Suppose you are given a random vector ( n x 1 matrix ) where n=9.

How many rows are in V(X)?

c002.p0180.q003

Question

Suppose you are given a random vector ( n x 1 matrix )     where n=9.

How many rows are in V(X)?

c002.p0180.q003

BlackTom AI · Accepted Answer

The covariance (variance–covariance) matrix V(X) of an n × 1 random vector X is an n × n matrix, so it has n rows (and n columns).

Relevant concepts:
- If X = [X1, X2, ..., Xn]' is a column random vector, its mean vector is μ = E[X] = [E[X1], ..., E[Xn]]' (an n × 1 vector).
- The covariance matrix is defined as V(X) = E[(X − μ)(X − μ)'].
- The outer product (X − μ)(X − μ)' produces an n × n matrix because an n × 1 vector multiplied by a 1 × n vector yields n × n.

Step-by-step reasoning:
1. Let n = 9 and write X = [X1, X2, ..., X9]'.
2. μ = E[X] is 9 × 1.
3. Compute (X − μ), which is 9 × 1. Its transpose (X − μ)' is 1 × 9.
4. The product (X − μ)(X − μ)' has dimension (9 × 1) × (1 × 9) = 9 × 9.
5. Taking expectation of that matrix elementwise gives V(X) which is still 9 × 9.

Intermediate/calculation detail for entries:
- The (i, j) entry of V(X) is Cov(Xi, Xj) = E[(Xi − E[Xi])(Xj − E[Xj])]. There are 9 choices for i (rows) and 9 for j (columns), giving 9 rows.

Why the correct answer is correct:
Because the covariance matrix records covariances for each pair of the n components, it must have one row per component. With n = 9 components, V(X) has 9 rows.

Why other possible misunderstandings are wrong:
- Thinking V(X) is 1 row (confusing with the original column vector) ignores the outer product transpose step.
- Thinking V(X) is 81 rows confuses number of elements (9×9 = 81 entries) with number of rows.

Conclusion: V(X) has 9 rows. Final answer: 9.

Suppose you are given a random vector ( n x 1 matrix ) where n=9. How many rows are in V(X)? c002.p0180.q003简答题

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