\hypertarget{chapter-4}{%
\section{Chapter 4}\label{chapter-4}}

\hypertarget{expected-value}{%
\subsection{Expected Value}\label{expected-value}}

\hypertarget{definition}{%
\subsubsection{Definition}\label{definition}}

Let \(X\) be a random variable with probability distribution \(f(x)\).
The mean, or expected value, of \(X\) is:

For a discrete distribution \[E[X] = \sum\limits_x xf(x)\]

For a continuous distribution:
\[E[X] = \int\limits_{-\infty}^{\infty} xf(x)dx\]

Given \(\{1, 2, 3, 3, 5\}\), the mean is: \[{1+2+3+3+5 \over 5} = 2.8\]
\[f(x) = \begin{cases}
{1\over5} & x=1 \\
{1\over5} & x=2 \\
{2\over5} & x=3 \\
{1\over5} & x=5 \\
\end{cases}\]

\[\sum\limits_x xf(x) = {1\over5}(1) + {1\over5}(2) + {1\over5}(3) + {1\over5}(5) = 2.8\]

\hypertarget{example}{%
\subsubsection{Example}\label{example}}

The probability distribution of a discrete random variable \(X\) is:
\[f(x) = {3 \choose x}\left({1 \over 4}\right)^x\left({3\over4}\right)^{3-x}, x \in \{0, 1, 2, 3\}\]
Find \(E[X]\): \[f(x) =
\begin{cases}
0 & x=0 \\
0.422 & x=1 \\
0.14 & x=2 \\
{1\over64} & x=3
\end{cases}\]

\[E[X] = \sum\limits_x x {3 \choose x}\left({1\over4}\right)^x \left({3\over4}\right)^{3-x}\]
\[E[X] = 0(0)+ 0.422(1) + 0.14(2) + {1\over64}(3) = 0.75\]

\hypertarget{example-1}{%
\subsubsection{Example}\label{example-1}}

Let \(X\) be the random variable that denotes the life in hours of a
certain electronic device. The PDF is: \[f(x) =
\begin{cases}
{20000\over x^3} & x > 100 \\
0 & elsewhere
\end{cases}\]

Find the expected life of this type of device:
\[E[X] = \int\limits_{-\infty}^{\infty} xf(x)dx = \int\limits_{100}^{\infty}x{20000 \over x^3}dx = 200 \text{[hrs]}\]

\textbf{Note:} \[E[x^2] = \int\limits_{\infty}^{\infty}x^2f(x)dx\]

\hypertarget{properties-of-expectations}{%
\subsubsection{Properties of
Expectations}\label{properties-of-expectations}}

\[E(b) = b\] Where \(b\) is a constant \[E(aX) = aE[X]\] Where \(a\) is
a constant \[E(aX + b) aE[X] + b\] \[E[X + Y] = E[X] + E[Y]\] Where
\(X\) and \(Y\) are random variables

\hypertarget{example-2}{%
\subsubsection{Example}\label{example-2}}

Given: \[f(x) = \begin{cases}
{x^2\over3} & -1 < x < 2 \\
0 & \text{elsewhere}
\end{cases}\] Find the expected value of \(Y = 4X + 3\):

\[E[Y] = E[4X + 3] = 4E[X] + 3\]
\[E[X] = \int\limits_{-1}^{3} {X^3 \over 3}dx = {1\over12}X^4 \Big|_{-1}^{3}={5\over4}\]

\hypertarget{variance-of-a-random-variable}{%
\subsubsection{Variance of a Random
Variable}\label{variance-of-a-random-variable}}

The expected value/mean is of special importance because it describes
where the probability distribution is centered. However, we also need to
characterize the variance of the distribution.

\hypertarget{definition-1}{%
\subsubsection{Definition}\label{definition-1}}

Let \(X\) be a random variable with probability distribution, \(f(x)\),
and mean, \(\mu\). The variance of \(X\) is given by:
\[\text{Var}[X] = E[(X-\mu)^2]\] Which is the average squared distance
away from the mean. This simplifies to:
\[\text{Var}[X] = E[X^2] - E[X]^2\] \textbf{Note:} Generally,
\[E[X^2] \ne E[X]^2\]

The standard deviation, \(\sigma\), is given by:
\[\sigma = \sqrt{\text{Var}[X]}\]

\textbf{Note}: The variance is a measure of uncertainty (spread) in the
data.

\hypertarget{example-3}{%
\subsubsection{Example}\label{example-3}}

The weekly demand for a drinking water product in thousands of liters
from a local chain of efficiency stores is a continuous random variable,
\(X\), having the probability density: \[F(x) = \begin{cases}
2(x-1) & 1 < x < 2 \\
0 & \text{elsewhere}
\end{cases}\]

Find the expected value:
\[E[X] = \int\limits_1^2 x (2(x-1)) dx = 2\int\limits_1^2 (x^2 - x)dx\]
\[E[X] = 2\left[{1\over3}x^3 - {1\over2}x^2 \Big|_1^2 \right] = {5\over3}\]

Find the variance: \[\text{Var}[X] = E[X^2] - E[X]^2\]
\[E[X^2] = \int\limits_1^2 2x^2(x-1)dx = 2\int\limits_1^2 (x^3 - x^2)dx\]
\[E[X^2] = {17\over6}\]
\[\text{Var}[X] = {17\over6} - \left({5\over3}\right)^2 = {1\over18}\]

Find the standard deviation:
\[\sigma = \sqrt{\text{Var}[X]} = {1\over3\sqrt{2}} = {\sqrt{2}\over6}\]

\hypertarget{example-4}{%
\subsubsection{Example}\label{example-4}}

The mean and variance are useful when comparing two or more
distributions.

\begin{longtable}[]{@{}lll@{}}
\toprule()
& Plan 1 & Plan 2 \\
\midrule()
\endhead
Avg Score Improvement & \(+17\) & \(+15\) \\
Standard deviation & \(\pm8\) & \(\pm2\) \\
\bottomrule()
\end{longtable}

\hypertarget{theorem}{%
\subsubsection{Theorem}\label{theorem}}

If \(X\) has variance, \(\text{Var}[X]\), then
\(\text{Var}[aX + b] = a^2\text{Var}[X]\).

\hypertarget{example-5}{%
\subsubsection{Example}\label{example-5}}

The length of time, in minutes, for an airplane to obtain clearance at a
certain airport is a random variable, \(Y = 3X - 2\), where \(X\) has
the density: \[F(x) = \begin{cases}
{1\over4} e^{x/4} & x > 0 \\
0 & \text{elsewhere}
\end{cases}\]

\[E[X] = 4\] \[\text{Var}[X] = 16\]

Find \(E[Y]\): \[E[Y] = E[3X-2] = 3E[X] - 2 = 10\]
\[\text{Var}[Y] = 3^2\text{Var}[X] = 144\]
\[\sigma = \sqrt{\text{Var}[Y]} = 12\]

\hypertarget{the-exponential-distribution}{%
\subsection{The Exponential
Distribution}\label{the-exponential-distribution}}

The continuous random variable, \(X\), has an exponential distribution
with parameter \(\beta\) if its density function is given by:
\[F(x) = \begin{cases}
{1\over\beta}e^{-x/\beta} & x > 0 \\
0 & \text{elsewhere}
\end{cases}\]

Where \(\beta > 0\).

\[E[X] = \beta\]
\[E[X] = \int\limits_0^{\infty} x{1\over\beta}e^{-x/\beta} dx\]

Aside: \[\Gamma(Z) = \int\limits_0^\infty x^{Z - 1}e^{-x}dx\] Where
\(\Gamma(Z) = (Z - 1)!\)

\[E[X] = \beta \int\limits_0^\infty \left({x\over\beta}\right)^{(2-1)} e^{-x/\beta} \left({dx\over\beta}\right) = \beta\Gamma(2)\]
\[E[X] = \beta(2-1)! = \beta\]

\[\text{Var}[X] = E[X^2] - E[X]^2\]
\[E[X^2] = \int\limits_0^\infty x^2{1\over\beta}e^{-x/\beta}dx = \beta^2 \int\limits_0^\infty \left({x\over\beta}\right)^{(2-1)} e^{-x/\beta} \left({dx\over\beta}\right)\]
\[E[X^2] = \beta^2\Gamma(3) = 2\beta^2\]
\[\text{Var}[X] = 2\beta^2 - \beta^2 = \beta^2\]

\hypertarget{application}{%
\paragraph{Application}\label{application}}

Reliability analysis: the time to failure of a certain electronic
component can be modeled by an exponential distribution.

\hypertarget{example-6}{%
\subsubsection{Example}\label{example-6}}

Let \(T\) be the random variable which measures the time to failure of a
certain electronic component. Suppose \(T\) has an exponential
distribution with \(\beta = 5\).

\[F(x) = \begin{cases}
{1\over5}e^{-x/5} & x > 0 \\
0 & \text{elsewhere}
\end{cases}\]

If 6 of these components are in use, what is the probability that
exactly 3 components are still functioning at the end of 8 years?

What is the probability that an individual component is still
functioning after 8 years?

\[P(T > 8) = \int\limits_8^\infty {1\over5}e^{-x/5}dx \approx 0.2\]

\[{6 \choose 3}(0.2)^3(0.8)^3 = 0.08192\]

\begin{Shaded}
\begin{Highlighting}[]
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\FloatTok{0.08192000000000003}
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\end{Shaded}

\hypertarget{the-normal-distribution}{%
\subsection{The Normal Distribution}\label{the-normal-distribution}}

The most important continuous probability distribution in the field of
statistics is the normal distribution. It is characterized by 2
parameters, the mean, \(\mu\), and the variance, \(\sigma^2\).
\[\text{mean} = \text{median} = \text{mode}\]

\[F(x|\mu,\sigma^2) = {1 \over \sqrt{2\pi} \sigma^2} e^{\left({1 \over 2\sigma^2}(x-\mu)^2\right)}\]
\[E[X] = \mu\] \[\text{Var}[X] = \sigma^2\]

For a normal curve:
\[P(x_1 < x < x_2) = \int\limits_{x_1}^{x_2} F(x)dx\]

\hypertarget{definition-2}{%
\subsubsection{Definition}\label{definition-2}}

The distribution of a normal variable with mean 0 and variance 1 is
called a standard normal distribution.

The transformation of any random variable, \(X\) into a standard normal
variable, \(Z\): \[Z = {X - \mu \over \sigma}\]

\hypertarget{example-7}{%
\subsubsection{Example}\label{example-7}}

Given a normal distribution with mean \(\mu = 30\) and standard
deviation, \(\sigma = 6\), find the normal curve area to the right of
\(x = 17\).

Transform to standard normal. \[Z = {17 - 30 \over 6} = -2.16\]

That is, \(x = 17\) on a normal distribution with \(\mu = 30\) and
\(\sigma = 6\) is equivalent to \(Z=-2.16\) on a normal distribution
with \(\mu = 0\) and \(\sigma = 1\).

\[P(X > 17) = P(Z > -2.16)\]

\[P(Z > -2.16) = 1 -P(Z \le -2.16) = 0.9846\]

\begin{Shaded}
\begin{Highlighting}[]
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\OperatorTok{\textgreater{}\textgreater{}\textgreater{}}\NormalTok{ norm.cdf(}\OperatorTok{{-}}\FloatTok{2.16}\NormalTok{)}
\FloatTok{0.015386334783925445}
\end{Highlighting}
\end{Shaded}

\hypertarget{example-8}{%
\subsubsection{Example}\label{example-8}}

The finished inside diameter of a piston ring is normally distributed
with mean, \(\mu = 10\){[}cm{]}, and standard deviation,
\(\sigma = 0.03\){[}cm{]}.

What is the probability that a piston ring will have inside diameter
between 9.97{[}cm{]} and 10.03{[}cm{]}?

\[Z_1 = {9.97 - 10 \over 0.03} = -1\] \[Z_2 = {10.03 - 10 \over 3} = 1\]
\[P(9.97 < x < 10.03) = 0.68\]

\begin{Shaded}
\begin{Highlighting}[]
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\OperatorTok{\textgreater{}\textgreater{}\textgreater{}}\NormalTok{ norm.cdf(}\DecValTok{1}\NormalTok{) }\OperatorTok{{-}}\NormalTok{ norm.cdf(}\OperatorTok{{-}}\DecValTok{1}\NormalTok{)}
\FloatTok{0.6826894921370859}
\end{Highlighting}
\end{Shaded}
