Calculus

Integration of Functions

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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 \(x = b.\)

    Probability that a random variable X takes on values in the interval [a,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 \(\large{\frac{1}{2}}\normalsize:\)

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

    Median of a probability distribution
    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

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

    Uniform probability distribution on the interval [a,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.

    Exponential probability density distribution for different rates lambda.
    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.

    Normal distribution for different values of the mean and variance.
    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 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. Probability density function f(x) = 2/(pi*(1+x^2)) for non-negative x
      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. \[{\mu = \int\limits_{ – \infty }^\infty {xf\left( x \right)dx} }={ \frac{2}{\pi }\int\limits_0^\infty {\frac{{xdx}}{{1 + {x^2}}}} }={ \frac{1}{\pi }\int\limits_0^\infty {\frac{{d\left( {1 + {x^2}} \right)}}{{1 + {x^2}}}} }={ \frac{1}{\pi }\mathop {\lim }\limits_{b \to \infty } \int\limits_0^b {\frac{{d\left( {1 + {x^2}} \right)}}{{1 + {x^2}}}} }={ \frac{1}{\pi }\mathop {\lim }\limits_{b \to \infty } \left[ {\left. {\ln \left| {1 + {x^2}} \right|} \right|_0^\infty } \right] }={ \frac{1}{\pi }\mathop {\lim }\limits_{b \to \infty } \left[ {\ln \left( {1 + {b^2}} \right) – \ln 1} \right] }={ \infty .}\]