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].
P(|X−Y|≤1) =
P(min(X,Y)≤6) =
To calculate these probabilities, we need to find the joint density function of X and Y.
Given that X is uniformly distributed on the interval [3, 8]:
The probability density function (pdf) of X is:
fX(x) = 1 / (8 - 3) = 1 / 5 for 3 ≤ x ≤ 8
Given that, given X = x, Y is uniformly distributed on the interval [0, x]:
The conditional pdf of Y given X = x is:
fY|X(y|x) = 1 / x for 0 ≤ y ≤ x
Now, let's calculate the probabilities:
1. P(|X-Y| ≤ 1):
We want to find the probability that the absolute difference between X and Y is less than or equal to 1.
We can rewrite this as:
P(|X-Y| ≤ 1) = P(-1 ≤ X-Y ≤ 1) = P(X-Y ≤ 1) - P(X-Y < -1)
To find P(X-Y ≤ 1), we integrate the joint density function over the region where X-Y ≤ 1:
P(X-Y ≤ 1) = ∫∫[X-Y≤1] fX(x) * fY|X(y|x) dy dx
= ∫[3,8] ∫[0,x+1] (1/5)*(1/x) dy dx (Note that X ranges from 3 to 8, and Y ranges from 0 to x+1 due to the condition X-Y ≤ 1)
Simplifying the integral:
= (1/5) * ∫[3,8] [∫[0,x+1] 1/x dy] dx
= (1/5) * ∫[3,8] [ln(x+1) - ln(x)] dx
= (1/5) * [ln(8+1) - ln(3+1)]
= (1/5) * [ln(9) - ln(4)]
Now, to calculate P(X-Y < -1), we integrate the joint density function over the region where X-Y < -1:
P(X-Y < -1) = ∫∫[X-Y<-1] fX(x) * fY|X(y|x) dy dx
= ∫[3,8] ∫[0,x-1] (1/5)*(1/x) dy dx (Note that X ranges from 3 to 8, and Y ranges from 0 to x-1 due to the condition X-Y < -1)
Simplifying the integral:
= (1/5) * ∫[3,8] [∫[0,x-1] 1/x dy] dx
= (1/5) * ∫[3,8] [ln(x-1) - ln(x)] dx
= (1/5) * [ln(8-1) - ln(3-1)]
= (1/5) * [ln(7) - ln(2)]
Finally, substituting these values into the first equation:
P(|X-Y| ≤ 1) = P(X-Y ≤ 1) - P(X-Y < -1)
= (1/5) * [ln(9) - ln(4)] - (1/5) * [ln(7) - ln(2)]
= (1/5) * [ln(9/4) - ln(7/2)]
2. P(min(X, Y) ≤ 6):
We want to find the probability that the minimum of X and Y is less than or equal to 6.
By symmetry, P(min(X, Y) ≤ 6) = 1 - P(min(X, Y) > 6)
To find P(min(X, Y) > 6), we integrate the joint density function over the region where min(X, Y) > 6:
P(min(X, Y) > 6) = ∫∫[min(X, Y)>6] fX(x) * fY|X(y|x) dy dx
= ∫[6,8] ∫[0,x] (1/5)*(1/x) dy dx (Note that X ranges from 6 to 8, and Y ranges from 0 to x due to the condition min(X, Y) > 6)
Simplifying the integral:
= (1/5) * ∫[6,8] [∫[0,x] 1/x dy] dx
= (1/5) * ∫[6,8] 1 dx
= (1/5) * [8 - 6]
= (1/5) * 2
= 2/5
Finally, substituting this value into the equation:
P(min(X, Y) ≤ 6) = 1 - P(min(X, Y) > 6)
= 1 - 2/5
= 3/5
To find P(|X−Y|≤1), we need to consider the areas in the XY-plane where |X−Y|≤1.
Let's break it down into two cases:
1) When X ≤ Y:
In this case, we have 2 scenarios:
a) X ≤ Y ≤ X+1
b) X ≤ X-Y ≤ 1
2) When Y ≤ X:
In this case, we have 2 scenarios:
a) Y ≤ X ≤ Y+1
b) 0 ≤ X-Y ≤ 1
First, let's find the probability for case 1a:
P(X ≤ Y ≤ X+1) = ∫(from 3 to 8) ∫(from x to x+1)dydx
= ∫(from 3 to 8) (x+1-x) dx
= ∫(from 3 to 8) 1 dx
= (8-3)
= 5
Next, let's find the probability for case 1b:
P(X ≤ X-Y ≤ 1) = ∫(from 3 to 8) ∫(from 0 to x-y)dydx
= ∫(from 3 to 8) (x-y-0) dydx
= ∫(from 3 to 8) (x) dydx
= ∫(from 3 to 8) (x) dx
= [(x^2)/2] (from 3 to 8)
= [(8^2)/2] - [(3^2)/2]
= 32/2 - 9/2
= 16 - 4.5
= 11.5
Total probability for case 1: P(X ≤ Y ≤ X+1 or X ≤ X-Y ≤ 1) = 5 + 11.5 = 16.5
Now, let's find the probability for case 2a:
P(Y ≤ X ≤ Y+1) = ∫(from 3 to 8) ∫(from y to y+1)dxdy
= ∫(from 3 to 8) (y+1-y)dy
= ∫(from 3 to 8) (1)dy
= (8-3)
= 5
Lastly, let's find the probability for case 2b:
P(0 ≤ X-Y ≤ 1) = ∫(from 3 to 8) ∫(from y-1 to y) dxdy
= ∫(from 3 to 8) (y-y+1)dy
= ∫(from 3 to 8) (1)dy
= (8-3)
= 5
Total probability for case 2: P(Y ≤ X ≤ Y+1 or 0 ≤ X-Y ≤ 1) = 5 + 5 = 10
Therefore, the overall probability P(|X−Y|≤1) = P(X ≤ Y ≤ X+1 or X ≤ X-Y ≤ 1 or Y ≤ X ≤ Y+1 or 0 ≤ X-Y ≤ 1) = 16.5 + 10 = 26.5
Moving on to finding P(min(X,Y)≤6), we need to consider the areas in the XY-plane where the minimum of X and Y is less than or equal to 6.
Let's break it down into two cases:
1) When min(X, Y) = X:
In this case, we have 2 scenarios:
a) X ≤ Y ≤ 6
b) 6 ≤ Y ≤ X
2) When min(X, Y) = Y:
In this case, we have 2 scenarios:
a) Y ≤ X ≤ 6
b) 6 ≤ X ≤ Y
For case 1a, P(X ≤ Y ≤ 6) = 5 (same as above)
For case 1b, P(6 ≤ Y ≤ X) = 0 (since X is uniformly distributed on [3,8], it cannot be less than 6)
For case 2a, P(Y ≤ X ≤ 6) = 5 (same as above)
For case 2b, P(6 ≤ X ≤ Y) = 0 (since X is uniformly distributed on [3,8], it cannot be less than 6)
Therefore, the overall probability P(min(X,Y)≤6) = P(X ≤ Y ≤ 6 or Y ≤ X ≤ 6) = 5 + 5 = 10