Question 1172514: Let X and Y be two continuous random variables whose joint distribution function is given by
F(x,y) ={(1−e^−x)(1−e^−y) for x >0 , y >0
0 elsewhere
ComputeP({X >1}∪{Y >1}) and P(1≤X≤2,1≤Y≤2).
Thank You.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's solve this problem step-by-step.
**Given:**
* Joint distribution function:
* F(x, y) = (1 - e^(-x))(1 - e^(-y)) for x > 0, y > 0
* F(x, y) = 0 elsewhere
**1. Compute P({X > 1} ∪ {Y > 1})**
We'll use the formula:
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
In our case:
* A = {X > 1}
* B = {Y > 1}
* **P(X > 1):**
* P(X > 1) = 1 - P(X ≤ 1) = 1 - lim(y→∞) F(1, y)
* P(X > 1) = 1 - (1 - e^(-1)) = e^(-1)
* **P(Y > 1):**
* P(Y > 1) = 1 - P(Y ≤ 1) = 1 - lim(x→∞) F(x, 1)
* P(Y > 1) = 1 - (1 - e^(-1)) = e^(-1)
* **P(X > 1, Y > 1) = P(X > 1 ∩ Y > 1):**
* P(X > 1, Y > 1) = 1 - P(X ≤ 1) - P(Y ≤ 1) + P(X ≤ 1, Y ≤ 1)
* P(X > 1, Y > 1) = lim(x→∞, y→∞) F(x,y) - lim(x→∞,y=1)F(x,y) - lim(x=1,y→∞) F(x,y) + F(1,1)
* P(X > 1, Y > 1) = 1 - (1-e^-1) - (1-e^-1) + (1-e^-1)(1-e^-1)
* P(X > 1, Y > 1) = 1 - 1 + e^(-1) - 1 + e^(-1) + 1 - 2e^(-1) + e^(-2)
* P(X > 1, Y > 1) = e^(-2)
* **P(X > 1 ∪ Y > 1):**
* P(X > 1 ∪ Y > 1) = P(X > 1) + P(Y > 1) - P(X > 1, Y > 1)
* P(X > 1 ∪ Y > 1) = e^(-1) + e^(-1) - e^(-2)
* P(X > 1 ∪ Y > 1) = 2e^(-1) - e^(-2)
**2. Compute P(1 ≤ X ≤ 2, 1 ≤ Y ≤ 2)**
* P(1 ≤ X ≤ 2, 1 ≤ Y ≤ 2) = F(2, 2) - F(1, 2) - F(2, 1) + F(1, 1)
* F(2, 2) = (1 - e^(-2))(1 - e^(-2))
* F(1, 2) = (1 - e^(-1))(1 - e^(-2))
* F(2, 1) = (1 - e^(-2))(1 - e^(-1))
* F(1, 1) = (1 - e^(-1))(1 - e^(-1))
* P(1 ≤ X ≤ 2, 1 ≤ Y ≤ 2) = (1 - e^(-2))(1 - e^(-2)) - (1 - e^(-1))(1 - e^(-2)) - (1 - e^(-2))(1 - e^(-1)) + (1 - e^(-1))(1 - e^(-1))
* P(1 ≤ X ≤ 2, 1 ≤ Y ≤ 2) = (1 - e^(-2) - (1-e^-1)) * (1-e^-2 - (1-e^-1))
* P(1 ≤ X ≤ 2, 1 ≤ Y ≤ 2) = (e^(-1) - e^(-2)) * (e^(-1) - e^(-2))
* P(1 ≤ X ≤ 2, 1 ≤ Y ≤ 2) = (e^(-1) - e^(-2))^2
**Summary**
* P({X > 1} ∪ {Y > 1}) = 2e^(-1) - e^(-2)
* P(1 ≤ X ≤ 2, 1 ≤ Y ≤ 2) = (e^(-1) - e^(-2))^2
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