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)
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.
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.
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).
To calculate P(|X-Y| ≤ 1), we need to find the joint probability density function (PDF) of X and Y first.
Since we have the PDF of Y, let's denote it as fY(y).
For y in the interval [0, 3], we have fY(y) = 0.2 ln(8).
For y in the interval [3, 8], we have fY(y) = 0.2 ln(8/y).
For all other values of y, fY(y) = 0.
To calculate P(|X-Y| ≤ 1), we need to integrate the joint PDF over the region where |X-Y| ≤ 1.
1. Let's consider the first case: y in [0, 3].
In this interval, the possible values for X that satisfy |X-Y| ≤ 1 are X in [y-1, y+1].
Hence, the joint PDF for X and Y is given by:
fX,Y(x, y) = fY(y), for x in [y-1, y+1]
= 0, otherwise
To calculate the probability, we integrate fX,Y(x, y) over the region:
P1 = ∫∫[y-1, y+1] fX,Y(x, y) dx dy
2. Now let's consider the second case: y in [3, 8].
In this interval, the possible values for X that satisfy |X-Y| ≤ 1 are X in [Y-1, Y+1].
However, we need to take into account that X cannot go below 0, so we have two scenarios:
a) If Y is less than or equal to 1, then X is in [0, Y+1].
In this case, the joint PDF for X and Y is given by:
fX,Y(x, y) = fY(y), for x in [0, y+1]
= 0, otherwise
P2a = ∫∫[0, y+1] fX,Y(x, y) dx dy
b) If Y is greater than 1, then X is in [Y-1, Y+1].
In this case, the joint PDF for X and Y is given by:
fX,Y(x, y) = fY(y), for x in [y-1, y+1]
= 0, otherwise
P2b = ∫∫[y-1, y+1] fX,Y(x, y) dx dy
The total probability in this case is:
P2 = P2a + P2b
3. Finally, let's consider the third case: all other values of Y.
In this case, the joint PDF for X and Y is always zero:
fX,Y(x, y) = 0, for all x and y
Hence, the probability is zero:
P3 = 0
The total probability P(|X-Y| ≤ 1) is the sum of the probabilities calculated in each case:
P(|X-Y| ≤ 1) = P1 + P2 + P3
Now let's calculate P(min(X, Y) ≤ 6).
Since we already have the joint PDF fX,Y(x, y), we can calculate this probability by integrating over the region where min(X, Y) ≤ 6.
P(min(X, Y) ≤ 6) = ∫∫[0, 6] fX,Y(x, y) dx dy
Let's calculate these probabilities step-by-step.