Question

For random variable W with probability density function (PDF) g(w) = (4/3) * (1 - w^2) for w in the interval [-1, 1], what is the expected value E(W)? Please provide the result rounded to 4 decimal places if necessary.

Answer 1

The expected value E(W) of the random variable W with the given PDF can be calculated using the integral formula. After performing the integration and substituting the limits of integration, the expected value is found to be 1/3.

Step-by-step explanation:

The expected value, denoted as E(W), of a random variable W with a probability density function (PDF) g(w) = (4/3) * (1 - w^2) for w in the interval [-1, 1] can be calculated using the formula:

E(W) = ∫w * g(w) dw

First, we need to calculate the integral of w * g(w) over the interval [-1, 1].

Let's integrate:

∫ (w * g(w)) dw = ∫ w * (4/3) * (1 - w^2) dw

Expanding the integral and integrating term by term, we get:

∫ (w * g(w)) dw = (4/3) * ∫ (w - w^3) dw

Integrating, we get:

∫ (w * g(w)) dw = (4/3) * [(w^2)/2 - (w^4)/4]

Now, substitute the limits of integration:

E(W) = [(4/3) * (1/2 - 1/4)] - [(4/3) * (-1/2 - 1/4)] = 4/3 * (1/4) = 1/3

Therefore, the expected value E(W) of the random variable W is 1/3.