Continuous Random Variables Over Infinite Domains

Extending continuous random variables to unbounded intervals such as [0,)[0,\infty) or (,)(-\infty,\infty). Students learn to evaluate the normalization condition and compute probabilities using improper integrals.

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PDFs on Infinite Domains

So far, we've worked with continuous random variables on a bounded interval [a,b][a,b], where the probability density function (PDF) f(x)f(x) satisfies abf(x)dx=1\int_a^b f(x)\,dx = 1. We now extend this idea to infinite domains such as [0,)[0,\infty), (,0](-\infty,0], or (,)(-\infty,\infty).

If XX takes values on an unbounded interval, its PDF f(x)f(x) must still satisfy the normalization condition

f(x)dx=1,\int_{-\infty}^{\infty} f(x)\,dx = 1,

where any unbounded part of the integral is treated as an improper integral, evaluated via a limit:

0g(x)dx=limb0bg(x)dx.\int_0^\infty g(x)\,dx = \lim_{b\to\infty} \int_0^b g(x)\,dx.

For example, consider f(x)=2e2xf(x) = 2e^{-2x} on [0,)[0,\infty). We check normalization:

02e2xdx=limb[e2x]0b=limb(1e2b)=10=1.\begin{align*} \int_0^\infty 2e^{-2x}\,dx &= \lim_{b\to\infty}\left[-e^{-2x}\right]_0^b \\ &= \lim_{b\to\infty}\left(1 - e^{-2b}\right) \\ &= 1 - 0 \\ &= 1. \end{align*}

Since the integral equals 11, f(x)=2e2xf(x) = 2e^{-2x} is a valid PDF on [0,)[0,\infty).

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