Question

Let Y1 and Y2 denote the proportion of time during which employees I and II actually performed their assigned tasks during a workday, The joint density of Y1 and Y2 is given by

f(y1,y2) = { y1+y2, 0<=y1<=1, 0<=y2<=1, 0, elsewhere

Required:
a. Find the marginal density function of Y1 and Y2
b. Find P(Y1 >= 1/2 | Y2 >= 1/2).
c. If employee II spends exactly 50% of the dayworking on assigned duties, find the probability that employee I spends more than 75% of the day working on similarduties.

Answer 1

Explanation:

From the information given:

The joint density of
`y_1` and
`y_2` is given by:

`f_((y_1,y_2)) \left \{ {{y_1+y_2, \ \ 0\ \le \ y_1 \ \le 1 , \ \ 0 \ \ \le y_2 \ \ \le 1} \atop {0, \ \ \ elsewhere \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \right.`

a)To find the marginal density of
`y_1`.

`f_(y_1) (y_1) = \int \limits ^(\infty)_(-\infty) f_(y_1,y_2) (y_1 >y_2) \ dy_2`

`=\int \limits ^(1)_(0)(y_1+y_2)\ dy_2`

`=\int \limits ^(1)_(0) \ \ y_1dy_2+ \int \limits ^(1)_(0) \ y_2 dy_2`

`= y_1 \ \int \limits ^(1)_(0) dy_2+ \int \limits ^(1)_(0) \ y_2 dy_2`

`= y_1[y_2]^1_0 + \bigg [ (y_2^2)/(2)\bigg]^1_0`

`= y_1 [1] + [(1)/(2)]`

`= y_1 + (1)/(2)`

i.e.

`f_((y_1)(y_1)}= \left \{ {{y_1+(1)/(2), \ \ 0\ \ \le \ y_1 \ \le , \ 1} \atop {0, \ \ \ elsewhere \ \\ \ \ \ \ \ \ \ \ } \right.`

The marginal density of
`y_2` is:

`f_(y_1) (y_2) = \int \limits ^(\infty)_(-\infty) fy_1y_1(y_1-y_2) dy_1`

`= \int \limits ^1_0 \ y_1 dy_1 + y_2 \int \limits ^1_0 dy_1`

`=\bigg[ (y_1^2)/(2) \bigg]^1_0 + y_2 [y_1]^1_0`

`= [ (1)/(2)] + y_2 [1]`

`= y_2 + (1)/(2)`

i.e.

`f_((y_1)(y_2)}= \left \{ {{y_2+(1)/(2), \ \ 0\ \ \le \ y_1 \ \le , \ 1} \atop {0, \ \ \ elsewhere \ \\ \ \ \ \ \ \ \ \ } \right.`

b)

`P\bigg[y_1 \ge (1)/(2)\bigg |y_2 \ge (1)/(2) \bigg] = (P\bigg [y_1 \ge (1)/(2) . y_2 \ge(1)/(2) \bigg])/(P\bigg[ y_2 \ge (1)/(2)\bigg])`

`= \frac{\int \limits ^1_{(1)/(2)} \int \limits ^1_{(1)/(2)} f_(y_1,y_1(y_1-y_2) dy_1dy_2)}{\int \limits ^1_{(1)/(2)} fy_1 (y_2) \ dy_2}`

`= \frac{\int \limits ^1_{(1)/(2)} \int \limits ^1_{(1)/(2)} (y_1+y_2) \ dy_1 dy_2}{\int \limits ^1_{(1)/(2)} (y_2 + (1)/(2)) \ dy_2}`

`= ((3)/(8))/((5)/(8))`

`= (3)/(8)}* {(8)/(5)}`

`= (3)/(5)}`

= 0.6

(c) The required probability is:

`P(y_2 \ge 0.75 \ y_1 = 0.50) = (P(y_2 \ge 0.75 . y_1 =0.50))/(P(y_1 = 0.50))`

`= (\int \limits ^1_(0.75) (y_2 +0.50) \ dy_2)/((0.50 + (1)/(2)))`

`= (0.34375)/(1)`

= 0.34375