1. **State the problem:** Given the cumulative distribution function (CDF) $$F(x) = \frac{1}{1 + e^{-x}}$$ for $$-\infty < x < \infty$$, find the probability density function (PDF) of the random variable. 2. **Recall the relationship between CDF and PDF:** The PDF $$f(x)$$ is the derivative of the CDF $$F(x)$$ with respect to $$x$$: $$f(x) = \frac{d}{dx}F(x)$$ 3. **Differentiate the given CDF:** Start with $$F(x) = \frac{1}{1 + e^{-x}}$$. Rewrite as $$F(x) = (1 + e^{-x})^{-1}$$. Using the chain rule: $$f(x) = -1 \times (1 + e^{-x})^{-2} \times \frac{d}{dx}(1 + e^{-x})$$ Calculate the derivative inside: $$\frac{d}{dx}(1 + e^{-x}) = 0 + (-e^{-x}) = -e^{-x}$$ So, $$f(x) = - (1 + e^{-x})^{-2} \times (-e^{-x}) = \frac{e^{-x}}{(1 + e^{-x})^{2}}$$ 4. **Simplify the expression:** The PDF is $$f(x) = \frac{e^{-x}}{(1 + e^{-x})^{2}}$$ This is the probability density function corresponding to the logistic distribution. 5. **Interpretation:** The PDF is positive for all real $$x$$ and integrates to 1 over $$(-\infty, \infty)$$, confirming it is a valid density function. **Final answer:** $$f(x) = \frac{e^{-x}}{(1 + e^{-x})^{2}}$$