Answer:
We can find the expected value of C(w) as follows:
E(C) = ∫[0,1] C(w) dw + ∫(1,∞) C(w) f(w) dw
where f(w) is the probability density function of w outside the interval [0,1].
Since C(w) is a constant function in the interval [0,1], we have:
∫[0,1] C(w) dw = -a ∫[0,1] dw = -a
Using the fact that the integral of a probability density function over its entire domain is equal to 1, we can find f(w) as:
∫(1,∞) f(w) dw = 1 - ∫[0,1] dw = 1 - 1 = 0
Therefore, we can write:
E(C) = -a (0 - 1) + ∫(1,∞) (w - 1 - a) f(w) dw
Simplifying, we get:
E(C) = a - ∫(1,∞) (w - 1 - a) f(w) dw
To find the value of a that makes E(C) = 0, we need to solve the equation:
a - ∫(1,∞) (w - 1 - a) f(w) dw = 0
Multiplying both sides by -1 and rearranging, we get:
∫(1,∞) (w - 1 - a) f(w) dw = -a
Expanding the integrand, we get:
∫(1,∞) wf(w) dw - ∫(1,∞) f(w) dw - a ∫(1,∞) f(w) dw = -a
Since the integral of f(w) over its entire domain is equal to 1, we can simplify further:
∫(1,∞) wf(w) dw - 1 - a = -a
Rearranging, we get:
∫(1,∞) wf(w) dw = 1
This means that f(w) is a probability density function over the entire real line, not just outside the interval [0,1].
To find the value of a that satisfies this condition, we need to find the probability density function f(w) that integrates to 1 over the entire real line.
Since f(w) is a probability density function, it must be nonnegative and integrate to 1 over its entire domain.
One possible choice for f(w) that satisfies these conditions is:
f(w) = (1 - a) e^(-w) for w ≥ 1
Using this choice for f(w), we can verify that:
∫(1,∞) f(w) dw = ∫(1,∞) (1 - a) e^(-w) dw = (1 - a) e^(-1) = 1
Therefore, a = 1 - e^(-1) ≈ 0.6321 is the value that makes E(C) = 0.