To determine the expected value of R in each case, we need to calculate the integral of R multiplied by the probability density function fX(x) with respect to x, and evaluate it from 0 to infinity.
1. In the first case, where fX(x) = λe^(-λx) when x ≥ 0 and x < 0 otherwise:
E[R] = ∫(0 to ∞) R * λe^(-λx) dx
To solve this integral, we use integration by parts. Let u = R and dv = λe^(-λx) dx. Then du = dR and v = -e^(-λx) / λ.
E[R] = [-R * e^(-λx) / λ] from 0 to ∞ + ∫(0 to ∞) e^(-λx) / λ dx
Since e^(-λx) approaches 0 as x approaches infinity, the first term on the right side of the equation is 0. Therefore, we are left with:
E[R] = ∫(0 to ∞) e^(-λx) / λ dx
To solve this integral, we can make use of the property of the exponential distribution:
∫(0 to ∞) e^(-μx) dx = 1 / μ
By replacing λ with μ, we can rewrite the equation as:
E[R] = 1 / λ
Therefore, the expected value of R in the first case is 1 / λ.
2. In the second case, where fX(x) = λ3x^2e^(-λx^2) when x ≥ 0 and x < 0 otherwise:
E[R] = ∫(0 to ∞) R * λ3x^2e^(-λx^2) dx
Since R is a constant, we can move it outside the integral:
E[R] = λ3 * ∫(0 to ∞) x^2e^(-λx^2) dx
This integral does not have a closed-form solution. However, it can be evaluated numerically using methods such as numerical integration or approximations.
Therefore, the expected value of R in the second case is λ3 * (∫(0 to ∞) x^2e^(-λx^2) dx).