a) The probability of the event that 0 <= Y <= X <= 1 is of the form int a ^ b ( int c ^ d f X,Y (x,y)dx)dy.

Question

Find the values of a, b, c, d. Each one of your answers should be one of the following: 0, x, y, or 1.
b) The probability of the event that 0 ≤ Y≤ X ≤ 1 is also of the form int a ^ b ( int c ^ d f X,Y (x,y)dy)dx. Note the different order of integration as compared to part (a)

Find the values of a, b, c, d. Each one of your answers should be one of the following: 0, x, y, or 1.

Answer 1

To find the values of \(a, b, c, d\) for both parts of the question, we need to analyze the region defined by the inequalities \(0 \leq Y \leq X \leq 1\).

Part (a)

We want to express the probability \(P(0 \leq Y \leq X \leq 1)\) in the form of:

\[ \int_a^b \left( \int_c^d f_{X,Y}(x,y) , dx \right) dy \]

Determining \(a\), \(b\), \(c\), and \(d\):

1. Outer Integral (with respect to \(y\)):

   • Since \(Y\) can take values from 0 to 1, we have:
     • \(a = 0\)
     • \(b = 1\)

2. Inner Integral (with respect to \(x\)):

   • For fixed \(y\), \(X\) can take values from \(y\) (since \(X \geq Y\)) to 1 (since \(X \leq 1\)):
     • \(c = y\)
     • \(d = 1\)

So for part (a):

• \(a = 0\)
• \(b = 1\)
• \(c = y\)
• \(d = 1\)

Part (b)

Now we need to change the order of integration to express it in the form:

\[ \int_a^b \left( \int_c^d f_{X,Y}(x,y) , dy \right) dx \]

Determining \(a\), \(b\), \(c\), and \(d\):

1. Outer Integral (with respect to \(x\)):

   • \(X\) can take values from 0 to 1, so:
     • \(a = 0\)
     • \(b = 1\)

2. Inner Integral (with respect to \(y\)):

   • For fixed \(x\), \(Y\) can take values from 0 to \(x\) (since \(Y \leq X\)):
     • \(c = 0\)
     • \(d = x\)

So for part (b):

• \(a = 0\)
• \(b = 1\)
• \(c = 0\)
• \(d = x\)

Summary of the Answers:

• For part (a): \(a = 0\), \(b = 1\), \(c = y\), \(d = 1\)
• For part (b): \(a = 0\), \(b = 1\), \(c = 0\), \(d = x\)