Question

If \(X\) is a nonnegative random variable, and \(g\) is a differential function with \(g(0)=0\), then $$ E[g(X)]=\int_{0}^{\infty} P(X>t) g^{\prime}(t) d t $$ Prove the preceding when \(X\) is a continuous random variable.

Short answer

To prove the given formula for a continuous non-negative random variable \(X\) and a differentiable function \(g\) with \(g(0) = 0\), we followed these steps: 1. Write down the formula for the expectation of a function of a random variable. 2. Differentiate the given function g(t) to find its derivative g'(t). 3. Apply integration by parts to the integral on the right-hand side of the formula. 4. Make sure the boundary terms vanish. 5. Find the probability density function (PDF) of X. 6. Show that the left-hand side and right-hand side of the formula are equal, completing the proof. After completing these steps, we have shown that the given formula holds: \(E[g(X)] = \int_{0}^{\infty} P(X > t) g'(t) dt\)

Step-by-step solution

Write down the formula for the expectation of a function of a random variable

Given a continuous random variable X with probability density function (PDF) f(x), the expectation of a function g(X) is given by: \(E[g(X)] = \int_{-\infty}^{\infty} g(x) f(x) dx\) As X is a nonnegative random variable, the integration limits can be written as [0,∞).

Differentiate the given function g(t) to find its derivative g'(t)

We are given that g is a differentiable function, and we are interested in its derivative g'(t). We don't have the explicit form of g(t), so we'll just write g'(t) as the derivative.

Apply integration by parts to the integral on the right-hand side of the formula

Using integration by parts, we have: \(uv - \int v du = \int u dv\), where in our case; u = g(t) and dv = P(X > t) dt Now, differentiate u with respect to t to find du and integrate dv with respect to t to find v: du = g'(t) dt and v = -\int_{0}^{t} f(u) du Here, we have used the fact that P(X > t) = 1 - P(X ≤ t) = 1 - \(\int_{0}^{t} f(u) du\). Now, we can plug in our u, v, du, and dv into the integration by parts formula: \(E[g(X)] = \int_{0}^{\infty} -g(t) f(t) dt = -\left[ \lim_{t \rightarrow \infty} g(t) \int_{0}^{t} f(u) du - g(0) \right] - \int_{0}^{\infty} \left[- \int_{0}^{t} f(u) du\right] g'(t) dt\)

Make sure the boundary terms vanish

The boundary term \(-\left[ \lim_{t \rightarrow \infty} g(t) \int_{0}^{t} f(u) du - g(0) \right]\) vanishes because: - \(g(0) = 0\) (given) - \(\lim_{t \rightarrow \infty} g(t) \int_{0}^{t} f(u) du\) must vanish as the integral of the PDF of X is 1 when t goes to ∞ and the product of g(t) and the integral will be bounded. So, we have: \(E[g(X)] = \int_{0}^{\infty} \left[ \int_{0}^{t} f(u) du\right] g'(t) dt\)

Find the PDF of X

The PDF of X, f(x), can be derived from the given information about X being a nonnegative continuous random variable. However, without any specific information on the distribution of X, we will use f(x) as the PDF.

Show that the left-hand side and right-hand side of the formula are equal

We're now ready to complete the proof by showing that the left-hand side and right-hand side of the formula are equal: \(E[g(X)] = \int_{0}^{\infty} g(x) f(x) dx = \int_{0}^{\infty} \left[ \int_{0}^{t} f(u) du\right] g'(t) dt\) Since the left-hand side and right-hand side of the formula are equal, we have proven the given formula for a continuous random variable X: \(E[g(X)] = \int_{0}^{\infty} P(X > t) g'(t) dt\)