# Calculus

## Applications of Integrals # Probability Density Function

### Definition of Probability Density Function

We call $$X$$ a continuous random variable if $$X$$ can take any value on an interval, which is often the entire set of real numbers $$\mathbb{R}.$$

Every continuous random variable $$X$$ has a probability density function $$\left( {PDF} \right),$$ written $$f\left( x \right),$$ that satisfies the following conditions:

1. $$f\left( x \right) \ge 0$$ for all $$x,$$ and
2. $$\int\limits_{ – \infty }^\infty {f\left( x \right)dx} = 1.$$

The probability that a random variable $$X$$ takes on values in the interval $$a \le X \le b$$ is defined as

$P\left( {a \le X \le b} \right) = \int\limits_a^b {f\left( x \right)dx} ,$

which is the area under the curve $$f\left( x \right)$$ from $$x = a$$ to $$x = b.$$ Figure 1.

### Mean and Median

If a random variable $$X$$ has a density function $${f\left( x \right)},$$ then we define the mean value (also known as the average value or the expectation) of $$X$$ as

$\mu = \int\limits_{ – \infty }^\infty {xf\left( x \right)dx}.$

The median of a continuous probability distribution $$f\left( x \right)$$ is the value of $$x = m$$ that splits the probability distribution into two portions whose areas are identical and equal to $$\large{\frac{1}{2}}\normalsize:$$

${\int\limits_{ – \infty }^m {f\left( x \right)dx} }={ \int\limits_m^\infty {f\left( x \right)dx} }={ \frac{1}{2}.}$ Figure 2.

Note that not all $$PDFs$$ have mean values. For example, the Cauchy distribution is an example of a probability distribution which has no mean.

### Variance

The variance of a continuous random variable is defined by the integral

${\sigma ^2} = \int\limits_{ – \infty }^\infty {{{\left( {x – \mu } \right)}^2}f\left( x \right)dx} ,$

where $$\mu$$ is the mean of the random variable $$X.$$

### Uniform Distribution

The simplest $$PDF$$ is the uniform distribution. The density of the uniform distribution is defined by

${f\left( x \right) = \frac{1}{{b – a}}\;\;}\kern0pt{\text{for}\;\;a \le x \le b.}$ Figure 3.

The mean value of the uniform distribution across the interval $$\left[ {a,b} \right]$$ is

${\mu = \int\limits_a^b {xf\left( x \right)dx} }={ \frac{{a + b}}{2}.}$

If a random variable $$X$$ is distributed uniformly in the interval $$\left[ {a,b} \right],$$ the probability to fall within a range $$\left[ {c,d} \right] \in \left[ {a,b} \right]$$ is expressed by the formula

${P\left( {c \le X \le d} \right) = \int\limits_c^d {f\left( x \right)dx} }={ \int\limits_c^d {\frac{{dx}}{{b – a}}} }={ \frac{{d – c}}{{b – a}}.}$

The variance of the distribution is

${{\sigma ^2} = \int\limits_a^b {{{\left( {x – \mu } \right)}^2}f\left( x \right)dx} }={ \frac{{{{\left( {b – a} \right)}^2}}}{{12}}.}$

### Exponential Distribution

The exponential distribution is a continuous distribution that is commonly used to describe the waiting time until some specific event occurs. For example, the amount of time until a hurricane or other dangerous weather event occurs obeys an exponential distribution law.

The one-parameter exponential distribution of the probability density function $$PDF$$ is described as follows:

${f\left( x \right) = \lambda {e^{ – \lambda x}},\;\;}\kern0pt{x \ge 0,}$

where the rate $$\lambda$$ represents the average amount of events per unit of time. Figure 4.

The mean value (or the average waiting for the next event) is $$\mu = \large{\frac{1}{\lambda }}\normalsize.$$ The median of the exponential distribution is $$m = \large{\frac{{\ln 2}}{\lambda }}\normalsize,$$ and the variance is given by $${\sigma ^2} = \large{\frac{1}{{{\lambda ^2}}}}\normalsize.$$

### Normal Distribution

The normal distribution is the most widely known probability distribution since it describes many natural phenomena.

The $$PDF$$ of the normal distribution is given by the formula

$f\left( x \right) = \frac{1}{{\sqrt {2\pi {\sigma ^2}} }}{e^{ – \frac{{{{\left( {x – \mu } \right)}^2}}}{{2{\sigma ^2}}}}},$

where $$\mu$$ is the mean of the distribution, and $${\sigma^2}$$ is the variance.

The two parameters $$\mu$$ and $${\sigma}$$ entirely define the shape and all other properties of the normal distribution function. Figure 5.

If a random variable $$X$$ follows the normal distribution with the parameters $$\mu$$ and $$\sigma,$$ we write $$X \sim N\left( {\mu ,\sigma } \right).$$

The normal distribution is said to be standard when $$\mu = 0$$ and $$\sigma = 1.$$ In this special case, the normal random variable $$X$$ is called a standard score or a $$Z-$$score. Thus, by definition, $$Z \sim N\left( {0 ,1} \right).$$

Every normal random variable $$X$$ can be transformed into a $$Z-$$score by using the substitution

$z = \frac{{x – \mu }}{\sigma }.$

Pay attention to the notations: $$X, Z$$ denote the random variables, and $$x,z$$ denote the possible values of the variables.

To compute probabilities for $$Z,$$ we use a standard normal table ($$Z-$$table) or a software tool.

To find the probability that a normally distributed random variable $$X$$ falls within a range $$\left[ {a,b } \right],$$ we rely on the $$Z-$$score and use the formula

${P\left( {a \le X \le b} \right) }={ P\left( {\frac{{a – \mu }}{\sigma } \le Z \le \frac{{b – \mu }}{\sigma }} \right).}$

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Calculate the mean value $$\mu$$ and the variance $${\sigma^2}$$ of the uniform distribution $$f\left( x \right) = \large{\frac{1}{{b – a}}}\normalsize$$ for $$a \le x \le b.$$

### Example 2

Let $$X$$ be a random variable distributed uniformly in the interval $$\left[ {{x_0} – L,{x_0} + L} \right].$$ Find the mean $$\mu$$ and variance $${\sigma^2}$$ of the random variable $$X.$$

### Example 3

Find the mean value $$\mu$$ and the median $$m$$ of the exponential distribution $$f\left( x \right) = \lambda {e^{ – \lambda x}}.$$

### Example 4

Assume that the waiting time for your next email is described by the exponential density function with rate $$\lambda = 3$$ (emails per hour). Determine the probability that you receive no email during the next hour.

### Example 5

Let $$X$$ be a normal random variable with the mean $$\mu = 20$$ and the variance $${\sigma ^2} = 25.$$ Find the probability that the random variable $$X$$ falls in the range $$\left[ {15,30} \right].$$

### Example 6

$$IQ$$ tests are designed to follow the distribution law $$N(100,15).$$ What is the percentage of individuals with an $$IQ$$-score higher than $$120?$$

### Example 7

A random variable $$X$$ is defined by the linear $$PDF$$ in the form $$f\left( x \right) = kx$$ on the interval $$\left[ {0,10} \right].$$
1. Find the value of $$k$$;
2. Determine the mean value $$\mu$$ of $$X$$;
3. Calculate the probability $$P\left( {2 \le X \le 5} \right);$$

### Example 8

A random variable $$X$$ is defined by the quadratic $$PDF$$ in the form $$f\left( x \right) = k{x^2}$$ on the interval $$\left[ {0,3} \right].$$
1. Find the value of $$k$$;
2. Determine the mean value $$\mu$$ of $$X$$;
3. Calculate the probability $$P\left( {1 \le X \le 2} \right);$$

### Example 9

The $$PDF$$ of a random variable $$X$$ is defined by

$f\left( x \right) = \begin{cases} {\frac{k}{{{x^2}}}}, & \text{if } {x \ge 1} \\ 0, & \text{if } x \lt 1 \end{cases}.$

1. Calculate the value of $$k$$;
2. Find the mean value $$\mu$$ of $$X.$$

### Example 10

Let $$X$$ be a continuous random variable with $$PDF$$ given by

$f\left( x \right) = \begin{cases} {\frac{k}{{1 + {x^2}}}}, & \text{if } {x \ge 0} \\ 0, & \text{if } x \lt 0 \end{cases}.$

1. Find the value of $$k$$;
2. Graph the $$PDF;$$
3. Calculate the probability $$P\left( {0 \le X \le 1} \right);$$
4. Determine the mean value $$\mu$$ of $$X.$$

### Example 1.

Calculate the mean value $$\mu$$ and the variance $${\sigma^2}$$ of the uniform distribution $$f\left( x \right) = \large{\frac{1}{{b – a}}}\normalsize$$ for $$a \le x \le b.$$

Solution.

First we find the mean $$\mu:$$

$\require{cancel}{\mu = \int\limits_a^b {xf\left( x \right)dx} }={ \int\limits_a^b {\frac{{xdx}}{{b – a}}} }={ \frac{1}{{b – a}}\int\limits_a^b {xdx} }={ \frac{1}{{b – a}}\left. {\left( {\frac{{{x^2}}}{2}} \right)} \right|_a^b }={ \frac{1}{{b – a}} \cdot \frac{{{b^2} – {a^2}}}{2} }={ \frac{{\cancel{\left( {b – a} \right)}\left( {b + a} \right)}}{{2\cancel{\left( {b – a} \right)}}} }={ \frac{{a +b}}{2}.}$

Now let’s derive the expression for the variance $${\sigma ^2}.$$ By definition,

${\sigma ^2} = \int\limits_a^b {{{\left( {x – \mu } \right)}^2}f\left( x \right)dx} .$

Expanding the square in the integrand, we can write:

${{\sigma ^2} \text{ = }}\kern0pt{\int\limits_a^b {\left( {{x^2} – 2\mu x + {\mu ^2}} \right)f\left( x \right)dx} }={ \int\limits_a^b {{x^2}f\left( x \right)dx} }-{ 2\mu \int\limits_a^b {xf\left( x \right)dx} }+{ {\mu ^2}\int\limits_a^b {f\left( x \right)dx} .}$

Recall that

${\int\limits_a^b {xf\left( x \right)dx} = \mu ,\;\;\;}\kern0pt{\int\limits_a^b {f\left( x \right)dx} = 1.}$

Then

${{\sigma ^2} = \int\limits_a^b {{x^2}f\left( x \right)dx} – 2{\mu ^2} + {\mu ^2} }={ \int\limits_a^b {{x^2}f\left( x \right)dx} – {\mu ^2} }={ \frac{1}{{b – a}}\int\limits_a^b {{x^2}dx} – {\left( {\frac{{a + b}}{2}} \right)^2} }={ \frac{1}{{b – a}}\left. {\frac{{{x^3}}}{3}} \right|_a^b – {\left( {\frac{{a + b}}{2}} \right)^2} }={ \frac{{{b^3} – {a^3}}}{{3\left( {b – a} \right)}} – {\left( {\frac{{a + b}}{2}} \right)^2} }={ \frac{{{b^2} + ab + {a^2}}}{3} }-{ \frac{{{a^2} + 2ab + {b^2}}}{4} }={ \frac{{{b^2} – 2ab + {a^2}}}{{12}} }={ \frac{{{{\left( {b – a} \right)}^2}}}{{12}}.}$

### Example 2.

Let $$X$$ be a random variable distributed uniformly in the interval $$\left[ {{x_0} – L,{x_0} + L} \right].$$ Find the mean $$\mu$$ and variance $${\sigma^2}$$ of the random variable $$X.$$

Solution.

Make sure that the mean value coincides with the middle of the interval:

$\require{cancel}{\mu = \int\limits_{{x_0} – L}^{{x_0} + L} {xf\left( x \right)dx} }={ \frac{1}{{2L}}\int\limits_{{x_0} – L}^{{x_0} + L} {xdx} }={ \frac{1}{{2L}}\left. {\frac{{{x^2}}}{2}} \right|_{{x_0} – L}^{{x_0} + L} }={ \left. {\frac{{{x^2}}}{{4L}}} \right|_{{x_0} – L}^{{x_0} + L} }={ \frac{1}{{4L}}\left[ {{{\left( {{x_0} + L} \right)}^2} – {{\left( {{x_0} – L} \right)}^2}} \right] }={ \frac{1}{{4L}}\left[ {\cancel{x_0^2} + 2{x_0}L + \cancel{L^2} }\right.}-{\left.{ \cancel{x_0^2} + 2{x_0}L – \cancel{L^2}} \right] }={ \frac{{\cancel{4}{x_0}\cancel{L}}}{{\cancel{4L}}} }={ {x_0}.}$

Compute the variance:

${{\sigma ^2} = \int\limits_{{x_0} – L}^{{x_0} + L} {{{\left( {x – \mu } \right)}^2}f\left( x \right)dx} }={ \frac{1}{{2L}}\int\limits_{{x_0} – L}^{{x_0} + L} {{{\left( {x – {x_0}} \right)}^2}dx} }={ \frac{1}{{2L}}\left. {\frac{{{{\left( {x – {x_0}} \right)}^3}}}{3}} \right|_{{x_0} – L}^{{x_0} + L} }={ \frac{1}{{6L}}\left[ {{L^3} – {{\left( { – L} \right)}^3}} \right] }={ \frac{{2{L^3}}}{{6L}} }={ \frac{{{L^2}}}{3}.}$

### Example 3.

Find the mean value $$\mu$$ and the median $$m$$ of the exponential distribution $$f\left( x \right) = \lambda {e^{ – \lambda x}}.$$

Solution.

The mean value $$\mu$$ is determined by the integral

${\mu = \int\limits_{ – \infty }^\infty {xf\left( x \right)dx} }={ \lambda \int\limits_0^\infty {x{e^{ – \lambda x}}dx} .}$

Integrating by parts, we have

${\mu = \lambda \int\limits_0^\infty {x{e^{ – \lambda x}}dx} = \left[ {\begin{array}{*{20}{l}} {u = x}\\ {dv = {e^{ – \lambda x}}dx}\\ {du = dx}\\ {v = – \frac{1}{\lambda }{e^{ – \lambda x}}} \end{array}} \right] }={ \lambda \left[ { – \left. {\frac{x}{\lambda }{e^{ – \lambda x}}} \right|_0^\infty – \int\limits_0^\infty {\left( { – \frac{1}{\lambda }{e^{ – \lambda x}}} \right)dx} } \right] }={ \int\limits_0^\infty {{e^{ – \lambda x}}dx} – \left. {x{e^{ – \lambda x}}} \right|_0^\infty }={ – \frac{1}{\lambda }\left. {{e^{ – \lambda x}}} \right|_0^\infty – \left. {x{e^{ – \lambda x}}} \right|_0^\infty .}$

We evaluate the second term with the help of l’Hopital’s Rule:

${\left. {x{e^{ – \lambda x}}} \right|_0^\infty = \lim \limits_{b \to \infty } \left[ {\left. {b{e^{ – \lambda b}}} \right|_0^b} \right] }={ \lim \limits_{b \to \infty } \frac{b}{{{e^{\lambda b}}}} }={ \left[ {\frac{\infty }{\infty }} \right] }={ \lim \limits_{b \to \infty } \frac{{b^\prime}}{{\left( {{e^{\lambda b}}} \right)^\prime}} }={ \lim \limits_{b \to \infty } \frac{1}{{\lambda {e^{\lambda b}}}} }={ 0.}$

Hence, the mean (average) value of the exponential distribution is

${\mu = – \frac{1}{\lambda }\left. {{e^{ – \lambda x}}} \right|_0^\infty }={ – \frac{1}{\lambda }\left( {0 – 1} \right) }={ \frac{1}{\lambda }.}$

Determine the median $$m:$$

${\int\limits_{ – \infty }^m {f\left( x \right)dx} = \frac{1}{2},}\;\; \Rightarrow {\lambda \int\limits_0^m {{e^{ – \lambda x}}dx} = \frac{1}{2},} \Rightarrow {\lambda \left. {\left( { – \frac{1}{\lambda }{e^{ – \lambda x}}} \right)} \right|_0^m = \frac{1}{2},}\;\; \Rightarrow {- {e^{ – \lambda m}} + {e^0} = \frac{1}{2},}\;\; \Rightarrow {{e^{ – \lambda m}} = \frac{1}{2},}\;\; \Rightarrow {{e^{\lambda m}} = 2,}\;\; \Rightarrow {\lambda m = \ln 2,}\;\; \Rightarrow {m = \frac{{\ln 2}}{\lambda }.}$

### Example 4.

Assume that the waiting time for your next email is described by the exponential density function with rate $$\lambda = 3$$ (emails per hour). Determine the probability that you receive no email during the next hour.

Solution.

The probability density function has the form

$f\left( t \right) = \lambda {e^{ – \lambda t}} = 3{e^{ – 3t}},$

where the time $$t$$ is measured in hours.

Let’s calculate the probability that you receive an email during the hour. Integrating the exponential density function from $$t = 0$$ to $$t = 1,$$ we have

${P\left( {0 \le t \le 1} \right) }={ \int\limits_0^1 {f\left( t \right)dt} }={ \int\limits_0^1 {3{e^{ – 3t}}dt} }={ 3\int\limits_0^1 {{e^{ – 3t}}dt} }={ 3 \cdot \left. {\left( { – \frac{1}{3}{e^{ – 3t}}} \right)} \right|_0^1 }={ 1 – {e^{ – 3}}.}$

So, the probability $${P^C}$$ of the opposite (complementary) event (that is, that you will not receive any email within an hour) is equal to

${{P^C} = 1 – P\left( {0 \le t \le 1} \right) }={ 1 – \left( {1 – {e^{ – 3}}} \right) }={ {e^{ – 3}} \approx 0.05 }={ 5\% }$

### Example 5.

Let $$X$$ be a normal random variable with the mean $$\mu = 20$$ and the variance $${\sigma ^2} = 25.$$ Find the probability that the random variable $$X$$ falls in the range $$\left[ {15,30} \right].$$

Solution.

We compute the probability by the formula

${P\left( {a \le X \le b} \right) }={ P\left( {\frac{{a – \mu }}{\sigma } \le Z \le \frac{{b – \mu }}{\sigma }} \right).}$

Substituting $$\mu = 20,$$ $$\sigma = 5,$$ $$a = 15,$$ $$b = 30,$$ we have

${P\left( {15 \le X \le 30} \right) }={ P\left( {\frac{{15 – 20}}{5} \le Z \le \frac{{30 – 20}}{5}} \right) }={ P\left( { – 1 \le Z \le 2} \right) }={ P\left( {z = 2} \right) – P\left( {z = – 1} \right).}$

Using a $$Z-$$table, we find that

${P\left( {z = 2} \right) = 0.9772,\;\;\;}\kern0pt{P\left( {z = – 1} \right) = 0.1587}$

Hence,

${P\left( {15 \le X \le 30} \right) = 0.9772 – 0.1587 }={ 0.8185}$

### Example 6.

$$IQ$$ tests are designed to follow the distribution law $$N(100,15).$$ What is the percentage of individuals with an $$IQ$$-score higher than $$120?$$

Solution.

We need to calculate the $$Z-$$score. Using the formula

${P\left( {a \le X \le b} \right) }={ P\left( {\frac{{a – \mu }}{\sigma } \le Z \le \frac{{b – \mu }}{\sigma }} \right),}$

where $$X \sim N\left( {100,15} \right),$$ we get

${P\left( {120 \le X \le \infty } \right) = P\left( {\frac{{120 – 100}}{{15}} \le Z \le \frac{{\infty – 100}}{{15}}} \right) }={ P\left( {\frac{4}{3} \le Z \le \infty } \right) }={ P\left( {Z \ge \frac{4}{3}} \right).}$

From a $$Z-$$table, we find that

${P\left( {z = \frac{4}{3}} \right) \approx P\left( {z = 1.33} \right) }={ 0.9082 }\approx{ 90.8\% }$

Thus, about $$9.2\%$$ of the population have an $$ID-$$score greater than $$120.$$

### Example 7.

A random variable $$X$$ is defined by the linear $$PDF$$ in the form $$f\left( x \right) = kx$$ on the interval $$\left[ {0,10} \right].$$
1. Find the value of $$k$$;
2. Determine the mean value $$\mu$$ of $$X$$;
3. Calculate the probability $$P\left( {2 \le X \le 5} \right);$$

Solution.

1. To find the value of $$k,$$ we integrate the $$PDF$$ on the interval from $$0$$ to $$10$$ and equate it to $$1:$$

${\int\limits_0^{10} {f\left( x \right)dx} = 1,}\;\; \Rightarrow {\int\limits_0^{10} {kxdx} = 1,}\;\; \Rightarrow {\left. {k\frac{{{x^2}}}{2}} \right|_0^{10} = 1,}\;\; \Rightarrow {\frac{k}{2}\left( {100 – 0} \right) = 1,}\;\; \Rightarrow {50k = 1,}\;\; \Rightarrow {k = \frac{1}{{50}}.}$

So, the probability distribution is given by the function $$f\left( x \right) = \large{\frac{x}{{50}}}\normalsize.$$
2. Calculate the mean value $$\mu$$ of the distribution:
3. ${\mu = \int\limits_0^{10} {xf\left( x \right)dx} }={ \int\limits_0^{10} {\frac{{{x^2}}}{{50}}dx} }={ \frac{1}{{50}}\int\limits_0^{10} {{x^2}dx} }={ \frac{1}{{50}}\left. {\frac{{{x^3}}}{3}} \right|_0^{10} }={ \frac{{1000}}{{150}} }={ \frac{{20}}{3}.}$

4. Find the probability $$P\left( {2 \le X \le 5} \right):$$
5. ${P\left( {2 \le X \le 5} \right) = \int\limits_2^5 {f\left( x \right)dx} }={ \frac{1}{{50}}\int\limits_2^5 {xdx} }={ \left. {\frac{{{x^2}}}{{100}}} \right|_2^5 }={ \frac{1}{{100}}\left( {25 – 4} \right) }={ 0.21}$

### Example 8.

A random variable $$X$$ is defined by the quadratic $$PDF$$ in the form $$f\left( x \right) = k{x^2}$$ on the interval $$\left[ {0,3} \right].$$
1. Find the value of $$k$$;
2. Determine the mean value $$\mu$$ of $$X$$;
3. Calculate the probability $$P\left( {1 \le X \le 2} \right);$$

Solution.

1. Since the integral of $$PDF$$ over the domain must equal one, we have

${\int\limits_0^3 {f\left( x \right)dx} = 1,}\;\; \Rightarrow {\int\limits_0^3 {k{x^2}dx} = 1,}\;\; \Rightarrow {\left. {k\frac{{{x^3}}}{3}} \right|_0^3 = 1,}\;\; \Rightarrow {\frac{k}{3}\left( {27 – 0} \right) = 1,}\;\; \Rightarrow {k = \frac{1}{9}.}$

Hence, the $$PDF$$ is given by the function $$f\left( x \right) = \large{\frac{{x^2}}{{9}}}\normalsize.$$
2. We can easily find the mean value $$\mu$$ of the probability distribution:
3. ${\mu = \int\limits_0^3 {xf\left( x \right)dx} }={ \int\limits_0^3 {\frac{{{x^3}}}{9}dx} }={ \frac{1}{9}\int\limits_0^3 {{x^3}dx} }={ \frac{1}{9}\left. {\frac{{{x^4}}}{4}} \right|_0^3 }={ \frac{{81}}{{36}} }={ \frac{9}{4}.}$

4. The probability $$P\left( {1 \le X \le 2} \right)$$ is also determined through integration:
5. ${P\left( {1 \le X \le 2} \right) = \int\limits_1^2 {f\left( x \right)dx} }={ \frac{1}{9}\int\limits_1^2 {{x^2}dx} }={ \left. {\frac{{{x^3}}}{{27}}} \right|_1^2 }={ \frac{1}{{27}}\left( {8 – 1} \right) }={ \frac{7}{{27}} \approx 0.26}$

### Example 9.

The $$PDF$$ of a random variable $$X$$ is defined by

$f\left( x \right) = \begin{cases} {\frac{k}{{{x^2}}}}, & \text{if } {x \ge 1} \\ 0, & \text{if } x \lt 1 \end{cases}.$

1. Calculate the value of $$k$$;
2. Find the mean value $$\mu$$ of $$X.$$

Solution.

1. We determine the value of $$k$$ from the condition

$\int\limits_1^\infty {f\left( x \right)dx} = 1.$

Then
2. ${\int\limits_1^\infty {\frac{k}{{{x^2}}}dx} = 1,}\;\; \Rightarrow {\left. {\left( { – \frac{k}{x}} \right)} \right|_1^\infty = 1,}\;\; \Rightarrow {k\lim\limits_{b \to \infty } \left( { – \frac{1}{b} + 1} \right) = 1,}\;\; \Rightarrow {k = 1.}$

3. To find $$\mu,$$ we take the integral

$\require{cancel}{\mu = \int\limits_1^\infty {xf\left( x \right)dx} }={ \int\limits_1^\infty {\frac{{\cancel{x}dx}}{{{x^{\cancel{2}}}}}} }={ \int\limits_1^\infty {\frac{{dx}}{x}} }={ \left. {\ln x} \right|_1^\infty }={ \infty .}$

As you can see, the probability density function $$f\left( x \right) = \large{\frac{1}{{{x^2}}}}\normalsize$$ defined on the domain $$\left[ {1,\infty } \right)$$ has no mean.

### Example 10.

Let $$X$$ be a continuous random variable with $$PDF$$ given by

$f\left( x \right) = \begin{cases} {\frac{k}{{1 + {x^2}}}}, & \text{if } {x \ge 0} \\ 0, & \text{if } x \lt 0 \end{cases}.$

1. Find the value of $$k$$;
2. Graph the $$PDF;$$
3. Calculate the probability $$P\left( {0 \le X \le 1} \right);$$
4. Determine the mean value $$\mu$$ of $$X.$$

Solution.

This is a variation of the well-known Cauchy distribution.

1. We integrate the given $$PDF$$ and equate it to $$1:$$

${\int\limits_{ – \infty }^\infty {f\left( x \right)dx} = 1,}\;\; \Rightarrow {\int\limits_0^\infty {\frac{{kdx}}{{1 + {x^2}}}} = 1,}\;\; \Rightarrow {k\int\limits_0^\infty {\frac{{dx}}{{1 + {x^2}}}} = 1,}\;\; \Rightarrow {k\left. {\arctan x} \right|_0^\infty = 1,}\;\; \Rightarrow {k\left[ {\frac{\pi }{2} – 0} \right] = 1,}\;\; \Rightarrow {k = \frac{2}{\pi }.}$

Hence, the $$PDF$$ has the form
2. $f\left( x \right) = \begin{cases} {\frac{2}{\pi\left({1 + {x^2}}\right)}}, & \text{if } {x \ge 0} \\ 0, & \text{if } x \lt 0 \end{cases}.$

3. The $$PDF$$ is sketched in Figure below:
4. Figure 6.
5. Compute the probability $$P\left( {0 \le X \le 1} \right):$$
6. ${P\left( {0 \le X \le 1} \right) = \int\limits_0^1 {f\left( x \right)dx} }={ \frac{2}{\pi }\int\limits_0^1 {\frac{{dx}}{{1 + {x^2}}}} }={ \frac{2}{\pi }\left. {\arctan x} \right|_0^1 }={ \frac{2}{\pi }\left( {\frac{\pi }{4} – 0} \right) }={ \frac{1}{2}.}$

7. Find the mean value $$\mu:$$
8. ${\mu = \int\limits_{ – \infty }^\infty {xf\left( x \right)dx} }={ \frac{2}{\pi }\int\limits_0^\infty {\frac{{xdx}}{{1 + {x^2}}}} }={ \frac{1}{\pi }\int\limits_0^\infty {\frac{{d\left( {1 + {x^2}} \right)}}{{1 + {x^2}}}} }={ \frac{1}{\pi }\mathop {\lim }\limits_{b \to \infty } \int\limits_0^b {\frac{{d\left( {1 + {x^2}} \right)}}{{1 + {x^2}}}} }={ \frac{1}{\pi }\mathop {\lim }\limits_{b \to \infty } \left[ {\left. {\ln \left| {1 + {x^2}} \right|} \right|_0^\infty } \right] }={ \frac{1}{\pi }\mathop {\lim }\limits_{b \to \infty } \left[ {\ln \left( {1 + {b^2}} \right) – \ln 1} \right] }={ \infty .}$