Continuous random variables — FAQ

Because probability is represented by the area under the PDF curve. The area of a region with zero width (at a single point 'c') is zero. Think of it as integrating from c to c: $\int_{c}^{c} f(x) dx = 0$.

The PDF, $f(x)$, is the 'density' function; its value is not a probability, but the area under it is. The CDF, $F(x)$, is a cumulative probability function; its value at $x$ is the actual probability $P(X \le x)$. The PDF is the derivative of the CDF.

Variance can never be negative. If you get a negative answer, you have made a calculation error. The most common mistakes are: (1) an error in the integration of $x^2 f(x)$, or (2) forgetting to square $E(X)$ in the formula $Var(X) = E(X^2) - [E(X)]^2$. The worked example above shows how to spot and correct this.

You need to define $F(x)$ for all real numbers. It will be 0 for $x$ less than the lower bound of the PDF's range. It will be 1 for $x$ greater than the upper bound. In between, you integrate the PDF from the lower bound up to a variable $x$. If the PDF has multiple pieces, the CDF will also have multiple corresponding pieces.