Continuous Expectation, Variance, And Transformations

Scope Label

Mixed scope. Integrating arbitrary pdfs to construct moments is conceptual support / enrichment under the current scope reference. The expectation, variance, transformation, and independent-sum laws are also prerequisites used directly in core normal-distribution questions. This note explains why those formulas behave as they do.

What This Note Assumes

You should already know:

• a pdf $f(x)$ gives probability by area;
• $P(a<X<b)={\int }_a^{b}f(x)dx$;
• $E(X)$ describes centre;
• $\mathrm{Var}(X)$ describes spread.

This note transfers the discrete expectation and variance ideas into the continuous setting.

Expectation as Continuous Weighted Average

For a discrete random variable,

$$E(X)=\sum xP(X=x).$$

For an absolutely continuous random variable, probability is spread continuously across the support. When $\int |x|f(x)dx<\infty$, the corresponding formula is

$$E(X)={\int }_{\text{all }x}xf(x)dx.$$

The pieces have direct meaning:

• $x$ is the value being weighted;
• $f(x)$ is the probability density near $x$;
• $f(x)dx$ represents a tiny amount of probability;
• $xf(x)dx$ represents a tiny contribution to the average.

Expectation is therefore still a weighted average. It is not necessarily the most likely value, and it does not have to be a value that the random variable can actually take.

Variance as Average Squared Spread

If

$$\mu =E(X),$$

then variance is

$$\mathrm{Var}(X)=E[(X-\mu {)}^2].$$

It measures average squared distance from the mean.

When the second moment is finite, the computational form is usually more efficient:

$$\mathrm{Var}(X)=E(X^2)-[E(X){]}^2,$$

where

$$E(X^2)={\int }_{\text{all }x}{x}^{2}f(x)dx.$$

The standard deviation is

$$\sigma =\sqrt{\mathrm{Var}(X)}.$$

It returns the measure of spread to the original unit.

In the examples below these integrals converge. The existence conditions matter because not every density has a finite mean or variance.

Worked Example: Expectation and Variance from a Pdf

Let $X$ have density

$$f(x)=2x,0\le x\le 1,$$

and $f(x)=0$ otherwise.

First check the pdf:

$${\int }_0^{1}2xdx=[x^2{]}_0^1=1.$$

Now calculate the expectation:

$$E(X)={\int }_0^{1}x(2x)dx={\int }_0^{1}2x^{2}dx=\frac{2}{3}.$$

Next calculate $E(X^2)$:

$$E(X^2)={\int }_0^1{x}^2(2x)dx={\int }_0^{1}2x^{3}dx=\frac{1}{2}.$$

So

$$\mathrm{Var}(X)=E(X^2)-[E(X){]}^2=\frac{1}{2}-{\left(\frac{2}{3}\right)}^2=\frac{1}{18}.$$

These values pass useful checks: $E(X)=2/3$ lies inside the support $[0,1]$, the variance is non-negative, and the standard deviation is

$$\sqrt{\frac{1}{18}}=\frac{1}{3\sqrt{2}}.$$

Variance has squared units; standard deviation returns to the original unit.

How to read this figure. The density $f(x)=2x$ puts more weight near $1$, so its weighted centre is $E(X)=2/3$. Expectation is a balance point of probability weight, not necessarily the highest-density value.

The workflow is:

1. check the total area is $1$;
2. calculate $E(X)$ using $\int xf(x)dx$;
3. calculate $E(X^2)$ using $\int x^{2}f(x)dx$;
4. use $\mathrm{Var}(X)=E(X^2)-[E(X){]}^2$.

Linear Transformations

For constants $a$ and $b$,

$$E(aX+b)=aE(X)+b.$$

This rule matches the meaning of average. If every value of $X$ is multiplied by $a$ and then shifted by $b$, the centre is multiplied by $a$ and shifted by $b$.

Variance behaves differently:

$$\mathrm{Var}(aX+b)=a^2\mathrm{Var}(X).$$

The shift $b$ does not affect spread. It moves the whole distribution left or right without changing distances from the mean.

The scale factor $a$ changes distances by $|a|$. Since variance uses squared distances, the variance changes by $a^2$.

How to read this figure. The positive-scale case shifts and stretches without reflection. The negative-scale case reverses orientation as well; in both cases centre follows $aE(X)+b$, spread follows $|a|\sigma$, and variance follows $a^2\mathrm{Var}(X)$.

The skewed piecewise-linear densities are illustrative. The expectation and variance laws in this section are general continuous-random-variable laws, not normal-distribution-only rules.

Worked Example: Linear Transformation After Finding Moments

Using the density from the earlier example,

$$f(x)=2x,0\le x\le 1,$$

we found

$$E(X)=\frac{2}{3},\mathrm{Var}(X)=\frac{1}{18}.$$

Suppose

$$Y=3X+2.$$

Then

$$E(Y)=E(3X+2)=3E(X)+2=3\left(\frac{2}{3}\right)+2=4.$$

For variance,

$$\mathrm{Var}(Y)=\mathrm{Var}(3X+2)=3^2\mathrm{Var}(X)=9\left(\frac{1}{18}\right)=\frac{1}{2}.$$

The shift $+2$ changes the centre but not the spread. The scale factor $3$ stretches distances from the mean by $3$, so the variance is multiplied by $9$.

For a negative-scale comparison, let $W=2-3X$. Since $0\le X\le 1$, the transformed support is $-1\le W\le 2$; the order reverses because the scale factor is negative. Moreover,

$$E(W)=2-3\left(\frac{2}{3}\right)=0,\mathrm{Var}(W)=(-3{)}^2\left(\frac{1}{18}\right)=\frac{1}{2}.$$

This is the same logic behind normal standardisation. If $X$ has mean $\mu$ and standard deviation $\sigma >0$, then

$$Z=\frac{X-\mu }{\sigma }$$

subtracts the centre and rescales the spread to standard-deviation unit size. The normal-distribution topic adds the special fact that this transformed variable follows the standard normal curve when $X$ is normal.

Equivalently, standardisation turns the mean into $0$ and the variance into $1$ when the original standard deviation is $\sigma$.

Useful Laws

For constants $a$ and $b$,

$$E(a)=a,$$ $$E(aX)=aE(X),$$

and

$$E(aX+b)=aE(X)+b.$$

For variance,

$$\mathrm{Var}(a)=0,$$ $$\mathrm{Var}(X+b)=\mathrm{Var}(X),$$

and

$$\mathrm{Var}(aX)=a^2\mathrm{Var}(X).$$

These laws are the same structural laws used for discrete random variables. The difference is how $E(X)$ and $E(X^2)$ may be computed when a pdf is given.

Sums and Differences of Random Variables

For random variables $X$ and $Y$,

$$E(X+Y)=E(X)+E(Y)$$

and

$$E(X-Y)=E(X)-E(Y).$$

These expectation laws do not require independence.

For variance, independence is the standard sufficient condition used in H2 questions. If $X$ and $Y$ are independent, then

$$\mathrm{Var}(X+Y)=\mathrm{Var}(X)+\mathrm{Var}(Y)$$

and

$$\mathrm{Var}(X-Y)=\mathrm{Var}(X)+\mathrm{Var}(Y).$$

The second formula often surprises students. Variance does not subtract for $X-Y$ because spread from independent sources accumulates.

Independence is sufficient rather than logically necessary: zero covariance also makes the cross-term vanish. If independence or equivalent information is not given, do not use the simple variance-addition rules.

Why These Laws Matter for Normal Distribution

In the normal-distribution topic, you often meet statements such as:

• $X\sim N(\mu ,{\sigma }^2)$;
• $aX+b$ is also normally distributed;
• sums of independent normal variables are normal;
• standardisation subtracts the mean and divides by the standard deviation.

Those procedures depend on the expectation and variance laws here.

For example, if

$$Y=aX+b,$$

then

$$E(Y)=a\mu +b$$

and

$$\mathrm{Var}(Y)=a^2{\sigma }^2.$$

So Continuous Random Variables explains what happens to centre and spread. Special Continuous Random Variables then applies those ideas to the special shape of a normal distribution.

Common Pitfalls

Mistake	Better thinking
Treating expectation as the most likely value	Expectation is a weighted average.
Forgetting the domain in an integral	Integrate over the support of the pdf.
Using $E(X{)}^2$ instead of $E(X^2)$	These are different quantities.
Writing $\mathrm{Var}(aX+b)=a\mathrm{Var}(X)+b$	Variance rescales by $a^2$ and is not shifted by $b$.
Subtracting variances for $X-Y$	For independent variables, variances add for both sums and differences.
Using variance-addition without justification	Independence is the usual sufficient condition supplied in H2 questions.

Revision Checklist

• Can you explain $E(X)=\int xf(x)dx$ as a continuous weighted average?
• Can you calculate $E(X)$ and $E(X^2)$ from a simple pdf?
• Can you use $\mathrm{Var}(X)=E(X^2)-[E(X){]}^2$?
• Can you explain why adding $b$ changes expectation but not variance?
• Can you explain why multiplying by $a$ changes variance by $a^2$?
• Can you state when variances of sums or differences may be added?
• Can you connect these laws to normal-distribution transformations?
• Can you explain why standardisation changes mean to $0$ and variance to $1$?