Paper 6. Total weights, differences in times, a scaled measurement: whenever random quantities are combined, the mean follows the obvious rule, the variance follows a less obvious one, and normal and Poisson shapes survive the combination.
\(\text{E}(aX + bY) = a\text{E}(X) + b\text{E}(Y)\) always. For independent \(X\) and \(Y\), \(\text{Var}(aX + bY) = a^{2}\text{Var}(X) + b^{2}\text{Var}(Y)\), so variances ADD even for a difference. Constants shift the mean but not the variance. Combinations of independent normals are normal; sums of independent Poissons are Poisson.
Fig. 1: Ann's time A ~ N(25, 9) and Ben's time B ~ N(23, 16). The difference A − B ~ N(2, 25); the shaded area P(A − B < 0) = 0.3446 is the probability that Ann is faster
Section 01
Scaling one variable: aX + b.
Multiplying by \(a\) stretches the spread by \(|a|\), so the variance is multiplied by \(a^{2}\). Adding \(b\) slides the whole distribution along without changing its spread.
Quantity
Rule
\(\text{E}(aX + b)\)
\(a\text{E}(X) + b\)
\(\text{Var}(aX + b)\)
\(a^{2}\text{Var}(X)\)
If \(X \sim N(\mu, \sigma^{2})\)
\(aX + b \sim N(a\mu + b,\ a^{2}\sigma^{2})\)
Worked example Mean and variance of a transformed variable
\(\text{E}(X) = 12\) and \(\text{Var}(X) = 4\). Find the mean and variance of \(Y = 3X - 5\).
\(\text{E}(Y) = 3 \times 12 - 5 = 31\).
\(\text{Var}(Y) = 3^{2} \times 4 = 36\). The \(-5\) has no effect on the spread.
The noon temperature \(C\) in degrees Celsius satisfies \(C \sim N(20, 9)\). Find the distribution of the temperature in Fahrenheit, \(F = 1.8C + 32\).
\(F\) is a linear function of a normal variable, so it is normal.
\(\text{Var}(3X - 5)\) is not \(3 \times 4 - 5\), and not \(3 \times 4\). The factor is squared and the constant disappears: \(9 \times 4 = 36\).
Section 02
Sums and differences: variances add.
For independent \(X\) and \(Y\), \(\text{Var}(X - Y) = \text{Var}(X) + \text{Var}(Y)\): taking away an uncertain amount adds uncertainty. Also be clear about the difference between doubling one value and adding two separate values.
\(2X\): one observation doubled. \(\text{Var}(2X) = 4\sigma^{2}\).
Both have mean \(2\mu\), but the sum of two separate items is less spread out, because a high value and a low value partly cancel.
Total of \(n\) items: \(\text{E} = n\mu\), \(\text{Var} = n\sigma^{2}\).
Worked example A total of several items
Apples have masses \(A \sim N(150, 20^{2})\) grams and empty boxes have masses \(B \sim N(500, 30^{2})\) grams, all independent. A box holds 6 apples. Find the probability that a full box has mass more than 1450 g.
Total \(T = A_1 + \cdots + A_6 + B\), which is normal.
\(\text{E}(T) = 6 \times 150 + 500 = 1400\).
\(\text{Var}(T) = 6 \times 400 + 900 = 3300\) (six separate apples, so \(6\sigma^{2}\), not \(36\sigma^{2}\)).
Ann's time to run a course is \(A \sim N(25, 9)\) minutes and Ben's is \(B \sim N(23, 16)\) minutes, independently. Find the probability that Ann is faster than Ben.
Ann is faster when \(A < B\), that is \(A - B < 0\).
Worked example Three of one against two of another
\(X \sim N(20, 4)\) and \(Y \sim N(28, 9)\) independently. Find the probability that the sum of three independent values of \(X\) is greater than twice a single value of \(Y\).
Let \(D = X_1 + X_2 + X_3 - 2Y\). We need \(\text{P}(D > 0)\).
\(X \sim N(10, 4)\). Find \(\text{P}(X_1 + X_2 > 23)\) for two independent values, and say why this differs from \(\text{P}(2X > 23)\).
\(X_1 + X_2 \sim N(20, 8)\): \(z = \dfrac{3}{\sqrt{8}} = 1.061\), probability \(1 - 0.8556 = 0.144\). But \(2X \sim N(20, 16)\): \(z = 0.75\), probability \(0.227\). One doubled value is more variable than two separate values.
Examiner trap
The commonest Paper 6 error: \(\text{Var}(A - B) = 9 - 16\). A variance can never be negative; for independent variables they always add.
Stage 1 · Learn
Check your understanding
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\(X\) and \(Y\) are independent. Which is \(\text{Var}(X - Y)\)?
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\(X \sim \text{Po}(2)\) and \(Y \sim \text{Po}(3)\) are independent. Which of these has a Poisson distribution?
Section 03
Sums of Poisson variables.
If \(X \sim \text{Po}(\lambda)\) and \(Y \sim \text{Po}(\mu)\) are independent, then \(X + Y \sim \text{Po}(\lambda + \mu)\). This only works for sums of the raw counts: \(2X\) and \(X - Y\) are not Poisson.
Worked example Combining two sources
Emails from colleagues arrive at random at a mean rate of 2 per hour, and independently emails from customers arrive at a mean rate of 3.5 per hour. Find the probability that exactly 4 emails arrive in an hour.
Total \(T \sim \text{Po}(2 + 3.5) = \text{Po}(5.5)\).
\(X \sim \text{Po}(3)\). Show that \(2X\) does not have a Poisson distribution.
\(\text{E}(2X) = 2 \times 3 = 6\).
\(\text{Var}(2X) = 2^{2} \times 3 = 12\).
A Poisson variable has mean equal to variance; \(6 \ne 12\). (Also \(2X\) can only take even values.)
Answer: Mean 6, variance 12, so \(2X\) is not Poisson.
Examiner trap
Before adding Poisson parameters, put them on the same interval. A rate of 2 per hour and 0.5 per 10 minutes give \(\text{Po}(2 + 3) = \text{Po}(5)\) for an hour, not \(\text{Po}(2.5)\).
Stage 2 · Exam
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\(X\) and \(Y\) are independent with \(\text{Var}(X) = 3\) and \(\text{Var}(Y) = 2\). Find \(\text{Var}(2X - 3Y)\).
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Bags of flour have masses \(N(0.5, 0.03^{2})\) kg, independently. Find the probability that the total mass of 4 bags exceeds 2.1 kg.
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\(X \sim \text{Po}(1.8)\) and \(Y \sim \text{Po}(2.7)\) are independent. Find \(\text{P}(X + Y = 3)\).
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\(X \sim N(60, 16)\) and \(Y \sim N(28, 9)\) are independent. Find \(\text{P}(X > 2Y)\).
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