Past exam of the mathematics course of the University of Cambridge / 2004 / ia / Paper 2 / 11F (Probability)

Table of contents
- Solution 11F

Solution

Take

λ > 0

as the rate of the exponential distribution. By independence, the joint probability density of the original pair is

λ^{2} e^{- λ (x_{1} + x_{2})}

on

x_{1}, x_{2} > 0

. Put

s = Y_{1}

,

r = Y_{2}

. The inverse change of variables is

x_{1} = \frac{sr}{1 + r}, x_{2} = \frac{s}{1 + r}, s > 0, r > 0.

Its Jacobian determinant is

det \frac{\partial ( x _{1} , x _{2} )}{\partial ( s , r )} = det (r / (1 + r) 1/ (1 + r) s / (1 + r)^{2} - s / (1 + r)^{2}) = - \frac{s}{( 1 + r ) ^{2}} .

Using the absolute value in the change of variables formula, the transformed joint probability density is

f_{Y_{1}, Y_{2}} (s, r) = \frac{λ ^{2} s e ^{- λ s}}{( 1 + r ) ^{2}} (s > 0, r > 0),

and zero otherwise. The support is the entire positive quadrant, and the factors are separately normalized:

\int_{0}^{\infty} λ^{2} s e^{- λ s} d s = 1, \int_{0}^{\infty} \frac{d r}{( 1 + r ) ^{2}} = 1.

Integrating out either variable therefore yields these factors as the marginal densities. Factorization of the joint probability density proves independence, with

f_{Y_{1}} (s) = λ^{2} s e^{- λ s} (s > 0), f_{Y_{2}} (r) = \frac{1}{( 1 + r ) ^{2}} (r > 0) .

The sum has a shape-two gamma distribution, and the ratio has a beta-prime distribution with parameters one and one, independent of

λ

. Its cumulative distribution function is

r / (1 + r)

for

r > 0

.

 Ancestors (10)

 View article source

 Discussion (0) New discussion

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