Suppose that X is uniformly distributed on the interval [3,8], and that 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.
The PDF fY(y) of Y is:
fY(y)=⎨(1/5)ln(8/3)), y∈[0,3]
(1/5)ln(8/y), y∈[3,8]
0, otherwise
Calculate P(|X−Y|≤1)
Calculate P(min(X,Y)≤6)
To calculate P(|X - Y| ≤ 1), we need to find the probability that the absolute difference between X and Y is less than or equal to 1.
1. We can start by finding the joint PDF of X and Y, denoted as f(X,Y). This can be calculated by multiplying the conditional PDF of Y given X=x and the PDF of X.
f(X,Y) = fY|X(Y|X) * fX(X)
For 3 ≤ X ≤ 8 and 0 ≤ Y ≤ X:
f(X,Y) = (1/X) * (1/5), where 0 ≤ Y ≤ X and 3 ≤ X ≤ 8
Otherwise, f(X,Y) = 0
2. Next, we need to find the range of Y values that satisfy |X - Y| ≤ 1. We can split it into two cases:
Case 1: X - Y ≤ 1
In this case, Y can range from X-1 to X. So, Y satisfies this condition when X-1 ≤ Y ≤ X.
Case 2: Y - X ≤ 1
In this case, Y can range from X to X+1. So, Y satisfies this condition when X ≤ Y ≤ X+1.
3. Finally, we can integrate the joint PDF over the range of Y values that satisfy |X - Y| ≤ 1 to find the probability.
P(|X - Y| ≤ 1) = ∫∫[f(X,Y)]dydx, where Y satisfies X-1 ≤ Y ≤ X and X-1 ≤ X ≤ 8 + ∫∫[f(X,Y)]dydx, where Y satisfies X ≤ Y ≤ X+1 and 3 ≤ X ≤ 8
P(|X - Y| ≤ 1) = ∫[X-1,X] ∫[0,X] [(1/X) * (1/5)]dydx + ∫[X,X+1] ∫[0,X] [(1/X) * (1/5)]dydx
Now, let's calculate P(min(X,Y) ≤ 6).
1. The minimum of X and Y will be less than or equal to 6 if either X or Y is less than or equal to 6.
2. Therefore, we need to calculate the probability that X ≤ 6 or Y ≤ 6.
P(min(X,Y) ≤ 6) = P(X ≤ 6 or Y ≤ 6)
3. Using the concept of marginal probability, we can calculate this probability by integrating the joint PDF over the range of X and Y values satisfying X ≤ 6 or Y ≤ 6.
P(min(X,Y) ≤ 6) = ∫[3,8] ∫[0,6] [(1/X) * (1/5)]dydx
Please let me know if you need further clarification or have any additional questions.
To calculate P(|X−Y|≤1), we first need to find the joint PDF of X and Y. Since X and Y are dependent, we need to use the conditional PDF to calculate the joint PDF:
fX,Y(x,y) = fY|X(y|x) * fX(x)
The marginal PDF of Y can then be found by integrating the joint PDF over all possible values of X:
fY(y) = ∫[3,8] fX,Y(x,y) dx
Given that fY|X(y|x) = 1/x for 0≤y≤x, and X is uniformly distributed on [3,8], we have:
fX(x) = 1/(8-3) = 1/5
Substituting these values into the joint PDF equation, we have:
fX,Y(x,y) = (1/x) * (1/5) = 1/(5x)
To find the marginal PDF of Y, we integrate the joint PDF over all possible values of X:
fY(y) = ∫[3,y] (1/(5x)) dx + ∫[y,8] (1/(5x)) dx
= (1/5) ln(y/3) + (1/5) (ln(8)-ln(y))
= (1/5) (ln(8/3) + ln(y/3))
= (1/5) ln(8y/9)
Now we can calculate P(|X−Y|≤1):
P(|X−Y|≤1) = P(X-Y ≤ 1) + P(Y-X ≤ 1)
For X-Y ≤ 1, we have X ≤ Y+1, so the limits of integration for X will change depending on the value of Y:
P(X-Y ≤ 1) = ∫[0,2] ∫[Y+3,Y+4] (1/(5x)) dx dy + ∫[2,8] ∫[Y+3,Y+4] (1/(5x)) dx dy
Calculating these integrals, we get:
P(X-Y ≤ 1) = (1/5) (∫[0,2] [(ln(Y+4) - ln(Y+3)] dy + ∫[2,6] [(ln(Y+4) - ln(Y+3)] dy + ∫[6,7] [(ln(8) - ln(Y+3)] dy
Simplifying, we get:
P(X-Y ≤ 1) = (1/5) [2ln(3/4) + 4ln(8/9) + (ln(8)-ln(9))]
To calculate P(min(X,Y) ≤ 6), we need to determine the region of the joint PDF where min(X,Y) ≤ 6. This occurs when Y ≤ 6, so we have:
P(min(X,Y) ≤ 6) = P(Y ≤ 6)
Using the marginal PDF of Y that we previously calculated, we integrate from 0 to 6:
P(Y ≤ 6) = ∫[0,6] (1/5) ln(8y/9) dy
Calculating this integral, we get:
P(Y ≤ 6) = (1/5) [6ln(4/3) - 9ln(8/9)]
I hope this helps! Let me know if you have any further questions.