To solve the given problem, we need to find an expression for \( E(X^2) \) using the provided options. We know that \( X \) is a continuous random variable with \( P(X \geq 0) = 1 \) and \( \text{Var}(X) < \infty \). The task is to identify the correct integral expression for \( E(X^2) \) from the given options.
Let's analyze the given options to identify the correct expression for \( E(X^2) \). For a non-negative continuous random variable \( X \) with cumulative distribution function (CDF) \( F(x) \), the expectation of \( X^2 \) is calculated as follows:
Using the property of integration by parts, the expectation of \( X^2 \) is given by:
\(E(X^2) = \int_0^\infty x^2 f(x) \, dx\)
where \( f(x) \) is the probability density function (PDF) of \( X \).
However, for cumulative distribution functions, the expectation can be rewritten using the survival function \( S(x) = P(X > x) = 1 - F(x) \), leading to:
\(E(X^2) = \int_0^\infty \left(2x P(X > x)\right) \, dx\)
Thus, integrating by parts, the formula for \( E(X^2) \) for a non-negative random variable X becomes:
\(E(X^2) = 2\int_0^\infty x P(X > x) \, dx\)
Now let's consider the given options:
The correct option matches the derived expression 2β«0β x P(X > x) dx, which is option 3.
Therefore, the correct answer is:
\(2\int_0^\infty x P(X > x) \, dx\)
This expression correctly evaluates the expectation of the square of a non-negative continuous random variable using integration by parts.