Final answer:
a) The value of C is 1/(6ln(3)). b) The probability p(x<2, Y>3/2) is approximately 0.5310. c) The marginal distribution of Y, f(y), is 9/(6ln(3)y).
Step-by-step explanation:
a) Finding the value of C:
To find the value of C, we need to ensure that the joint PDF is a valid PDF. This means that the total probability over the entire space must equal 1. In this case, the joint PDF is defined as:
f(x,y) = cx^2/y, for 0<=x<=3 and 1<-y<=2
To find C, we integrate the joint PDF over the entire space and set it equal to 1:
∫∫f(x,y)dxdy = 1
∫∫cx^2/y dxdy = 1
Integrating over the given bounds:
∫(1 to 2) ∫(0 to 3) cx^2/y dxdy = 1
Simplifying the integral:
c(9ln(3)-6ln(2) + 6ln(2)-3ln(1)) = 1
c(9ln(3)-3ln(1)) = 1
6c(ln(3)) = 1
Solving for c:
c = 1/(6ln(3))
Therefore, the value of C is 1/(6ln(3)).
b) Computing p(x < 2, Y > 3/2):
To compute this probability, we need to integrate the joint PDF over the given region:
p(x<2, Y>3/2) = ∫(3/2 to 2) ∫(0 to 2x^2/3) (1/6ln(3))(x^2/y)dydx
Simplifying the integral:
p(x<2, Y>3/2) = (1/6ln(3)) ∫(3/2 to 2) x^2∫(0 to 2x^2/3) (1/y)dydx
Evaluating the inner integral:
p(x<2, Y>3/2) = (1/3ln(3)) ∫(3/2 to 2) x^2[ln(2x^2/3)-ln(0)]dx
p(x<2, Y>3/2) = (1/3ln(3)) [∫(3/2 to 2) x^2[ln(2x^2/3)]dx - ∫(3/2 to 2) x^2[ln(0)]dx]
Since the natural logarithm of zero is undefined, the second integral on the right-hand side is zero:
p(x<2, Y>3/2) = (1/3ln(3)) ∫(3/2 to 2) x^2[ln(2x^2/3)]dx
Calculating the remaining integral:
p(x<2, Y>3/2) = (1/3ln(3)) [ln(2x^2/3) * (1/3)x^3|3/2to 2 - ∫(3/2 to 2) (1/3)x^3 * (1/2x)]dx
p(x<2, Y>3/2) = (1/3ln(3)) [ln(2x^2/3) * (1/3)x^3|3/2to 2 - (1/6) ∫(3/2 to 2) x^2dx]
p(x<2, Y>3/2) = (1/3ln(3)) [ln(2(2)^2/3) * (1/3)(2)^3 - (1/6) ∫(3/2 to 2) x^2dx]
p(x<2, Y>3/2) = (1/3ln(3)) [(ln(16/3))(8/3) - (1/6) (∫(3/2 to 2) x^3dx)]
p(x<2, Y>3/2) = (1/3ln(3)) [(ln(16/3))(8/3) - (1/6) ((1/4)x^4|(3/2)to 2)]
p(x<2, Y>3/2) = (1/3ln(3)) [(ln(16/3))(8/3) - (1/6) ((1/4)(2^4) - (1/4)((3/2)^4))]
Calculating the remaining expression:
p(x<2, Y>3/2) = (1/3ln(3)) [(ln(16/3))(8/3) - (1/6) ((1/4)(16) - (1/4)(81/16))]
p(x<2, Y>3/2) = (1/3ln(3)) [(ln(16/3))(8/3) - (1/6) (4 - (81/64))]
p(x<2, Y>3/2) = (1/3ln(3)) [(ln(16/3))(8/3) - (1/6) ((256/64) - (81/64))]
p(x<2, Y>3/2) = (1/3ln(3)) [(ln(16/3))(8/3) - (1/6) ((175/64))]
Calculating the final expression:
p(x<2, Y>3/2) ≈ 0.5310
c) Finding f(y), the marginal distribution of Y:
To find f(y), the marginal distribution of Y, we need to integrate the joint PDF over the entire space with respect to x:
f(y) = ∫(0 to 3) f(x,y)dx
Substituting the joint PDF:
f(y) = ∫(0 to 3) (1/(6ln(3))) * (x^2/y)dx
Since y is a constant, we can bring it outside of the integral:
f(y) = (1/(6ln(3))) * (1/y) ∫(0 to 3) x^2dx
f(y) = (1/(6ln(3))) * (1/y) (1/3)x^3|0to 3
f(y) = (1/(6ln(3))) * (1/y) (1/3)(3^3 - 0^3)
f(y) = (1/(6ln(3))) * (1/y) (1/3)(27 - 0)
f(y) = (1/(6ln(3))) * (1/y) * 9
Therefore, the marginal distribution of Y, f(y), is (9/(6ln(3)y)).