Mean and variance of a continuous random variable | CAIE 9709 Probability & Statistics 2 Notes

Chapters

Probability & Statistics 2

Mathematics9709

Use x f(x) for the mean and x squared f(x) next

Change the weight, not the density

Probability

∫ f(x) dx

total area

Mean

∫ x f(x) dx

weight by x

Second moment

∫ x² f(x) dx

weight by x²

Var(X) = E(X²) − [E(X)]²

The same density is weighted differently for total probability, mean and variance.

E(X)=\int_L^U x f(x)\,dx

Symbols

$L,U$ = ends of the support (same as X)
$x f(x)$ = value multiplied by its density (dimensionless inside the integral)

E(X^2)=\int_L^U x^2 f(x)\,dx

\operatorname{Var}(X)=E(X^2)-[E(X)]^2

Worked example

Build both moments from one density

Let $f(x)=\tfrac38x^2$ for $0\le x\le2$ .

E(X)=\int_0^2\frac38x^3\,dx=\frac32

E(X^2)=\int_0^2\frac38x^4\,dx=\frac{12}{5}

\operatorname{Var}(X)=\frac{12}{5}-\left(\frac32\right)^2

Answer

E(X)=\frac32

\operatorname{Var}(X)=\frac3{20}

Keep exact values through harder integration

Worked example

Current-paper mean with integration by parts

X has PDF $f(x)=1+\cos(\pi x)$ for $0\le x\le1$ . Show that $E(X)=\tfrac12-\tfrac2{\pi^2}$ .

E(X)=\int_0^1x(1+\cos(\pi x))\,dx

For the integration-by-parts term, let

I=\int x\cos(\pi x)\,dx

u=x,\qquad v=\frac{\sin(\pi x)}{\pi}

I=uv-\int v\,du

I=\frac{x\sin(\pi x)}{\pi}+\frac{\cos(\pi x)}{\pi^2}

E(X)=\frac12-\frac2{\pi^2}

(9709/61/M/J/25 Q7(b))

Key idea

Keep exact fractions and constants until the final line. Rounding the mean early can damage the variance because the mean must be squared.

Use the mean in reverse when a parameter is unknown

Worked example

Current-paper support parameter

The density simplifies to $f(x)=\tfrac{3x^2}{a^3}$ for $0\le x\le a$ . Given $E(X)=1$ , find a.

E(X)=\int_0^a x\left(\frac{3x^2}{a^3}\right)\,dx

E(X)=\frac{3a}{4}=1

Answer

a=\frac43

(9709/62/M/J/25 Q7(b))

Examiner note

If p is the probability that one item meets a condition, the expected number among N items is

Np

. This is different from the mean value

E(X)

of the continuous measurement.

Your turnIndependent mean-and-variance check [5 marks]

The PDF is $f(x)=2x$ for $0\le x\le1$ . Find the mean and variance of X.

Show worked answer

E(X)=\int_0^1 2x^2\,dx=\frac23

E(X^2)=\int_0^1 2x^3\,dx=\frac12

Answer

E(X)=\frac23

\operatorname{Var}(X)=\frac12-\frac49=\frac1{18}

Common mistake

Variance is

E(X^2)-[E(X)]^2

, not

E(X^2)-E(X)

. A negative result signals an error.

Use x f(x) for the mean and x squared f(x) next

Change the weight, not the density

Probability

∫ f(x) dx

total area

Mean

∫ x f(x) dx

weight by x

Second moment

∫ x² f(x) dx

weight by x²

Var(X) = E(X²) − [E(X)]²

The same density is weighted differently for total probability, mean and variance.

E(X)=\int_L^U x f(x)\,dx

Symbols

$L,U$ = ends of the support (same as X)
$x f(x)$ = value multiplied by its density (dimensionless inside the integral)

E(X^2)=\int_L^U x^2 f(x)\,dx

\operatorname{Var}(X)=E(X^2)-[E(X)]^2

Worked example

Build both moments from one density

Let $f(x)=\tfrac38x^2$ for $0\le x\le2$ .

E(X)=\int_0^2\frac38x^3\,dx=\frac32

E(X^2)=\int_0^2\frac38x^4\,dx=\frac{12}{5}

\operatorname{Var}(X)=\frac{12}{5}-\left(\frac32\right)^2

Answer

E(X)=\frac32

\operatorname{Var}(X)=\frac3{20}

Keep exact values through harder integration

Worked example

Current-paper mean with integration by parts

X has PDF $f(x)=1+\cos(\pi x)$ for $0\le x\le1$ . Show that $E(X)=\tfrac12-\tfrac2{\pi^2}$ .

E(X)=\int_0^1x(1+\cos(\pi x))\,dx

For the integration-by-parts term, let

I=\int x\cos(\pi x)\,dx

u=x,\qquad v=\frac{\sin(\pi x)}{\pi}

I=uv-\int v\,du

I=\frac{x\sin(\pi x)}{\pi}+\frac{\cos(\pi x)}{\pi^2}

E(X)=\frac12-\frac2{\pi^2}

(9709/61/M/J/25 Q7(b))

Key idea

Keep exact fractions and constants until the final line. Rounding the mean early can damage the variance because the mean must be squared.

Use the mean in reverse when a parameter is unknown

Worked example

Current-paper support parameter

The density simplifies to $f(x)=\tfrac{3x^2}{a^3}$ for $0\le x\le a$ . Given $E(X)=1$ , find a.

E(X)=\int_0^a x\left(\frac{3x^2}{a^3}\right)\,dx

E(X)=\frac{3a}{4}=1

Answer

a=\frac43

(9709/62/M/J/25 Q7(b))

Examiner note

If p is the probability that one item meets a condition, the expected number among N items is

Np

. This is different from the mean value

E(X)

of the continuous measurement.

Your turnIndependent mean-and-variance check [5 marks]

The PDF is $f(x)=2x$ for $0\le x\le1$ . Find the mean and variance of X.

Show worked answer

E(X)=\int_0^1 2x^2\,dx=\frac23

E(X^2)=\int_0^1 2x^3\,dx=\frac12

Answer

E(X)=\frac23

\operatorname{Var}(X)=\frac12-\frac49=\frac1{18}

Common mistake

Variance is

E(X^2)-[E(X)]^2

, not

E(X^2)-E(X)

. A negative result signals an error.