Question

Suppose that we are waiting for two events A and B to occur. X = the time until event A occurs and Y = the time until event B occurs. We model the probability distribution of the two times with the joint density

f(x, y) = e^(−2x− y/2) [3/4 e^−x + e^−y/2 ]

for x > 0 and y > 0.

(a) What is the probability that X < 1 but Y > 1?
(b) What is the probability that X > Y ?

Answer 1

a) The probability that X < 1 but Y > 1 is
`\approx` 0.45

b) The probability that X > Y is
`\approx` 0.24

Explanation:

Given a set of random variables X, Y, ... , the joint probability distribution for of them is a probability distribution that gives the probability that each of X,Y, ... variables are in any given interval or discrete set of values specified for that variable. For the case of two random variables X and Y and the given joint probability density function
`f(x,y)` this is:

`P(x\in [x_0, x_1], y\in [y_0, y_1])=\int_(x_0)^(x_1)\int_(y_0)^(y_1)f(x,y)dydx` (1)

In the exercise the joint probability function is given and the interval of the variables goes from
`x, y \in (0, \infty)`.

a) Applying the definition given in (1):

See the first attached figure for a visualization of the area we need to integrated. It is depicted with a grey rectangle, while the colorful area is a density plot of the joint probability density function.

`P(x<1, x>1) = \int_(0)^(1)\int_(1)^(\infty) e^(-2x-y/2)\cdot \left((3)/(4)e^(-x)+e^(-y/2) \right)dydx\\\boxed{P(x<1, x>1) = 0.447\approx0.45}`

b) In this, case the integration limits for the random variable
`x` are given by the function
`y=x` and notice (second figure attached) that the region where
`x>y` is below the function. Therefore
`x\in[y, \infty)`:

`P(x>y) = \int_(0)^(\infty)\int_(y)^(\infty) e^(-2x-y/2)\cdot \left((3)/(4)e^(-x)+e^(-y/2) \right)dydx\\P(x>y) = \int_(0)^(\infty)  (1)/(4)  e^(-7 y/2) \cdot(1 + 2 e^(y/2))dy\\\boxed{P(x>y) = 0.238 \approx 0.24}`