Final answer:
To find a valid bivariate pdf, we determined that C must be 6. The marginal pdf of X is f(x) = 2x, and the marginal cdf of Y is F(Y=y) = 1 - e⁻³²y. The expected values of X and Y are 2/3 and 1/9 respectively, and the covariance of X and Y is 0 due to their independence.
Step-by-step explanation:
Finding a Valid Joint Probability Density Function and Marginals
To ensure the given bivariate pdf f(x,y) = Cxe⁻³²y for X > 0 and Y > 0 is valid, we must find the constant C such that the integral of f(x,y) over all x and y equals 1. Since X and Y are independent, we can separate the integrals:
∫ f(x,y) dy dx = ∫ (∫ Cxe⁻³²y dy) dx = ∫ Cx dx (∫ e⁻³²y dy)
Solving the inner integral:
∫ e⁻³²y dy = [-1/3 e⁻³²y]⁰∞ = 1/3
Now, integrating with respect to x:
∫ Cx dx = C[⅓x²]⁰∞ = C/2
So, multiplying these results, we get:
(1/3)(C/2) = 1/6 C = 1, hence C = 6.
For part b, the marginal pdf of X, f(x), is found by integrating the joint pdf over y:
f(x) = ∫ 6xe⁻³²y dy = 6x[-1/3 e⁻³²y]⁰∞ = 2x
For part c, the marginal cumulative distribution function, F(Y=y), is the integral of the pdf of Y:
F(Y=y) = ∫ 6xe⁻³²y dy = [e⁻³²y]⁰y = 1 - e⁻³²y
The expected value of X, E[X], is given by:
E[X] = ∫ x * f(x) dx = ∫ x * 2x dx = [2/3x³]⁰∞ = 2/3
The expected value of Y, E[Y], is given by the integral of y times the marginal of Y, but since X and Y are independent and the marginal of Y is a constant multiple of the exponential distribution, E[Y] = 1/(3²).
Finally, for part e, since X and Y are independent, their covariance, Cov(X,Y), is 0.