   Chapter 15.4, Problem 27E

Chapter
Section
Textbook Problem

The joint density function for random variables X and Y is f ( x ,   y )   = { C x   ( 1 + y ) if   0  ≤   x   ≤   1 ,   0   ≤   y   ≤   2 0 otherwise (a) Find the value of the constant C.(b) Find P(X ≤ 1, Y ≤ 1).(c) Find P(X + Y ≤ 1).

(a)

To determine

To find: The value of C.

Explanation

Given:

The joint density function, f(x,y)={Cx(1+y) , if 0x1,0y2       0       ,  otherwise .

The region D is  0x1,0y2 .

Property used:

The double integral value of the given function over D is always 1 that is f(x,y)dA=1 .

Formula used:

If g(x) is the function of x and h(y) is the function of y then,

abcdg(x)h(y)dydx=abg(x)dxcdh(y)dy (1)

Calculation:

By the property mentioned above, the value of f(x,y)dA becomes,

f(x,y)dA=00f(x,y)dydx+0102f(x,y)dydx+12f(x,y)dydx=1

From the given condition, the value of f(x,y)dA is,

(b)

To determine

To find: The value of P(X1,Y1) .

(c)

To determine

To find: The value of P(X+Y1) .

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