle \frac{25}{18}\).

(d)

\(\displaystyle (\frac{m}{5})^{2}= \frac{1}{2}\implies m= 52^{- \frac{1}{2}}\)

Answer: \(3.53553390593274\).

Common mistakes

Practise with answer checking

Original exam-style problems

Write complete solutions before opening the worked solution. The marking guidance rewards reasoning as well as answers.

Examination practice · 2020–2026

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