Combine independent random variables and normal distributions

Start from transformations and independent sums

A single transformation aX+c has mean aE(X)+c and variance a²Var(X). Here the variables may have different distributions, means and variances

Linearity of expectation has no independence requirement

For variables with finite means and fixed real coefficients, the mean of a linear combination is the same combination of the means. A constant shift adds to the mean. This rule also holds for dependent variables

Independent variances add after squaring coefficients

For mutually independent variables with finite variances, first square each coefficient, then multiply by that variable’s variance and add. A fixed constant contributes zero variance. Subtraction does not subtract variance. Standard deviations themselves do not add

Work a signed combination

Let independent X and Y have means 4 and -2, and variances 3 and 5. Combine them as follows. If the input quantities are in centimetres and the coefficients are dimensionless, the mean and standard deviation are in centimetres and variance is in square centimetres

Let X and Y be independent with Var(X)=3 and Var(Y)=5. For W=2X-3Y+7, give Var(W)

Recover a missing variance from a combination

Suppose independent X and Y satisfy Var(X)=2 and Var(3X-2Y+5)=34. Then 9×2+4Var(Y)=34, so Var(Y)=4. The variance of X-Y is 2+4=6. A proposed negative recovered variance would show that the supplied assumptions or data are inconsistent

Independent X and Y satisfy Var(X)=2 and Var(3X-2Y+5)=34. Give Var(X-Y)

Extend the same law to several distinct variables

Let independent X,Y,Z have means 2,-1,4 and variances 1,3,2. For T=3X+2Y-Z, E(T)=6-2-4=0 and Var(T)=9+12+2=23. Mean zero does not mean the outcome is always zero. For independent identically distributed Xᵢ with mean μ and variance σ², the special unweighted sum has mean nμ and variance nσ²

Independent X, Y and Z have variances 1, 3 and 2 respectively. Give the variance of T=3X+2Y-Z.

Build a finite distribution from joint outcomes

Take independent X and Y with these distributions. For T=X+Y, the ordered pairs (0,-1), (0,1), (2,-1) and (2,1) give sums -1, 1, 1 and 3 respectively. Add the two contributions to the sum of 1.

Independent X and Y satisfy Pr(X=0)=1/3, Pr(X=2)=2/3, Pr(Y=-1)=1/4 and Pr(Y=1)=3/4. For T=X+Y, what explains the mass at T=1 in this graph?

Check a distribution using its moments

The three masses sum to1. Directly, E(T)=(-1)/12+5/12+3/2=11/6. Also E(X)=4/3 and E(Y)=1/2, giving the same sum. Var(X)=8/9 and Var(Y)=3/4, so Var(T)=59/36. Mapping to aX+bY+c can merge more outcomes, including a single constant when both coefficients are zero

Independent X and Y satisfy Pr(X=0)=1/3, Pr(X=2)=2/3, Pr(Y=-1)=1/4 and Pr(Y=1)=3/4. For T=X+Y, give Pr(T=1)

Normal combinations have a full distribution

If the inputs are mutually independent normal variables, their linear combination is normal when its calculated variance is positive. State the mean and variance as N(μ,σ²). If every coefficient is zero, the result is constant; an ordinary positive-variance normal model and standardisation are not available. Independent non-normal variables need not have a normal sum

Work a weighted normal combination

For independent normal X and Y, form the weighted combination below. Its mean and variance determine its normal distribution. The stated boundary is one standard deviation above the mean, giving an upper-tail probability of approximately 0.1587. The independent normal theorem applies to different nonzero real coefficients, not only to an unweighted sum

Independent X and Y have Var(X)=9 and Var(Y)=16. Give the variance of T=2X-3Y+1.

Work a normal difference

Let independent X~N(50,9) and Y~N(30,16), and W=X-Y+2. The mean is22 and variance9+16=25, so W~N(22,25) and its standard deviation is5. For Pr(W>27), standardise to z=(27-22)/5=1. The probability is1-Φ(1)≈0.1587, using a calculator and rounding only at the end

Independent X and Y have Var(X)=9 and Var(Y)=16. Give the standard deviation of W=X-Y+2.

Standardise each bound and choose the region

For the same W~N(22,25), Pr(17<W<29.5)=Pr(-1<Z<1.5)=Φ(1.5)-Φ(-1)≈0.7745. The probability outside that interval is1-0.774537…≈0.2255. Keep guard digits before subtraction or complementation. Equality at a boundary does not change a probability for this continuous variable

For W~N(22,25), give the standard score of 29.5

Repeated symbols can refer to the same source

If X and Y are independent, with means4,-2 and variances3,5, then W=2X-X+Y is X+Y: its mean is2 and variance8. In contrast V=2X₁-X₂+Y, for mutually independent copies X₁,X₂ and Y, has mean2 and variance4×3+3+5=20. Copy labels require a stated joint assumption; notation alone does not create independence

The two occurrences of X are the same variable. Independent X and Y have Var(X)=3 and Var(Y)=5. Give Var(2X-X+Y)

Turn a comparison into one random variable

With independent X~N(50,9) and Y~N(30,16), Pr(X>Y+25) equals Pr(X-Y>25). The difference has mean20 and variance25, so the event standardises to Z>1 and has probability≈0.1587. It is not the product Pr(X>25)Pr(Y<25), and replacing random Y by its mean would lose its variability

Independent X and Y have variances 9 and 16 respectively. Give the variance of X-Y.

Choose a coefficient to meet a target mean

Let independent X,Y have means3,-2 and variances4,9. For T=(1-r)X+rY, a zero target mean requires3(1-r)-2r=0, giving r=3/5. Then Var(T)=(2/5)²×4+(3/5)²×9=97/25. The coefficient on each mean must stay attached to the same variable in the variance calculation

For E(X)=3, E(Y)=-2 and T=(1-r)X+rY, give r such that E(T)=0.

Check what is and is not justified

Independence is sufficient for the variance sum, while expectation linearity needs no independence. Equal means, equal variances or matching marginal distributions do not establish independence. Normal marginals alone are not the stated independent-normal combination theorem. In applications, shared sources, common conditions and repeated measurements can create dependence

Derive the two combination laws, ungraded

For mutually independent variables with finite variances, derive the mean and variance of T=c+ΣaᵢXᵢ. State exactly where independence enters, explain why signs are squared for variance, and distinguish the case of a constant result

Model: expand centred variables before discarding cross terms

Linearity gives E(T)=c+Σaᵢμᵢ without an independence assumption. Subtract this mean to obtain T-E(T)=Σaᵢ(Xᵢ-μᵢ). Squaring and taking expectation yields Σaᵢ²Var(Xᵢ) plus twice the cross products aᵢaⱼE[(Xᵢ-μᵢ)(Xⱼ-μⱼ)]. Independence factors each cross expectation into two zero means, so those terms vanish. Each remaining coefficient is squared. If the sum of these nonnegative variance contributions is zero, T equals its mean with probability1; positive standard deviation is required for standardisation. Independence is a sufficient condition here, not a necessary condition for every possible variance-addition identity

Construct and check a non-identical sum, ungraded

Independent X and Y satisfy Pr(X=0)=1/3, Pr(X=2)=2/3, Pr(Y=-1)=1/4 and Pr(Y=1)=3/4. Construct the entire distribution of X+Y. Verify its total mass, mean and variance directly, then compare with the independent-sum laws. Explain the role of duplicate sums

Model: aggregate the ordered pairs and verify moments

For independent X with Pr(X=0)=1/3 and Pr(X=2)=2/3, and Y with Pr(Y=-1)=1/4 and Pr(Y=1)=3/4, let T=X+Y. The products are1/12,1/4,1/6,1/2 for pairs (0,-1),(0,1),(2,-1),(2,1). The sums -1,1,1,3 yield masses1/12,5/12,1/2, totalling1. The mean is11/6 and second moment is1/12+5/12+9/2=5, giving variance5-(11/6)²=59/36. Separately, E(X)+E(Y)=4/3+1/2=11/6 and Var(X)+Var(Y)=8/9+3/4=59/36. Duplicate sums correspond to different disjoint joint outcomes; their probabilities must be added

Explain the normal comparison model, ungraded

For independent measurement errors X~N(4,9) and Y~N(-1,16), find Pr(X>Y+10) and explain both the distribution argument and its modelling assumptions. Discuss what changes if Y is instead defined to be X.

Model: retain both uncertainty contributions

Independence and normality imply X-Y~N(5,25). The event is X-Y>10, so its standard score is1 and its probability is≈0.1587. The inputs may have different means and variances; independence concerns their joint errors and needs support from the measurement design, not only the marginal normal fits. If Y is defined to be X, its supplied marginal must also change to match X. Then X-Y=0 identically and Pr(X>Y+10)=0; the independent variance sum and positive-variance standardisation would be invalid

Distinguish matched marginals from independence, ungraded

Let X be equally likely to be -1 or1. Compare Y=X with an independent copy Y of X. In each case calculate E(X+Y), Var(X+Y) and Var(X-Y), and explain why marginal information alone is insufficient

Model: two different joint laws give different spreads

Both cases have E(X)=E(Y)=0 and Var(X)=Var(Y)=1. If Y=X, then X+Y=2X has variance4 and X-Y=0 has variance0; the sum mean is0. For independent copies, the sum and difference each have variance2 and mean0. The marginals are identical in both constructions, but the joint outcomes differ. This example also shows why a negative coefficient cannot simply be used to subtract independent variances

Assessment boundary

The marked activities check specified values, finite masses, normal probabilities and explicit assumption choices. Full derivations, joint-model explanations and evidence for independence remain ungraded. Sample means and statistical inference build on these combination laws in the next section