SR4–SR6 - Expectation and variance of combinations

SR4–SR6 - Expectation and variance of combinations

Work out how changing a scale, adding a constant or combining independent observations changes a distribution. SR4 and SR5 transform its mean and variance; SR6 combines variables and explains why averaging fresh observations reduces variation.

Transform the expected value

Two instruments report averages of two readings. One takes two fresh readings; the other accidentally uses its first reading twice. Can their reported averages have the same expected value but different variances? We will calculate both.

You already know

Expectation is the probability-weighted mean. Variance measures squared deviations from that mean, and standard deviation is its non-negative square root. A constant shift moves every value; a scale factor multiplies every value.

If each possible score xx becomes 10+2x10+2x, its contribution to the mean changes in the same linear way. The probabilities do not change. This gives a quick route to the new mean without rebuilding the entire table.

Expectation under a linear transformation

For constants a,ba,b and a discrete random variable XX,

E(a+bX)=a+bE(X)E(a+bX)=a+bE(X)

A shift adds aa to the mean, and a scale factor multiplies the mean by bb.

Not in the formula booklet: learn it.

For a finite table, the reason is ∑(a+bxi)pi=a∑pi+b∑xipi=a+bE(X)\sum(a+bx_i)p_i=a\sum p_i+b\sum x_i p_i=a+bE(X). The constant contributes once, since the probabilities sum to one.

Worked example 1: an offset and a scale

Question. A random variable XX has mean 33. Define Y=10+2XY=10+2X. Find E(Y)E(Y).

Plan. Apply the same transformation to the mean.

E(Y)=E(10+2X)identify the transformed variable=10+2E(X)linearity=10+2(3)=16substitute the mean\begin{aligned} E(Y)&=E(10+2X) &&\text{identify the transformed variable}\\ &=10+2E(X) &&\text{linearity}\\ &=10+2(3)=16 &&\text{substitute the mean} \end{aligned}

Why this step? Adding ten affects every observation once; it contributes ten, rather than ten times the number of possible outcomes.

Answer. E(Y)=16E(Y)=16.

Check. If XX happened to be constant at 33, every YY would be 1616; the same mean rule works when XX varies.

Change only the offset from 1010 to −10-10. The whole distribution and its mean shift down by 2020.

Worked example 2: a negative offset

Question. With E(X)=3E(X)=3, define Z=−10+2XZ=-10+2X. Find E(Z)E(Z).

Plan. Retain the multiplier and replace the constant term.

E(Z)=−10+2E(X)linear transformation=−10+2(3)=−4substitute\begin{aligned} E(Z)&=-10+2E(X) &&\text{linear transformation}\\ &=-10+2(3)=-4 &&\text{substitute} \end{aligned}

Why this step? Expected values can be negative when the variable represents a signed quantity such as net gain.

Answer. E(Z)=−4E(Z)=-4.

Check. E(Z)−E(Y)=−4−16=−20E(Z)-E(Y)=-4-16=-20, matching the change in offset.

The expectation rule concerns a variable's values, not probabilities: probabilities must remain between zero and one, but a random variable's mean need not.

Watch out

Wrong: E(−10+2X)=2E(X)E(-10+2X)=2E(X). The constant shifts every value and therefore shifts the mean too. Correct: E(−10+2X)=−10+2E(X)=−4E(-10+2X)=-10+2E(X)=-4 when E(X)=3E(X)=3.

Keep the offset when applying the next transformation.

Transform variance and standard deviation

Adding a constant changes where the values are, but not how far apart they are. Multiplying by bb multiplies every deviation from the new mean by bb; the squared deviations are therefore multiplied by b2b^2.

Variance under a linear transformation

For constants a,ba,b,

Var⁡(a+bX)=b2Var⁡(X),sd⁡(a+bX)=∣b∣sd⁡(X)\operatorname{Var}(a+bX)=b^2\operatorname{Var}(X),\qquad \operatorname{sd}(a+bX)=|b|\operatorname{sd}(X)

The offset contributes no variance. A negative multiplier reflects the distribution and still gives non-negative variance and standard deviation.

Not in the formula booklet: learn it.

Indeed, (a+bX)−E(a+bX)=b(X−E(X))(a+bX)-E(a+bX)=b(X-E(X)). This shows exactly why the offset disappears and why the multiplier must be squared in the variance.

Worked example 3: scale the spread

Question. A random variable XX has variance 99. For Y=10+2XY=10+2X, find the variance and standard deviation.

Plan. Square the scale factor for variance, then take a square root.

Var⁡(Y)=22Var⁡(X)offset does not affect spread=4(9)=36squared multipliersd⁡(Y)=36=6standard deviation\begin{aligned} \operatorname{Var}(Y)&=2^2\operatorname{Var}(X) &&\text{offset does not affect spread}\\ &=4(9)=36 &&\text{squared multiplier}\\ \operatorname{sd}(Y)&=\sqrt{36}=6 &&\text{standard deviation} \end{aligned}

Why this step? The distances double, so their squares become four times as large.

Answer. Variance 3636; standard deviation 66.

Check. The original standard deviation is 33, and multiplying distances by 22 gives 66.

Change only the multiplier's sign. The values are reflected, but the spread is unchanged.

Worked example 4: a negative multiplier

Question. With Var⁡(X)=9\operatorname{Var}(X)=9, define Z=10−2XZ=10-2X. Find its variance and standard deviation.

Plan. Include the minus sign inside the square, and report a non-negative standard deviation.

Var⁡(Z)=(−2)2(9)=36square the signed factorsd⁡(Z)=∣−2∣9=6absolute scale factor\begin{aligned} \operatorname{Var}(Z)&=(-2)^2(9)=36 &&\text{square the signed factor}\\ \operatorname{sd}(Z)&=|-2|\sqrt9=6 &&\text{absolute scale factor} \end{aligned}

Why this step? A reflected distribution has the same distances from its reflected mean; its variance cannot become negative.

Answer. Variance 3636; standard deviation 66.

Check. Y+Z=20Y+Z=20, so one distribution is a reflection of the other about 1010.

In money problems, state the requested measure. A variance in square pounds is not a standard deviation in pounds.

Watch out

Wrong: Var⁡(10−2X)=10−2Var⁡(X)\operatorname{Var}(10-2X)=10-2\operatorname{Var}(X). Right: Var⁡(10−2X)=4Var⁡(X)\operatorname{Var}(10-2X)=4\operatorname{Var}(X). Constants shift the centre; squared multipliers change variance.

Try a new multiplier and variance.

Combine independent random variables

Suppose XX and YY are fresh, independent quantities. A total adds their contributions; a difference subtracts their centres. For independent quantities, uncertainty from either one still contributes to the spread of a difference.

Linear combinations

For independent random variables X,YX,Y and constants a,b,ca,b,c,

E(c+aX+bY)=c+aE(X)+bE(Y)Var⁡(c+aX+bY)=a2Var⁡(X)+b2Var⁡(Y)\begin{aligned} E(c+aX+bY)&=c+aE(X)+bE(Y)\\ \operatorname{Var}(c+aX+bY)&=a^2\operatorname{Var}(X)+b^2\operatorname{Var}(Y) \end{aligned}

The expectation rule does not require independence. The variance rule here does. The same rules extend to any finite number of mutually independent variables.

Not in the formula booklet: learn it.

Use these results without proving the general variance rule. In particular, a difference uses a negative coefficient whose square is positive.

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