Combining Normals

Means add

X and Y are independent normal variables: X ~ N(1, 1) and Y ~ N(3, 1). What can be said about their sum, X + Y?

The mean of a sum is the sum of the means: E(X + Y) = E(X) + E(Y) = 1 + 3 = 4. On average, X contributes 1 and Y contributes 3, so together they average 4. This holds for any two random variables, independent or not.

Variances add

For independent variables the variances add as well: Var(X + Y) = Var(X) + Var(Y) = 1 + 1 = 2. And a sum of independent normal variables is normal again, so X + Y ~ N(4, 2).

The sum is more spread out than either part. Sometimes X and Y are both above their means, and then the sum is further above 4 than either part is above its own mean. Sometimes both are below, and the sum is further below. So the sum has the widest curve of the three.

It is the variances that add, not the standard deviations. The standard deviation of X + Y is √2 ≈ 1.41, not 1 + 1 = 2: a high X often comes with a low Y, and part of the spread cancels.

The two plain curves are X ~ N(1, 1) and Y ~ N(3, 1). The gold curve is their sum, N(4, 2): centered at 1 + 3 = 4, and wider and lower than either, with variance 1 + 1 = 2.

Subtracting adds the variances too

For the difference, the means subtract: E(Y − X) = E(Y) − E(X) = 3 − 1 = 2. The variances still add: Var(Y − X) = Var(Y) + Var(X) = 1 + 1 = 2. So Y − X ~ N(2, 2).

The reason is that −X is just as spread out as X. Its curve is the mirror image of the curve of X, the same width facing the other way, so Var(−X) = Var(X), and Y − X is the sum of Y and −X. A difference is uncertain because each of its two parts is, so the spreads add. Subtracting the variances, 1 − 1 = 0, would say Y − X never varies at all.

The plain curves are X ~ N(1, 1) and Y ~ N(3, 1) again. The gold curve is the difference Y − X, N(2, 2): centered at 3 − 1 = 2, and as wide as the sum was.

Will the cup overflow?

A coffee machine pours X ml of coffee, with X ~ N(240, 6²), into a cup that holds Y ml, with Y ~ N(250, 8²), independently. The cup overflows when the coffee is more than the cup holds, which is when Y − X < 0.

Let D = Y − X, the room left in the cup. Its mean is 250 − 240 = 10 ml, and its variance is 6² + 8² = 36 + 64 = 100, so D ~ N(10, 10²), with standard deviation 10 ml.

Standardize 0: z = (0 − 10)/10 = −1. So P(D < 0) = P(Z < −1) = 1 − 0.8413 = 0.1587. About 16% of cups overflow, although the average cup has 10 ml to spare.

The room left in a cup, N(10, 10²) in ml. The shaded part, below 0, is the cups that overflow: one standard deviation below the mean, 0.1587 of the area.

Only when independent

Means always add. Variances add only when the variables are independent, and dependence can change the spread in either direction.

The extreme case shows it. Take Y to be X itself. Then X + X = 2X, every value doubled, so the spread doubles too: the standard deviation is 2σ and the variance is 4σ², not σ² + σ² = 2σ². And X − X = 0 every time, with variance 0, not 2σ².

Two independent cups of coffee from the same machine are different. Their total varies less than one cup doubled, because a full cup is often paired with a short one: Var(X₁ + X₂) = 2σ², against Var(2X) = 4σ².

The usual mistakes

Subtracting the variances for a difference. Var(X − Y) is Var(X) + Var(Y), whether the variables are added or subtracted.

Adding the standard deviations. With standard deviations 6 and 8, the difference has standard deviation √(36 + 64) = 10, not 6 + 8 = 14.

Leaving one variable out. Var(X − Y) is not Var(X) alone: subtracting an independent Y still brings in its spread.

Worked example: Filled Jam Jars on a Packing Line: the Total Mass of Jar and Jam, and Lids That Must Be Wider Than the Neck

Question At a jam factory the mass of an empty jar is J ∼ N(200, 3²) and the mass of jam put into it is M ∼ N(450, 4²), in grams, independently. (a) Find the probability that a filled jar has a total mass under 640 g. (b) The inside diameter of a lid is L ∼ N(71.25, 0.4²) and the outside diameter of a jar's rim is R ∼ N(70.0, 0.3²), in millimeters, independently. A lid fits only if it is wider than the rim. Find the probability that a lid chosen at random does not fit a jar chosen at random.

1.The total mass is T = J + M, with E(T) = 200 + 450 = 650. For independent variables the variances add: Var(T) = 3² + 4² = 9 + 16 = 25, so T ∼ N(650, 5²).

The means add and, for independent masses, so do the variances: T ∼ N(650, 5²).
2.(a) z = 640 − 6505 = −2, so P(T < 640) = 1 − Φ(2) = 1 − 0.9772 = 0.0228.

(a) 640 g is at z = −2, and the shaded tail is 1 − Φ(2) = 0.0228.
3.The lid does not fit when L − R < 0. Let D = L − R; its mean is E(D) = 71.25 − 70.0 = 1.25 mm.

A lid does not fit when D = L − R is negative. The mean of D is 1.25 mm.
4.The variances add for a difference too: Var(D) = 0.4² + 0.3² = 0.16 + 0.09 = 0.25, so D ∼ N(1.25, 0.5²).

Var(D) = 0.4² + 0.3² = 0.25, so D ∼ N(1.25, 0.5²) and 0 sits at z = −2.5.
5.(b) z = 0 − 1.250.5 = −2.5, so P(D < 0) = 1 − Φ(2.5) = 1 − 0.9938 = 0.0062. About 6 lids in 1000 will not fit a jar picked at random.

(b) The shaded tail is P(D < 0) = 1 − Φ(2.5) = 0.0062: about 6 lids in 1000 do not fit.

Answer: (a) 0.0228; (b) 0.0062

Common mistakes

Subtracting the variances for L − R: 0.16 − 0.09 = 0.07. A difference is as uncertain as both of its parts together, so the variances add.
Adding the standard deviations, 3 + 4 = 7 g. Only variances add; the standard deviation of the total is √25 = 5 g.

More probability distributions problems, worked step by step →