Let X₁ and X₂ be independent, X₁ gamma with mean 10 and variance 10, and X₂ ~ N(3, 4). Let f₁, f₂ be their densities. Define Y with density f(y) = 0.4 f₁(y) + q f₂(y). Which of the following are true?

A, B. Densities integrate to 1 ⇒ q = 0.6. Mixture mean 0.4·10 + 0.6·3 = 5.8. E[Y²] = 0.4(10 + 100) + 0.6(4 + 9) = 51.8, so Var(Y) = 51.8 − 5.8² = 18.16. A mixture is not a linear combination of the variables.

Part CCSIR NET June 2023mixture-vs-linear-combination

Mixture vs linear combination

Let $X_{1}$ and $X_{2}$ be independent, $X_{1}$ gamma with mean 10 and variance 10, and $X_{2}$ ~ N(3, 4). Let $f_{1}, f_{2}$ be their densities. Define Y with density $f(y) = 0.4 f_{1}(y) + q f_{2}(y)$ . Which of the following are true?

A.q = 0.6✓ Correct
B.E[Y] = 5.8✓ Correct
C.Var(Y) = 3.04
D. $Y = 0.4X_{1} +$ qX $_{2}$

The idea, in plain words

The two weights in a mixture must add to one, which fixes the unknown weight immediately. After that the mixture's average is the weighted average of the two, but the spread is not — you have to go through the average of the squares.

Solution

Densities integrate to 1 ⇒ q = 0.6.

Mixture mean 0.4·10 + 0.6·3 = 5.8.

$E[Y^{2}] = 0.4(10 + 100) + 0.6(4 + 9) = 51.8$

so $Var(Y) = 51.8 - 5.8^{2} = 18.16$ .

A mixture is not a linear combination of the variables.

Why each option is right or wrong

Checked against 5 symbolic computations

A.Correct
True.

A density must integrate to 1, and $f_{1}$ and $f_{2}$ each already do

so 0.4 + q = 1 and q = 0.6.

(For the record, SymPy solves the gamma parameters from mean 10 and variance 10 as shape 10, scale 1.) The trigger: the mixing weights of a density mixture must sum to 1 — that alone fixes q before any moment is computed.
B.Correct
True.

For a mixture, moments mix with the same weights: $E[Y] = 0.4\cdot{}E[X_{1}] + 0.6\cdot{}E[X_{2}] = 0.4\cdot10 + 0.6\cdot3 = 5.8 ($ verified).

Note that the mean of a mixture does coincide with the mean of the corresponding linear combination — which is exactly why (C) and (D) are tempting.
C.WrongExecution slip
False, and the number offered is revealing.

Variances do not mix with the weights.

Correctly, $E[Y^{2}] = 0.4(10 + 100) + 0.6(4 + 9) = 51.8$

so $Var(Y) = 51.8 - 5.8^{2} = 18.16 ($ verified).

SymPy also computes $0.4^{2}\cdot10 + 0.6^{2}\cdot4 = 3.04$ — precisely the option offered — which is the variance of the linear combination $0.4X_{1} + 0.6X_{2}$ .

This option is the right formula applied to the wrong object.
D.WrongExecution slip
False, and it is the conceptual error behind (C).

A mixture is a random choice between two distributions — with probability 0.4 you observe a gamma variable, otherwise a normal one — not a weighted sum of both.

The distinction shows up immediately in the variance (18.16 versus 3.04) and in the shape: a mixture here is bimodal

while a linear combination is not.

2 of these 2 wrong options correspond to a named reasoning error — the same errors recur across subjects.

The trap it tests

Invariants don't determine

Two objects sharing an invariant were treated as the same object.

Drill statements like this

Related counterexample: The maximum of independent exponentials is exponential

Mixture vs linear combination

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