Question

Given the PDF fY(y) of Y:

fY(y)= 0.2 ln(8) if y in [0, 3]
0.2 ln(8/y) if y in [3, 8]
0 otherwise
Calculate P(|X−Y|≤1)
Calculate P(min(X,Y)≤6)

Answer 1

X is uniformly distributed on the interval [3,8]. Given X=x, Y is uniformly distributed on the interval [0,x]. That is, the conditional PDF of Y given X=x is

fY|X(y|x)=1/x, 0≤y≤x.

Answer 2

To calculate P(|X−Y|≤1), we need to find the joint PDF f(x, y) of X and Y.

Since X and Y are independent, the joint PDF can be calculated by multiplying the PDFs of X and Y:
f(x, y) = fX(x) * fY(y)

Given the PDF fY(y) of Y, we also need the PDF fX(x) of X to calculate the joint PDF.

The PDF fX(x) of X was not provided in the question. Please provide the PDF fX(x) of X, or let me know if X follows a specific distribution.

Answer 3

Now that we have the conditional PDF fY|X(y|x), we can calculate P(|X−Y|≤1).

To calculate this probability, we integrate the joint PDF f(x, y) over the region where |x - y| ≤ 1.

The region where |x - y| ≤ 1 can be split into two regions: y ≤ x+1 and y ≥ x-1.

Therefore, we have:
P(|X−Y|≤1) = ∫∫[f(x, y)] dy dx over the region y ≤ x+1 and y ≥ x-1

Since X is uniformly distributed on the interval [3, 8], we have fX(x) = 1/(8-3) = 1/5 for 3 ≤ x ≤ 8.

Using the conditional PDF fY|X(y|x) = 1/x, 0 ≤ y ≤ x, and the PDF fX(x) = 1/5, we have the joint PDF:
f(x, y) = fY|X(y|x) * fX(x) = (1/x) * (1/5) = 1/(5x), 0 ≤ y ≤ x and 3 ≤ x ≤ 8.

Now, let's calculate P(|X−Y|≤1):

P(|X−Y|≤1) = ∫∫[f(x, y)] dy dx over y ≤ x+1 and y ≥ x-1

P(|X−Y|≤1) = ∫[∫[f(x, y) dy] from x-1 to x+1] dx over 3 ≤ x ≤ 8

P(|X−Y|≤1) = ∫[∫[1/(5x) dy] from x-1 to x+1] dx over 3 ≤ x ≤ 8

P(|X−Y|≤1) = ∫[(1/x) dy] from x-1 to x+1 dx over 3 ≤ x ≤ 8

P(|X−Y|≤1) = ∫[1/x dy] from x-1 to x+1 dx over 3 ≤ x ≤ 8

P(|X−Y|≤1) = ∫[ln(x+1) - ln(x-1) ] dx over 3 ≤ x ≤ 8

Integrating this expression should give you the value of P(|X−Y|≤1).