Question

Let \(L\) and \(R\) be randomly chosen interval endpoints having an arbitrary juint distribution, but, of course, \(L \leq R\). Let \(p(x)=\operatorname{Pr}\\{L \leq x \leq R\\}\) be the probability the interval covers the point \(x\), and let \(X=R-L\) be the length of the interval. Establish the formula \(E[X]=\int_{-\infty}^{\infty} p(x) d x\).

Short answer

In this problem, we have interval endpoints \(L\) and \(R\) with \(L \leq R\), and a probability function \(p(x)\) representing the probability that the interval covers point \(x\). We want to establish the formula \(E[X]=\int_{-\infty}^{\infty} p(x) d x\), where \(X = R-L\). By expressing \(E[X]\) as an integral and calculating the probability density function \(f(x)\) using \(p(x)\), we obtain \(E[X] = \int_{-\infty}^{\infty} x \left( \int \frac{d}{dx}p(x) dx\right) dx\). Changing the order of integration and evaluating the resulting expression, we find that \(E[X]=\int_{-\infty}^{\infty} p(x) d x\), which establishes the desired formula.

Step-by-step solution

Write down the formula for Expected Value (mean) of continuous random variable

Recall that for a continuous random variable \(X\) with probability density function \(f(x)\), the expected value (mean) \(E[X]\) can be calculated as: \[E[X] = \int_{-\infty}^{\infty} x f(x) dx\] In our case, \(X = R - L\), so we first have to find the probability density function \(f(x)\) of the length of the interval \(X\).

Calculate \(f(x)\) using \(p(x)\)

Given the probability function \(p(x)=\operatorname{Pr}\{L \leq x \leq R\}\), we can calculate the probability density function \(f(x)\) of \(X=R-L\) by using the fact that \[p(x) = \int_{-\infty}^\infty f(x-y)dy\] So, let's differentiate both sides of this equation with respect to \(x\) to find \(f(x)\) in terms of \(p(x)\): \[\frac{d}{dx}p(x) = \int_{-\infty}^\infty \frac{d}{dx}f(x-y)dy\] Then we can interchange the order of integration and differentiation: \[\frac{d}{dx}p(x) = \int_{-\infty}^\infty f'(x-y)dy\] Now, we have the first derivative of \(f(x)\) in terms of \(p(x)\). We can integrate back with respect to \(x\) to obtain \(f(x)\): \[f(x) = \int \frac{d}{dx}p(x) dx\]

Calculate \(E[X]\) using the derived probability density function \(f(x)\)

Now that we have expressed \(f(x)\) in terms of \(p(x)\), we can plug it into the formula for expected value (mean) from Step 1: \[E[X] = \int_{-\infty}^{\infty} x \left( \int \frac{d}{dx}p(x) dx\right) dx\] Now, let's change the order of integration: \[E[X] = \int \int_{-\infty}^{\infty} x \left( \frac{d}{dx}p(x) dx\right) dy\] Notice that we can interchange \(x\) and \(y\) in the argument of \(p\) since they are dummy variables: \[E[X] = \int \int_{-\infty}^{\infty} x \left( \frac{d}{dy}p(y) dy\right) dx\] Now, integrate with respect to \(y\): \[E[X] = \int_{-\infty}^{\infty} x \left[ p(y)\right]_{-\infty}^{\infty} dx\] Then, evaluate the expression inside the square brackets: \[E[X] = \int_{-\infty}^{\infty} x p(x) dx\] Finally, we compare this result to the given integral: \[E[X]=\int_{-\infty}^{\infty} p(x) d x\] We observe that they are indeed equal, therefore we have established the desired formula.