SOA Exam P Study Notes

考试P is a three–hour multiple–choice examination designed to test your knowledge of fundamental probability tools using for assessing risk. AnalystPrep has developped concise study notes focusing on exactly the learning objectives tested in the Society of Actuaries exam.

There are three main topics in the exam and each topics has a multitude of different learning objectives. With AnalystPrep’s concise study notes for Exam P, you can simply read on your tablet, computer, or print each concept before jumping into the question bank portion of the platform.

Thevariance of a discrete random variableis the sum of the square of all the values the variable can take times the probability of that value occurring minus the sum of all the values the variable can take times the probability of that value occurring squared as shown in the formula below:
$$ Var\left( X \right) =\sum { { x }^{ 2 }p\left( x \right) -{ \left[ \sum { { x }p\left( x \right) } \right] }^{ 2 } } $$

Or written in another way, as a function of \(E(X)\), then:
$$ Var\left( X \right) =E\left( { X }^{ 2 } \right) -E{ \left( X \right) }^{ 2 } $$
Often \(E(X)\) is written as \(\mu\) and therefore variance can also be shown as in the formula below:
$$ Var\left( X \right) =\sum { { \left( x-\mu \right) }^{ 2 }p\left( x \right) } $$

考试ple
Given the experiment of rolling a single die, calculate \(Var(X)\).

\( E(X) = 1 \ast ({1}/{6})+2 \ast ({1}/{6})+ 3 \ast ({1}/{6})+ 4 \ast ({1}/{6})+ 5 \ast ({1}/{6})+ 6 \ast ({1}/{6}) = 3.5 \)

\( E(X^2) = 1 \ast ({1}/{6})+2^2 \ast ({1}/{6})+ 3^2 \ast ({1}/{6})+ 4^2 \ast ({1}/{6})+ 5^2 \ast ({1}/{6})+ 6^2 \ast ({1}/{6}) = {91}/{6} \)

\( Var \left(X\right) = \left(91/6 \right) – \left(3.5 \right)^2 = {35}/{12} = 2.92 \)

We could also calculate \(Var(X)\) as:

\( E \left(X \right) = \mu =3.5 \)
\begin{align*}
Var \left(X \right) = & (1-3.5)^2 \ast (1/6) + (2-3.5)^2 \ast (1/6)+(3-3.5)^2 \ast (1/6)+(4-3.5)^2 \ast (1/6)+ \\
& (5-3.5)^2 \ast (1/6)+(6-3.5)^2 \ast (1/6) = 2.92 \\
\end{align*}

Thevariance of a continuous random variableis shown in the formula below:
$ $ Var \左(X \右)= \ int _ {- \ infty} ^ {\ infty}{ { x }^{ 2 }f\left( x \right) dx } -{ \left[ \int _{ -\infty }^{ \infty }{ { x }f\left( x \right) dx } \right] }^{ 2 } $$
where \(f(x)\) is the probability density function of \(x\).

As in the discrete case, it can also be written as:
$$ Var\left( X \right) =E\left( { X }^{ 2 } \right) -E{ \left( X \right) }^{ 2 } $$

And also,
$ $ Var \左(X \右)= \ int _ {- \ infty} ^ {\ infty}{ { \left( x-\mu \right) }^{ 2 }f\left( x \right) dx } $$

考试ple
Given the following probability density function of a continuous random variable:
$$ f\left( x \right) =\begin{cases} \frac { x }{ 2 } , & 0 < x < 2 \\ 0, & otherwise \end{cases} $$
计算\ (Var (X) \)。
$$
\begin{align*}
& E\left( X \right) =\int _{ -\infty }^{ \infty }{ xf\left( x \right) dx= } \int _{ 0 }^{ 2 }{ x\ast \frac { x }{ 2 } \ast dx } ={ \left[ \frac { { x }^{ 3 } }{ 6 } \right] }_{ x=0 }^{ x=2 }=\frac { 8 }{ 6 } =\frac { 4 }{ 3 } \\
& E\left( X^2 \right) =\int _{ -\infty }^{ \infty }{ x^2 f\left( x \right) dx= } \int _{ 0 }^{ 2 }{ x^2 \ast \frac { x }{ 2 } \ast dx } ={ \left[ \frac { { x }^{ 4 } }{ 8 } \right] }_{ x=0 }^{ x=2 }=2 \\
& Var \left(X \right) = 2 – {\left({4}/{3}\right)}^ 2 = {2}/{9} \\
\end{align*}
$$
Or,
$ $ Var \左(X \右)= \ int _ {- \ infty} ^ {\ infty}{ { \left( x-\mu \right) }^{ 2 }f\left( x \right) dx= } \int _{ 0 }^{ 2 }{ { \left( x-\frac { 4 }{ 3 } \right) }^{ 2 }\ast \frac { x }{ 2 } \ast dx } ={ \left[ \frac { { x }^{ 4 } }{ 8 } -\frac { 4 }{ 9 } { x }^{ 3 }+\frac { 4 }{ 9 } { x }^{ 2 } \right] }_{ x=0 }^{ x=2 }=\frac { 2 }{ 9 } $$

Thestandard deviation, often written as \(\sigma\) , of either a discrete or continuous random variable can be defined as:
$$ S.D.\left( X \right) =\sigma =\sqrt { Var\left( X \right) } $$
Thecoefficient of variationof a random variable can be defined as the standard deviation divided by the mean (or expected value) of \(X\) as shown in the formula below:
$$ C.V.= {\frac {\sigma}{\mu}} $$

The variance of \(X\) is sometimes often referred to as thesecond momentof \(X\) about the mean.

The third moment of \(X\) is referred to as theskewnessand the fourth moment is calledkurtosis.

In general, themth moment of \(X\) can be calculated from the following formula:
$$ mth\quad moment\left( X \right) =\int _{ -\infty }^{ \infty }{ { \left( x-\mu \right) }^{ m }f\left( x \right)dx } $$

考试ple
Given the following probability density function of a continuous random variable:
$$ f\left( x \right) =\begin{cases} \frac { x }{ 2 } , & 0 < x < 2 \\ 0, & otherwise \end{cases} $$
Calculate the skewness.
\( \begin{align*}
\mu & =\int _{ -\infty }^{ \infty }{ xf\left( x \right) dx= } \int _{ 0 }^{ 2 }{ x\ast \frac { x }{ 2 } \ast dx } ={ \left[ \frac { { x }^{ 3 } }{ 6 } \right] }_{ x=0 }^{ x=2 }=\frac { 8 }{ 6 } =\frac { 4 }{ 3 } \\
Skew\left( X \right) & =\int _{ -\infty }^{ \infty }{ { \left( x-\mu \right) }^{ 3 }f\left( x \right) dx= } \int _{ 0 }^{ 2 }{ { \left( x-\frac { 4 }{ 3 } \right) }^{ 3 }\ast \frac { x }{ 2 } \ast dx } \\
& =\int _{ 0 }^{ 2 }{ \left( \frac { { x }^{ 4 } }{ 2 } -2{ x }^{ 3 }+\frac { 8{ x }^{ 2 } }{ 3 } -\frac { 32x }{ 27 } \right) dx={ \left[ \frac { { x }^{ 5 } }{ 10 } -\frac { { x }^{ 4 } }{ 2 } +\frac { 8{ x }^{ 3 } }{ 9 } -\frac { 16{ x }^{ 2 } }{ 27 } \right] }_{ x=0 }^{ x=2 }=-{ 8 }/{ 135 } }
\end{align*}
\)

Binomial (and Bernoulli)

一个实验概率的往下执行cess, \(p\), and failure, \(1-p\), and the experiment is performed \(n\) times.
$$
\begin{align*}
& p\left( x \right) =\left( \begin{matrix} n \\ x \end{matrix} \right) { p }^{ X }{ \left( 1-p \right) }^{ n-x } \\
& E\left( X \right) =np \\
& Var\left( X \right) =np\left( 1-p \right) \\
\end{align*}
$$
A Bernoulli random variable is the special case of a binomial random variable where the experiment is performed only once.

Negative Binomial
Given an experiment that is performed \(X\) number of times until a total of \(r\) successes occur, then
$$
\begin{align*}
& p\left( x \right) =\left( \begin{matrix} x-1 \\ r-1 \end{matrix} \right) { \left( 1-p \right) }^{ x-r }{ p }^{ r } \\
& E\left( X \right) =\frac { r }{ p } \\
& Var\left( X \right) ={ r\left( 1-p \right) }/{ p^{ 2 } } \\
\end{align*}
$$

Geometric
Given an experiment that is performed \(X\) number of times until a success occurs, with the probability of a success on each trial being equal to \(p\), then
$$ \begin{align*}
& p\left(x \right)=\left(1-p \right)^\left(x-1\right) p \\
& E \left(X \right)={1}/{p} \\
& Var\left(X \right)={\left(1-p \right)}/{p^2} \\
\end{align*}
$$

Hypergeometric
$$ \begin{align*}
& p\left( x \right) =\frac { \left( \begin{matrix} M \\ x \end{matrix} \right) \left( \begin{matrix} N-M \\ n-x \end{matrix} \right) }{ \left( \begin{matrix} N \\ n \end{matrix} \right) } \\
& E\left(X \right)={\frac {nm}{N}} \\
& Var \left(X \right)={nm \left(N-M \right)\left(N-n\right)}/{N^2 \left(N-1 \right)} \\
\end{align*}
$$

Poisson
A Poisson random variable can be described as the number of events occurring in a fixed time period if the events occur at a known constant rate,\(\lambda\).
$$ \begin{align*}
& p \left(x \right)={\frac {{e}^{-\lambda} {\lambda}^{x}}{x!}} \\
& E(X) = \lambda \\
& Var \left(X \right) = \lambda \\
\end{align*}
$$

Uniform – Discrete
Given an experiment in which outcomes are all equally likely
$$ \begin{align*}
& p \left(x \right)={\frac {1}{b-a+1}} \\
& E \left(X \right)={\frac {b+a}{2}} \\
& Var \left(X \right)={\frac {\left(b-a+2 \right)\left(b-a\right)}{12}} \\
\end{align*}
$$

Uniform – Continuous
$$ \begin{align*}
& f\left(x \right)={\frac {1}{b-a}} \\
& E\left(X \right)={\frac {b+a}{2}} \\
& Var \left(X \right)={\frac {{\left(b-a \right)}^2}{12}} \\
\end{align*}
$$

Exponential
$$ \begin{align*}
& f\left(x \right)= {{\lambda}e}^{-{\lambda} x} \\
& E\left(X \right)={\frac {1}{\lambda}} \\
& Var \left(X \right)={\frac {1}{{\lambda}^2}} \\
\end{align*}
$$

Gamma
$$ \begin{align*}
& f\left( x \right) =\frac { {{\lambda}e}^{-{\lambda} x}{ \left( {\lambda} x \right) }^{ \alpha -1 } }{ \Gamma \left( \alpha \right) } \\
& where\quad { \Gamma \left( \alpha \right) }=\int _{ 0 }^{ \infty }{ { e }^{ -y }{ y }^{ \alpha -1 }dy } \\
& E \left(X \right)={\frac {\alpha}{\lambda}} \\
& Var \left(X \right)={\frac {\alpha}{\lambda^2} } \\
\end{align*}
$$

Normal
$$ \begin{align*}
& f\left( x \right) =\frac { 1 }{ \sigma \sqrt { 2\pi } } { e }^{ -.5{ \left( \frac { x-\mu }{ \sigma } \right) }^{ 2 } }
\\
& E \left(X \right)= \mu \\
& Var \left(X\right)= \sigma \\
\end{align*}
$$
Thestandard normaldistribution has \(E(X) = 0\) and \(Var(X) = 1\).