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 \(x = b.\)

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 \(\large{\frac{1}{2}}\normalsize:\)

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

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

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

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.

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.

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

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

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Solved Problems

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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

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

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:

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

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 \(m:\)

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

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

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

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,

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

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 \(120.\)

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:

\[{\mu = \int\limits_0^{10} {xf\left( x \right)dx} }={ \int\limits_0^{10} {\frac{{{x^2}}}{{50}}dx} }={ \frac{1}{{50}}\int\limits_0^{10} {{x^2}dx} }={ \frac{1}{{50}}\left. {\frac{{{x^3}}}{3}} \right|_0^{10} }={ \frac{{1000}}{{150}} }={ \frac{{20}}{3}.}\]

3. Find the probability \(P\left( {2 \le X \le 5} \right):\)

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

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:

\[{\mu = \int\limits_0^3 {xf\left( x \right)dx} }={ \int\limits_0^3 {\frac{{{x^3}}}{9}dx} }={ \frac{1}{9}\int\limits_0^3 {{x^3}dx} }={ \frac{1}{9}\left. {\frac{{{x^4}}}{4}} \right|_0^3 }={ \frac{{81}}{{36}} }={ \frac{9}{4}.}\]

3. The probability \(P\left( {1 \le X \le 2} \right)\) is also determined through integration:

\[{P\left( {1 \le X \le 2} \right) = \int\limits_1^2 {f\left( x \right)dx} }={ \frac{1}{9}\int\limits_1^2 {{x^2}dx} }={ \left. {\frac{{{x^3}}}{{27}}} \right|_1^2 }={ \frac{1}{{27}}\left( {8 – 1} \right) }={ \frac{7}{{27}} \approx 0.26}\]

Solution.

1. We determine the value of \(k\) from the condition

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

   Then

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

2. 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.

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

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

2. The \(PDF\) is sketched in Figure below:



3. Compute the probability \(P\left( {0 \le X \le 1} \right):\)

\[{P\left( {0 \le X \le 1} \right) = \int\limits_0^1 {f\left( x \right)dx} }={ \frac{2}{\pi }\int\limits_0^1 {\frac{{dx}}{{1 + {x^2}}}} }={ \frac{2}{\pi }\left. {\arctan x} \right|_0^1 }={ \frac{2}{\pi }\left( {\frac{\pi }{4} – 0} \right) }={ \frac{1}{2}.}\]

4. Find the mean value \(\mu:\)

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