Independent copies versus one scaled draw
State a joint model before calculating the sum distribution
- Independent pair Multiply its two marginal probabilities
- A sum value Add the probabilities of every disjoint ordered pair giving that sum
- n independent copies The mean and variance each scale with the number of copies; the standard deviation scales with its square root
- One draw times n For a positive integer multiplier, the mean and standard deviation scale with the multiplier and the variance with its square
- Assumptions Identical marginal distributions do not establish independence
Worked example
Each of two independent copies of X takes zero or two with probabilities three quarters and one quarter. Compare their sum with twice one draw
\[\mu=0\left(\frac34\right)+2\left(\frac14\right)=\frac12,\quad E(X^2)=0^2\left(\frac34\right)+2^2\left(\frac14\right)=1,\quad\sigma^2=1-\left(\frac12\right)^2=\frac34\]
- \(\displaystyle \Pr(S_2=0)=\frac34\cdot\frac34=\frac9{16},\quad\Pr(S_2=2)=2\left(\frac34\right)\left(\frac14\right)=\frac6{16},\quad\Pr(S_2=4)=\frac14\cdot\frac14=\frac1{16},\quad\Pr(T=0)=\frac34,\quad\Pr(T=4)=\frac14\)Group the ordered independent pairs by their sum and add their probabilities
- \(\displaystyle E(S_2)=E(T)=1,\quad\operatorname{Var}(S_2)=\frac32,\quad\operatorname{Var}(T)=3\)S₂ combines independent draws; T scales one draw. Their distributions differ
Common mistakes
Treating copies as one draw
✕The sum of two independent copies and twice one draw always have the same distribution✓Independence must be represented in the joint outcomes
Mixed pairs contribute extra possible sums, and their variances generally differForgetting duplicate sums
✕keep only one pair for each sum✓add all pair probabilities that give that sum
Different ordered pairs are disjoint outcomes, even when their totals matchDifficulty levels
challengeCompare independent copies with a scaled single draw
Questions about this topic
Are independent sums necessarily equally likely?+
No. Even equal-probability draws can give unequal sum probabilities because several ordered outcomes produce the same sum
Does expectation addition need independence?+
No. For finite variables it follows from the joint weighted sum. Independence is sufficient for variance addition, and is explicitly assumed here
Can I use a calculator?+
A calculator is allowed unless a question is marked no calculator
Which year level and chapter is this?+
Year 12, VCE Units 3 & 4, Specialist Mathematics, chapter 8: Probability and statistical inference
In this chapter
Standardise tails of three independent normal variablesCombine independent random variables and normal distributionsConstruct finite independent sums and compare repeated copies with scalingCalculate probabilities for normal distributions8B Distribution of sample means: Use normal sampling distributions and standard errorsConstruct sample means and interpret finite sampling distributionsConstruct and interpret confidence intervals for a population mean8C Confidence intervals for population means: Use normal sampling distributions and standard errorsPredict interval coverage and plan a width reductionForm and interpret hypothesis tests for a population mean8D Hypothesis tests for population means: Use normal sampling distributions and standard errorsCalculate theoretical test errors and supplied-tail boundaries