Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 2 / 4F (Probability)

Table of contents
- a 4F
  - Solution a
- b 4F
  - Solution b
- c 4F
  - Solution c

a

Solution

The bivariate normal distribution has density

f (x_{1}, x_{2}) = \frac{1}{2 π σ _{1} σ _{2} 1 - ρ ^{2}} exp {- \frac{1}{2 ( 1 - ρ ^{2} )} [\frac{( x _{1} - μ _{1} ) ^{2}}{σ _{1}^{2}} - \frac{2 ρ ( x _{1} - μ _{1} ) ( x _{2} - μ _{2} )}{σ _{1} σ _{2}} + \frac{( x _{2} - μ _{2} ) ^{2}}{σ _{2}^{2}}]} .

b

Solution

Completing the square in

x_{1}

in the joint density shows that the conditional distribution of a bivariate normal variable is

X_{1} ∣ X_{2} = x_{2} \sim N (μ_{1} + ρ \frac{σ _{1}}{σ _{2}} (x_{2} - μ_{2}), σ_{1}^{2} (1 - ρ^{2})) .

Thus its conditional density is

\frac{1}{2 π σ _{1}^{2} ( 1 - ρ ^{2} )} exp - \frac{( x _{1} - μ _{1} - ρ \frac{σ _{1}}{σ _{2}} ( x _{2} - μ _{2} ) ) ^{2}}{2 σ _{1}^{2} ( 1 - ρ ^{2} )} .

c

Solution

Let

U = X_{1} + X_{2}, V = X_{1} - X_{2} .

They are jointly normal because they are linear combinations of a Gaussian vector. When

σ_{1} = σ_{2} = σ

,

cov (U, V) = var (X_{1}) - var (X_{2}) = 0.

Hence uncorrelated jointly normal variables are independent. Their means and variances are

E U = μ_{1} + μ_{2}, var U = 2 σ^{2} (1 + ρ),

and

E V = μ_{1} - μ_{2}, var V = 2 σ^{2} (1 - ρ) .

Therefore

U \sim N (μ_{1} + μ_{2}, 2 σ^{2} (1 + ρ)),

V \sim N (μ_{1} - μ_{2}, 2 σ^{2} (1 - ρ)),

independently.

 Ancestors (9)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version