If x and y are independent Gaussian random variables with mean 0 and the same…
2009
If x and y are independent Gaussian random variables with mean 0 and the same variance, their joint probability density function is:
- A.
\(p(x,y) = p(x) \cdot p(y)\)
- B.
\(p(x,y) = p(x) + p(y)\)
- C.
\(p(x,y) = p(x+y)\)
- D.
\(p(x,y) = p(x) \cdot p(y) + p(x)\)
Show answer & explanation
Correct answer: A
Concept
For independent continuous random variables, the joint probability density factorizes into the product of the marginal densities.
Thus, independence determines the factorization rule; the common mean and variance specify the marginals but do not alter that rule.
Application
Here x and y are explicitly independent, so substitute their marginal densities into the factorization identity: p(x,y) = p(x) p(y).
Contrast
p(x) + p(y) adds marginal densities; addition is not the independence factorization.
p(x+y) is a density evaluated at the sum and is not a joint density in the two variables.
p(x)p(y) + p(x) adds an extra marginal-density term to the product.
Cross-check
The product is normalized because integrating p(x)p(y) over both variables separates into two marginal integrals, each equal to 1. Therefore the joint density is p(x,y) = p(x)p(y).