Question

Consider two random variables X and Y be the proportion of time that a person travels to work by bus and MTR respectively (there are other modes as well). Moreover, X and Y has the following joint distribution f(x,y) 24xy, 0 x f(x,y)-0 otherwise 1,0 y 1, x + y 1 () Find the marginal distributions g(x) and h(y) respectively. (ii) Find the conditional density function fxly) (ii Ifit is known that a person has 0.75 chance of using bus, find the probability that the person is using MTR with less than 0.1 chance.

Answer 1

The region in which f(x,y) is defined is shown as

The marginal PDF is given by

For X

$f(x)=\int_{0}^{1-x}24xy dy \\ ~~~~~~~~~~~=24x\left ( y^2\over 2 \right )_{0}^{1-x}\\ ~~~~~~~~~~~=12x(1-x)^2$

For Y

$f(y)=\int_{0}^{1-y}24xy dy \\ ~~~~~~~~~~~=24y\left ( x^2\over 2 \right )_{0}^{1-y}\\ ~~~~~~~~~~~=12y(1-y)^2$

ii) By definition we have

$f(x|y)={f(x,y)\over f(y)}\\ ~~~~~~~~~~~~~={24xy \over 12y(1-y)^2}\\ ~~~~~~~~~~~~~={2x\over (1-y)^2}$

iii) Similarly to ii) we have

$f(y|x)={2y\over (1-x)^2}$

Hence the required probability is

$\int_{0}^{0.1}f(y|0.75) dy={2\over (1-0.75)^2}\int_{0}^{0.1}ydy \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~={1\over 0.0625}(0.1)^2\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~=0.16$