Mean and variance of a CRVAQA A-Level Further Maths: Revision notes

For a discrete variable $E(X)=\sum xP(X=x)$. For a continuous variable the sum becomes an integral over the range of $f$: $E(X)=\int_{-\infty}^{\infty}x\,f(x)\,dx.$ Only integrate over the interval where $f$ is non-zero. If $f$ is symmetric, the mean is the line of symmetry, which saves integration. Example: $f(x)=\frac{3x^2}{8}$ on $[0,2]$ gives $E(X)=\frac38\int_0^2x^3\,dx=\frac38\times4=\frac32$.

First find $E(X^2)=\int x^2f(x)\,dx$. Then $\text{Var}(X)=E(X^2)-[E(X)]^2,\qquad \text{SD}(X)=\sqrt{\text{Var}(X)}.$ Variance can never be negative; if you get a negative value, you have a mistake in $E(X^2)$ or $E(X)$. Continuing the example: $E(X^2)=\frac38\int_0^2x^4\,dx=\frac38\times\frac{32}{5}=\frac{12}{5}$, so $\text{Var}(X)=\frac{12}{5}-\frac94=\frac{3}{20}$ and $\text{SD}=\sqrt{0.15}\approx0.387$.

For constants $a$ and $b$: $E(aX+b)=aE(X)+b,\qquad \text{Var}(aX+b)=a^2\,\text{Var}(X).$ Adding $b$ shifts every value, so it changes the mean but not the spread. Multiplying by $a$ scales the spread by $|a|$, so the variance is multiplied by $a^2$. A negative $a$ never gives a negative variance. Example: $E(X)=5$, $\text{Var}(X)=4$. For $Y=2-3X$: $E(Y)=2-15=-13$ and $\text{Var}(Y)=9\times4=36$.

For any function $g$: $E(g(X))=\int_{-\infty}^{\infty}g(x)\,f(x)\,dx.$ For example $E(5X^3)=5\int x^3f(x)\,dx$ and $E(18X^{-3})=\int 18x^{-3}f(x)\,dx$. You are not allowed to substitute $E(X)$ into $g$: in general $E(g(X))\ne g(E(X))$. To find $\text{Var}(g(X))$, use $\text{Var}(g(X))=E([g(X)]^2)-[E(g(X))]^2$.

Let $f(x)=\frac{2}{x^2}$ on $[1,2]$. $E(X^{-1})=\int_1^2\frac{2}{x^3}\,dx=\left[-\frac1{x^2}\right]_1^2=\frac34$. $E(X^{-2})=\int_1^2\frac{2}{x^4}\,dx=\left[-\frac{2}{3x^3}\right]_1^2=\frac{7}{12}$. $\text{Var}(X^{-1})=\frac{7}{12}-\left(\frac34\right)^2=\frac{1}{48}$. $\text{Var}(6X^{-1})=36\times\frac{1}{48}=\frac34$.