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Subsections


Improper integrals and probability density functions

Introduction

Improper integrals like the ones we have been considering in class have many applications, for example in thermodynamics and heat transfer. In this lab we will consider the role of improper integrals in probability, which also has many applications in science and engineering.

Background

The first concept we need is that of a random variable. Intuitively, a random variable is used to measure an outcome whose value is not certain. For example, the number of hours that a hard disk can run before failing is a random variable because it is not the same for every drive, even if we only consider identical drives from the same production run. A few other examples of random variables that are important in science, engineering, or manufacturing are given below.

You may be more familiar with what are called discrete random variables, for example the number of heads obtained in ten tosses of a coin, which can only take a finite number of discrete values. In the case of a discrete random variable, the probability of a single outcome can be positive. For example, the probability that a single flip of a coin produces tails is 50%. The situation is very different when we consider a random variable like the number of miles a tire can be driven before failure, which can take any value from zero to something over 100,000 miles. Since there are an infinite number of possible outcomes, the probability that the tire fails at exactly some number of miles, for example 50,000 miles, is zero. However, we would expect that the probability that the tire would fail between 40,000 miles and 100,000 miles would not be zero, but would be a positive number.

A random variable that can take on a continuous range of values is called a continuous random variable. There turn out to be lots of applications of continuous random variables in science, engineering, and business, so a lot of effort has gone into devising mathematical models. These mathematical models are all based on the following definition.

Definition 281

We say that a random variable X is continuous if there is a function f(x), called the probability density function, such that

1.
$f(x) \geq 0$, for all x
2.
$\int_{- \infty}^{\infty} f(x) \, dx = 1$
3.
$P(a \leq X \leq b) = \int_{a}^{b} f(x) \, dx$ where $P(a \leq X
\leq b)$ represents the probability that the random variable X is greater than or equal to a but less than or equal to b.

For example, consider the following function.

\begin{displaymath}
f(x) = \left\{ \begin{array}
{ll}
 e^{-x} & \mbox{ if $x \gt 0$} \\  0 & \mbox{ otherwise} 
 \end{array} \right.\end{displaymath}

This function is non-negative, and also satisfies the second condition, since

\begin{displaymath}
\int_{- \infty}^{\infty} f(x) \, dx = \int_{0}^{\infty} e^{-x} \, dx
= 1\end{displaymath}

which is pretty easy to show. So this could be a probability density function for a continuous random variable X.

A lot of the effort involved in modeling a random process, that is, a process whose outcome is a random variable, is in finding a suitable probability density function. Over the years, lots of different functions have been proposed and used. One thing that they all have in common, though, is that they depend on parameters. For example, the general exponential probability density function is defined as

\begin{displaymath}
f(x) = \left\{ \begin{array}
{ll}
 \frac{1}{\lambda}e^{-x/\l...
 ...{ if $x \gt 0$} \\  0 & \mbox{ otherwise} 
 \end{array} \right.\end{displaymath}

where $\lambda$ is a parameter that can be adjusted to get the best fit to any particular situation. In the exercises, you will be asked to show that only positive values of $\lambda$ make sense.

The process of deciding what probability density function to use and how to determine the parameters is very complicated and can involve very sophisticated mathematics. However, in the simple approach we are taking here, the problem of determining the parameter value(s) often depends on quantities that can be determined experimentally, for example by collecting data on tire failure. For our purposes, the two most important quantities are the mean, $\mu$ and the standard deviation $\sigma$. The mean is defined by

\begin{displaymath}
\mu = \int_{- \infty}^{\infty} x f(x) \, dx \end{displaymath}

and the standard deviation is the square root of the variance, V, which is defined by

\begin{displaymath}
V = \int_{- \infty}^{\infty} (x- \mu)^2 f(x) \, dx \end{displaymath}

In practice, the variance V is often computed as follows,

\begin{displaymath}
V = \int_{- \infty}^{\infty} x^2 f(x) \, dx - \mu^2 \end{displaymath}

which can be easily be obtained by expanding $(x-\mu)^2$ and writing V as the sum of three integrals.

Probably the most important distribution is the normal distribution, widely referred to as the bell-shaped curve. The probability density function for a normal distribution with mean $\mu$ and standard deviation $\sigma$ is given by the following equation.

\begin{displaymath}
f(x) = \frac{1}{\sigma \sqrt{2 \pi}} e^{- \frac{(x- \mu)^2}{2
\sigma^2}} , \mbox{ for $- \infty < x < \infty$\space } \end{displaymath}

This distribution has a tremendous number of applications in science, engineering, and business. The exercises provide a few simple ones.

In applications, one generally has to know in advance that the random variable you want to model has, approximately, a certain kind of distribution. How one would determine this is way beyond the scope of this course, so we won't really discuss it. On the other hand, once you know, for example, that your random variable has a normal distribution you only need the values of the mean and the standard deviation to be able to model it. The exponential distribution is even simpler, since it only has one parameter, and you only need to know the mean of your random variable to use this distribution to model it.

One thing to keep in mind when you are using the normal distribution as a model is that calculations can involve values of your random variable that don't make physical sense. For example, suppose that a machining operation produces steel shafts whose diameters have a normal distribution, with a mean of 1.005 inches and a standard deviation of 0.01 inch. If you were asked to compute the percentage of the shafts in a certain production run had diameters less than 0.9 inches you would use the following integral

\begin{displaymath}
\int_{- \infty}^{0.9} \frac{1}{0.01 \sqrt{2 \pi}} e^{- \frac{(x- 1.005)^2}{
0.0002}} \end{displaymath}

even though negative values for the shaft diameters don't make sense.

Exercises

1.
Show that the probability density function given for the exponential distribution,

\begin{displaymath}
f(x) = \left\{ \begin{array}
{ll}
 \frac{1}{\lambda}e^{-x/\l...
 ...{ if $x \gt 0$} \\  0 & \mbox{ otherwise} 
 \end{array} \right.\end{displaymath}

satisfies the condition

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

as long as $\lambda$ is a positive number. What would happen if $\lambda$ was negative?
2.
Show that the mean and the standard deviation of the exponential distribution are both equal to $\lambda$.

3.
The time (in hours) required to repair a machine is exponentially distributed with a mean $\lambda = 1/2$. What is the probability that a repair takes longer than 2 hours?

4.
Suppose X is a continuous random variable. Explain why it is true that $P(X \geq a) = P(X \gt a)$, where a is any number. You may want to use the fact that $P(X \geq a) = P(X=a) + P(X \gt a)$.

5.
The median m of a random variable X is the value of the random variable such that P(X > m) = 1/2. If X has an exponential distribution with mean $\lambda$, find the median m.

6.
Suppose that X is a continuous random variable with probability density function given by

\begin{displaymath}
f(x) = \left\{ \begin{array}
{ll}
 c(1-x^2) & \mbox{ if $-1 < x < 14$} \\  0 & \mbox{otherwise}
 \end{array} \right.\end{displaymath}

First, compute the value of c. Then find the mean $\mu$.

7.
The width, in inches, of a slot of a certain production run of a forged machine part is normally distributed with $\mu = 0.9$ and $\sigma =
0.003$. The specifications for the slot are $0.9 \pm 0.005$. That is, for a part to be acceptable, the width of the slot must be between 0.895 and 0.905 inches. Find the percentage of the parts that are defective.

8.
The life of a certain type of automobile tire is normally distributed with a mean of 55,000 miles and a standard deviation of 6,000 miles.
(a)
What is the probability that such a tire lasts more than 65,000 miles?
(b)
What is the probability that it lasts between 50,000 and 60,000 miles?

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William W. Farr
10/29/1999