Let X = (X₁, X₂)ᵀ be bivariate normal with mean (0, 0)ᵀ and covariance Σ = [[5, −3], [−3, 10]]. The mean vector and covariance matrix of Y = (X₁, 5 − 2X₂)ᵀ are

D. E[5 − 2X₂] = 5; Var(5 − 2X₂) = 4·10 = 40; Cov(X₁, 5 − 2X₂) = −2·(−3) = 6.

Part BCSIR NET June 2023

Multivariate normal distribution

Let $X = (X_{1}, X_{2})^{T}$ be bivariate normal with mean (0, 0)ᵀ and covariance $\sum = [[5, -3], [-3, 10]]$ . The mean vector and covariance matrix of $Y = (X_{1}, 5 - 2X_{2})^{T}$ are

A.(0, 5)ᵀ, [[5, −3], [−3, 40]]
B.(0, 5)ᵀ, [[5, −6], [−6, 20]]
C.(0, 5)ᵀ, [[5, 3], [3, 20]]
D.(0, 5)ᵀ, [[5, 6], [6, 40]]✓ Correct

The idea, in plain words

A linear recipe applied to jointly normal variables stays normal, and its centre, spread, and overlap follow from three mechanical rules. Only remember that scaling multiplies the spread by the square of the factor, not the factor.

Solution

$E[5 - 2X_{2}] = 5; Var(5 - 2X_{2}) = 4\cdot10 = 40; Cov(X_{1}, 5 - 2X_{2}) = -2\cdot(-3) = 6$ .

Why each option is right or wrong

Checked against 2 symbolic computations

A.WrongExecution slip
False on the off-diagonal.

The variance 40 is right

but the covariance must also be rescaled: $Cov(X_{1}, 5 - 2X_{2}) = -2\cdot{}Cov(X_{1}, X_{2}) = -2(-3) = 6 ($ verified), not −3.

Scaling one coordinate scales every entry of the covariance matrix that involves it, not just the diagonal one.
B.WrongExecution slip
False on both entries.

The variance should be $(-2)^{2}\cdot10 = 40$ , not 20 — the factor is squared.

And the covariance should be +6, not −6: multiplying by −2 flips the sign of the original −3 to give a positive value.
C.Wrong
False on the variance.

The sign of the covariance is now right at +3

but it has not been scaled by |−2| = 2, and the variance uses 2·10 rather than $2^{2}\cdot10$ .

Both errors come from applying the factor to the first power instead of squaring where required.
D.Correct
True.

The mean is $E[(X_{1}, 5 - 2X_{2})] = (0, 5)$ since $E[X_{2}] = 0$ .

For the covariance, apply Cov(AX + b) = A·Cov(X)·Aᵀ with A = diag(1, −2): the (1,1) entry is unchanged at 5, the (2,2) entry becomes $(-2)^{2}\cdot10 = 40 ($ verified), and the off-diagonal becomes 1·(−2)·(−3) = 6 (verified).

The additive constant 5 shifts the mean and leaves the covariance alone.

The trigger: constants move the mean only; multipliers enter the covariance once per index

so squared on the diagonal and once each off it.

2 of these 3 wrong options correspond to a named reasoning error — the same errors recur across subjects.

Multivariate normal distribution

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