1 Let T Be a Nonnegative Continuous Random Variable Show Dobrow
Description
The number of days that it rains in a period of \(n\) days is a discrete random variable but let's consider the daily rainfall at a specified geographical point. Theoretically, with measuring equipment of perfect accuracy, the amount of rainfall could take on any value between 0 and 5 inches. As a result, each of the uncountably infinite number of point values in the interval \((0, 5)\) represents a distinct possible value of the amount of rainfall in a day.
A random variable that can take on any real value in an interval is called continuous.
Suppose the length of time required by students to complete a 2-hour exam is a continuous random variable with a density function given by is a random variable \(X.\)
Probability density function
A nonnegative function \(f(x)\) is called a probability density function (PDF) of the continuous random variable \(X\) if the total area bounded by its curve and the x-axis is equal to 1: \[\int_{-\infty}^{\infty} f(x)\ dx = 1.\]
For any two outcomes \(a\) and \(b\) with \(a < b\): \[\Pr(a < X < b) = \int_{a}^{b} f(x)\ dx.\]
Thus the probability that the continuous random variable takes a value in the interval \((a, b)\) is given by the area under the PDF curve between \(a\) and \(b\).
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\(\Pr(X = x) = 0.\)
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For \(a < b:\; \Pr(a < X < b) = \Pr(X < b) - \Pr(X < a) = \Pr(a \le X \le b)\).
Cumulative distribution function
\[ F(x) = \Pr(X \le x) = \int_{-\infty}^{x} f(u)\ du, \qquad -\infty < x < \infty. \]
Differentiating \(F(x)\) we find the inverse relationship between the PDF and CDF: \[ f(x) = \frac{dF(x)}{dx} = F^{\prime}(x). \]
The PDF \(f(x)\) is defined as the first derivative of the CDF \(F(x),\) and it is therefore not representing a probability value: \[f(x) = \lim_{\delta x \rightarrow 0^+}\dfrac{\Pr(x<X<x + \delta x)}{\delta x}.\]
Variance
The Variance of \(X\) is given by: \[ \begin{aligned} \sigma^{2} = V(X) &= E((X - \mu)^{2}) \\ &= \int_{-\infty}^{\infty} (x - \mu)^{2}\ f(x)\ dx. \end{aligned} \]
By an argument similar to the one used in the discrete case, an alternative formula for the variance is given by: \[ \begin{aligned} \sigma^{2} = V(X) &= E(X^{2}) - \mu^{2} \\ &= \int_{-\infty}^{\infty} x^{2}\ f(x)\ dx - \mu^{2}. \end{aligned} \]
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We may also refer to \(\mu\) and \(\sigma^{2}\) as the mean and variance, respectively, of the continuous probability distribution with PDF \(f(x) \ (-\infty < x < \infty)\).
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\(\sigma\), the square root of the variance \(\sigma^{2}\), is the standard deviation.
Example
Suppose the length of time required by students to complete a 2-hour exam is a random variable with a density function given by is a random variable \(X\) with PDF: \[ f(x) = \frac{1}{6}(x+x^3), \qquad 0 \leq x \leq 2. \] The corresponding CDF of \(X\) is: \[ \begin{aligned} F(x) &= \int_{-\infty}^{x} f(u) \; du \\ &= \int_{0}^{x} \frac{1}{6}(u+u^3) \; du \\ &= \frac{1}{12} x^2 + \frac{1}{24} x^4 , \qquad 0 \leq x \leq 2. \\ \end{aligned} \]
If \(f(x)\) is a PDF we have that: \[ \int_{-\infty}^{+\infty} f(x) \; dx = \int_{0}^{2} \frac{1}{6}(x+x^3) \; dx = F(2) - F(0) = 1 - 0 = 1.\\ \]
Let's find the probability that a student requires less than 1.7 hours to complete the 2-hour exam.
\[ \begin{aligned} \Pr(X < 1.7) = F(1.7) &= \int_{-\infty}^{1.7} f(x)\ dx\\ &= \int_{0}^{1.7} \frac{1}{6}(x+x^3)\ dx\\ &= F(1.7) - F(0)\\ &\approx 0.59. \end{aligned} \] that is equivalent of calculating the skyblue area of the curve displayed in following plot: 
Let's find the probability that a student requires more than 1 hour to to complete the 2-hour exam.
\[ \begin{aligned} \Pr(X > 1) &= 1 - \Pr(X \le 1) \\ &= 1 - \int_{-\infty}^{1} f(x)\ dx \\ &= 1 - \int_{0}^{1} \frac{1}{6}(x+x^3)\ dx \\ &= 1 - \left[ \frac{1}{12} x^2 + \frac{1}{24} x^4 \right]_0^1 \\ &= 1 - \left[F(1) - F(0)\right] \\ &= 1 - \left[\dfrac{1}{8} - 0\right] \\ &= \dfrac{7}{8}. \end{aligned} \]
Let's calculate the expected amount of time required by a student to complete the 2-hour exam. \[ \begin{aligned} E(X) &= \int_{-\infty}^{+\infty} x\ f(x) \; dx \\ &= \int_{0}^{2} \frac{1}{6}x(x+x^3) \; dx = 1.5111. \end{aligned} \]
Let's calculate the variance of the amount of time required by a student to complete the 2-hour exam. \[ \begin{aligned} V(X) &= \int_{-\infty}^{+\infty} x^2\ f(x) \; dx - E(X)^2 \\ &= \int_{0}^{2} \frac{1}{6}x^2(x+x^3) \; dx - 1.5111^2 \\ &= 2.444 - 2.2834 = 0.1606. \end{aligned} \]
Source: https://acaimo.github.io/teaching/statistics_1/home/notes_6.html
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