Basic Calculus

Integrations

• \(\dfrac{d}{dx}a^x=a^x\ln(a)\)
• \(\int{{{}\over{}}a^xdx}=\dfrac{a^x}{\ln(a)}\text{ for }a>0\)

 

Logarithmic Differentiation

\(\dfrac{df(x)}{dx}=f(x)(\dfrac{d\ln f(x)}{dx})\), since \(\dfrac{d\ln f(x)}{dx}=\dfrac{df(x)/dx}{f(x)}\)

 

Partial Fraction Decomposition

\(\int{{{x}\over{1+x}}dx}=\int{(1-{{1}\over{1+x}})dx}\)

 

Integration by Parts

\(\int{udv}=uv-\int{vdu}\)

Special Cases:

• • \(\int_{0}^{\infty }{xe^{-ax}dx}=\dfrac{1}{a^2}\), for \(a>0\)
  • \(\int_{0}^{\infty }{x^2e^{-ax}dx}=\dfrac{2}{a^3}\), for \(a>0\)

 

Sets

Set Properties

• Associative Property
  \((A\cup B)\cup C=A\cup (B\cup C)\) and \((A\cap B)\cap C=A\cap (B\cap C)\)
• Distributive Property
  \(A\cup (B\cap C)=(A\cup B)\cap (A\cup C))\) and \(A\cap (B\cup C)=(A\cap B)\cup (A\cap C)\)
• Distributive Property for Complement
  \((A\cup B)’=A’\cap B’\) and \((A\cap B)’=A’\cup B’\)

 

Basic Probability Relationships

• \(Pr[A\cup B]=Pr[A]+Pr[B]-Pr[A\cup B]\)
• \(Pr[A\cup B\cup C]=Pr[A]+Pr[B]+Pr[C]-Pr[A\cap B]-Pr[B\cap C]-Pr[A\cap C]+Pr[A\cap B\cap C]\)
• If A and B are independent, then:
  • \(Pr(A\cap B)=Pr(A)Pr(B)\)
  • \(Pr(A\cap B’)=Pr(A)Pr(B’)=Pr(A)(1-Pr(B))\)
  • \(Pr(A’\cap B)=Pr(A’)Pr(B)=(1-Pr(A))Pr(B)\)

 

Combinatorics

Number of Permutations

\(n!\)

 

Number of Distinct Permutations

\(_{n}P_k=\dfrac{n!}{(n-k)!}=\dfrac{n!}{{{k}_{1}}!{{k}_{2}}!\cdots {{k}_{j}}!}\)

 

Number of Combinations

\(_{n}C_k=\left( \begin{matrix}n\\k \\\end{matrix} \right)=\dfrac{n!}{k!(n-k)!}=\dfrac{n(n-1)\cdots (n-k+1)}{k!}\)

 

Conditional Probabilities

\(P[A|B]=\dfrac{P[A\cap B]}{P[B]}\)

 

Bayes’ Theorem

Law of Total Probabilities

\(P[A]=\sum\limits_{i=1}^{n}{P[B_i]P[A|B_i]}\)

 

Bayes’ Theorem

\(P[A_j|B]=\dfrac{P[A_j]P[B|A_j]}{\sum\limits_{i=1}^{n}{P[A_i]P[B|A_i]}}\)

 

Random Variables

Probability Mass Function (PMF)

\(p(x)=Pr(X=x)\)

 

Probability Density Function (PDF)

\(f(x)=\dfrac{dF(x)}{dx}\) or logarithmic differentiation:

1. Given \(F(x)=a\)
2. Taking log: \(\ln F(x) = ln(a)\)
3. Differentiate: \(\dfrac{d\ln F(x)}{dx}=\dfrac{d\ln a}{dx}=\dfrac{d F(x) / dx)}{F(x)}\)
4. Replace the \(F(x)\) in the differentiated formula: \(\dfrac{d \boxed{F(x)}}{dx}=\boxed{F(x)}\times \dfrac{d\ln a}{dx}\)
5. i.e. \(\dfrac{d \boxed{F(x)}}{dx}=\boxed{F(x)}(\dfrac{d\ln \boxed{F(x)}}{dx})\)
6. Similarly, \(\dfrac{d \boxed{f(x)}}{dx}=\boxed{f(x)}(\dfrac{d\ln \boxed{f(x)}}{dx})\)

The pdf \(f(x)\) must satisfy:

1. \(f(x)\ge0\text{ for all }x\)
2. \(\int_{-\infty}^{\infty}{f(x)dx}=1\)

 

Insurance Concepts

• Deductible: Amount \(d\) subtracted from each loss. Payment is \(max(X-d,0)\).

\(E[Y^L]=E(X-d)_+=\int_{d}^{\infty }{(x-d)f_X(x)dx}\) or \(=\int_{d}^{\infty }{S_X(x)}\)

• Benefit Limit: Highest amount \(u\) that will be paid. Payment is \(min(X,u)\).
• Coinsurance: Insurance pays \(c<1\) times the loss.
• Inflation: Losses increase from \(X\) to \((1+r)X\).

 

Percentiles

The \(100𝑝^\text{th}\) percentile is the smallest value of \(\pi_p\) where \(F_X(\pi_q) \ge p\).

 

Mode

Given PDF, the mode is calculated by:

1. \(p(n)=cp(n-1)\)
2. The mode is \(c\) when \(c<1\)

Or, by differentiating the PDF, the mode is calculated by:

1. The mode is \(x\) when \(\dfrac{d}{dx}f_X(x)=0\)

 

Conditional Probability for Random Variables

Let \(\gamma=X|\text{cond}\). \(\gamma\) is defined by its distribution function:

\(F_\gamma(y)=Pr(\gamma\le y|\text{cond})\)

 

Mean

The Mean of Discrete R.V.

\(E[X]=\sum{xp(x)}\)

 

The Mean of Continuous R.V.

\(E[X]=\int_{-\infty}^{\infty}{xf(x)dx}\)

\(E[X]=\int_{0}^{\infty}{S(x)dx}=\int_{0}^{\infty}{(1-F(x))dx}\)

\(E[g(X)]=\int_{-\infty}^{\infty}{g(x)f_X(x)dx}\)

\(E[g(X)]=\int_{0}^{\infty}{g'(x)S_X(x)dx}\) for domain \(x\ge 0\)

\(E[g(X)|j\le X\le k]=\dfrac{\int_{j}^{k}{g(x)f_X(x)dx}}{Pr(j\le X\le k)}\)

 

The Mean of Continuous R.V. and a Constant

\(E[\min(X,k)]=\int_{-\infty}^{k}{x(x)dx}+k(1-F(k))\)

If \(f(x)=0\) for \(x<0\), then

• \(E[X]=\int_{0}^{\infty}{(1-F(x))dx}\)
• \(E[\min(X,k)]=\int_{0}^{k}{(1-F(x))dx}\)

 

Variance and Other Moments

Variance

\(Var(X)=E[X^{2}]-\mu^{2}\)=\(E[(X-\mu)^2]\)

\(Var(aX+b)=a^2Var(X)\)

\(CV[X] = \dfrac{SD[X]}{E[X]}\)

 

Bernoulli Shortcut

If \(Pr(X=a)=1-p\) and \(Pr(X=b)=p\), then \(Var(X)=(b-a)^{2}p(1-p)\)

 

Coefficient of Variance: \(\dfrac{\sigma}{\mu}\)

 

Skewness: \(\dfrac{E[(X-\mu)^3]}{\sigma^3}\)

 

Kurtosis: \(\dfrac{E[(X-\mu)^4]}{\sigma^4}\)

 

Covariance and Correlation Coefficient

Covariance

\(Cov(𝑋,Y)=𝐸[(X-\mu_X)(Y-\mu_Y)]\)

\(Var(X+Y)=Var(X)+Var(Y)+2Cov(X,Y)\)

\(Var(aX+bY)=a^2Var(X)+b^2Var(Y)+2abCov(X,Y)\)

• \(Var(X-Y)=Var(X)+Var(Y)-2Cov(X,Y)\)

When \(a=1\), \(b=1\), \(Cov(XY)=E[XY]-E[X]E[Y]\)

 

Independence

Independence implies covariance of 0, but not conversely.

If two variables are independent, then the expected value of their product equals the product of their expectations.

\(E[XY]=E[X]E[Y]\)

 

Rules

• \(Cov(X,X)=Var(X)\)
• \(Cov(X,Y)=Cov(Y,X)\)
• \(Cov(aX,bY)=abCov(X,Y)\)
• \(Cov(X,aY+bZ)=aCov(X,Y)+bCov(X,Z)\)

 

Covariance Matrix

\(\sum =(\begin{matrix}
\sigma _{X}^{2} & {{\sigma }_{XY}} \\
{{\sigma }_{XY}} & \sigma _{Y}^{2} \\
\end{matrix})\)

 

Correlation Coefficient

\(\rho_{X,Y}=Corr(X,Y)=\dfrac{Cov(X,Y)}{\sqrt{Var(X)}\sqrt{Var(Y)}}\)

 

Joint Distribution

Joint Density Function

\(\Pr(a<X\leq b,c<Y\leq d)=\int_{a}^{b}{\int_{c}^{d}{f(x,y)dydx}}\)

 

Distributions

 

Marginal Distribution

The distributions of the individual variables are called marginal distributions.

\(f_X(x)=\int_{-\infty}^{\infty}{f(x,y)dy}\)

\(f_Y(y)=\int_{-\infty}^{\infty}{f(x,y)dx}\)

 

Independence

A necessary and sufficient condition for independence is that the joint density function can be expressed as the product of two functions \(f(x,y)=g(x)h(y)\) where

• \(g(x)\) is not a function of \(y\) and \(h(y)\) is not a function of \(x\), and in addition,
• the domain must be a rectangle.

 

Joint Moments

\(E[g(x,y)]=\int{\int{g(x,y)f(x,y)dydx}}\)

 

Independence

If the variables are independent, then \(E[g(X)h(Y)] = E[g(X)]E[h(Y)]\)

 

Conditional Distribution

Conditional Density

\(f_{X|Y}(x|y)=\dfrac{f(x,y)}{f_Y(y)}\)

 

Law of Total Probabilities

\(f_X(x)=\int_{-\infty}^{\infty}{f_{X|Y}(x|y)f_Y(y)dy}\)

 

Bayes’ Theorem

\(f_{X|Y}(x|y)=\dfrac{f_{X|Y}(x|y)f_Y(y)dy}{f_X(x)}=\dfrac{f_{X|Y}(x|y)f_Y(y)dy}{\int_{-\infty}^{\infty}{f_{Y|X}(y|w)f_X(w)dw}}\)

 

Conditional Moments

\(E[g(X)|Y=y]=\dfrac{\int_{-\infty}^{\infty}{g(x)f(x,y)dx}}{f_Y(y)}\)

\(E[g(X)|Y>y]=\dfrac{\int_{-\infty}^{\infty}{g(x)f(x,y)dx}}{S_Y(y)}\)

\(Var(X|Y>y)=\dfrac{E[X^2|Y>y]}{S_Y(y)}-(\dfrac{E[X|Y>y]}{S_Y(y)})^2\)

 

Double Expectation

\(E[g(X)]=E_Y[E_X[g(X)|Y]]\)

Specifically,

• \(E[X]=E_Y[E_X[X|Y]]\)
• \(E[X^2]=E_Y[E_X[X^2|Y]]\)

 

Conditional Variance

\(Var(X)=E[Var(X|Y)]+Var(E[X|Y])\)

 

Central Limit Theorem

The Central Limit Theorem says that:

If \(X_i,i=1,…,n\) are a set of independent identically distributed random variables not necessarily normal, then as \(n->\infty\), the distribution of the sample mean approaches a normal distribution.

Suppose the underlying distribution has mean \(\mu\) and variance \(\sigma^2\). Then:

• \(E[\bar{X}]=[\dfrac{\sum_{i=1}^{n}E[X_i]}{n}]=\dfrac{n\mu}{n}=\mu\)
• \(Var(\bar{X}]=Var(\dfrac{\sum_{i=1}^{n}E[X_i]}{n})=\dfrac{1}{n^2}\sum_{i=1}^{n}Var(X_i)=\dfrac{\sigma^2}{n}\)

 

Central Limit Theorem

Let \(\bar{X}\) be the sample mean of a random sample taken from a distribution with mean \(\mu\) and finite non-zero variance \(\sigma^2\). Then:

\(\dfrac{(\bar{X}-\mu)}{\sigma/\sqrt{n}}\) converges to a standard normal distribution (\(Z\sim N(0,1)\))

 

Order Statistics

Maximum & Minimum

\(X_{(1)}=min(X_1,X_2,…,X_n)\)

\(X_{(n)}=max(X_1,X_2,…,X_n)\)

For IID r.v’s:

For minimum, \(S_{X_{(1)}}(x)=[S_X(x)]^n\), and the pdf is \(f_{X_{(1)}}(x)=f_Y(x)=n(1-F_X(x))^{n-1}f_X(x)\)

For Maximum, \(F_{X_{(n)}}(x)=[F_X(x)]^n\), and the pdf is \(f_{X_{(n)}}(x)=f_Y(x)=n(F_X(x))^{n-1}f_X(x)\)

 

\(k^{th}\) Order Statistic

\(f_{Y_{(k)}}(x)=\dfrac{n!}{(k-1)!1!(n-k)!}F_X(x)^{k-1}f_X(x)(1-F_X(x))^{n-k}\)

For Uniform r.v. on \([0,\theta]\), \(E_{Y_{(k)}}(x)=\dfrac{k\theta}{n+1}\)

Median of a sample size of 3

Given \(f_X(x)\), we want \(f_Y(y)\), where \(Y\) is the median of the value from the sample of 3 randomly selected from the distribution.

1. 1. Obtain \(F_X(x)\)
   2. \(f_Y(y)=\dfrac{3!}{(2-1)!1!(3-1)!}F_X(x)^{2-1}f_X(x)S_X(x)^{3-2}\), which means
      • 1 sample is smaller than \(f_X(x)\), represented by \(F_X(x)\)
      • 1 sample is right at the value, represented by \(f_X(x)\)
      • 1 sample is larger than \(f_X(x)\), represented by \(S_X(x)=1-F_X(x)\)

 

Moment Generating Function

The moment generating function \(M_X(t)\) of a random variable X is defined by:

\(M_X(t)=E[e^{tX}]\)

 

MGF Features

1. The moment generating function generates moments. The n^th derivative of the moment generating function evaluated at 0 is the n^th raw moment:
   \(M_X^{(n)}(t=0)=E[X^{n}]\)
2. The moment generating function of the sum of independent random variables is the product of their moment generating functions:
   \(M_{X+Y}(t)=M_{X}(t)M_{Y}(t)\)

 

MGF for Common Distributions

• Gamma: \(M(t)=\dfrac{1}{(1-\theta t)^n}\)
• Normal: \(M(t)=e^{\mu t+\sigma^2t^2}\)

 

MGF for Two Variables

\(M_{X_1,X_2}(t_1,t_2)=E[e^{t_1 X_1+t_2 X_2}]\)

• \(\dfrac{\partial^n}{\partial t_i^n}M_{X_1,X_2}(0,0)=E[X_i^n]\)
• \(\dfrac{\partial^2}{\partial t_1 \partial t_2}M_{x_1,X_2}(t_1=0,t_2=0)=E[X_1X_2]\)

 

Probability Generating Function

The probability generating function \(P_N(z)\) of a random variable N is defined by:

\(P_N(z)=E[z^{N}]\)

 

PGF and Probability:

\(p_n=\dfrac{P^{(n)}(0)}{n!}\)

 

PGF and MGF

\(P(z)=E[z^N]=E[e^{N\ln z}]=M(\ln z)\)

 

PGF Features

1. It generates factorial moments; the n^th derivative of the probability generating function evaluated at 1 is the n^th factorial moment.:
   \(\mu_{(n)}=P^{(n)}(1)=E[X(X-1)…(X-n+1)]\)
2. It is multiplicative.
   \(P_{X+Y}(z)=P_X(z)P_Y(z)\)

 

Transformations

CDF

For a random variable \(X\) and another random variable that is a function of \(X\), \(Y=g(X)\), we may use CDF to determine the distribution of \(Y\):

\(F_Y(x)=Pr(Y\le x)=Pr(g(X)\le x)\)

If \(g(x)\) is a monotonically strictly increasing function, then it has an inverse function \(h(y)=g^{-1}(y)\) for which \(h(g(x))=x\):

\(F_Y(y)=Pr(X\le g^{-1}(y))=F_X(g^{-1}(y))\)

If \(g(x)\) is a monotonically strictly decreasing function, then it has an inverse function as well, and:

\(F_Y(y)=Pr(X\ge g^{-1}(y))=1-F_X(g^{-1}(y))+Pr(X=g^{-1})(y)\), where the last term may be dropped if \(X\) is a continuous random variable:

\(F_Y(y)=Pr(X\ge g^{-1}(y))=1-F_X(g^{-1}(y))\)

 

PDF

Assume \(g(x)\) is one-to-one, and let \(h(x)=g^{-1}(x)\). The chain rule requires multiplication by the absolute value of the derivative of \(h(x)\):

\(f_Y(x)=f_X(h(x))|h'(x)|\)

The ranges in which \(f_X(x)\) is nonzero must be transformed to the ranges in which \(f_Y(y)\) is nonzero by applying \(g(x)\) to those ranges.

For a many-to-one transformation, you must sum up densities for all values mapping to the region you are calculating the probability of.

 

Transformations of Two or More Variables

Given X and Y and the density function of \(Z=g(x,y)\) for example, then here are the steps:

1. Add another target variable. For example, let \(W=X\)
2. Determine the inverse transformation by expressing \(X\) and \(Y\) as functions of \(W\) and \(Z\).
   • \(X=h_1(W,Z)\)
   • \(Y=h_2(W,Z)\)
3. Calculate the Jacobian of the inverse transformation.
   Jacobian: \(J=\left| \begin{matrix}
   \dfrac{\partial x}{\partial z} & \dfrac{\partial x}{\partial w}  \\
   \dfrac{\partial y}{\partial z} & \dfrac{\partial y}{\partial w}  \\
   \end{matrix} \right|\)
4. The joint density function of the transformed variables is:
   \(f_{Z,W}(z, w)=f_{X,Y}(h_1(z, w), h_2(z, w))|J|\)
5. The domain of the variables must be transformed as well. Since we are interested in \(Z\), the domain would be expressed as a range of \(z\) and a range of was a function of \(z\).
6. Integrate \(f_{Z,W}(z,w)\) over \(w\) to obtain the marginal density function of \(Z\).

 

Life Insurance

For a(n) (ordinary) deductible \(d\),

Expected payment paid by policyholder:

\(E[X]=\int_{-\infty}^{d}{xf_X(x)dx}+\int_{d}^{\infty}{df_X(x)dx}=\int_{-\infty}^{d}{xf_X(x)dx}+dS_X(d)\)

Expected payment paid by insurer:

\(E[X]=\int_{d}^{\infty}{(x-d)f_X(x)dx}\)

Variance of payment paid by policyholder:

\(Var(Y)=E[Y^2]-E[Y]^2\), where

• • • \(E[Y]=\int_{-\infty}^{d}{xf_X(x)dx}+\int_{d}^{\infty}{df_X(x)dx}=\int_{-\infty}^{d}{xf_X(x)dx}+dS_X(d)\)
    • \(E[Y^2]=\int_{-\infty}^{d}{x^2 f_X(x)dx}+d^2 S_X(d)\)

Variance of payment paid by insurer:

\(Var(Y)=E[Y^2]-E[Y]^2\), where

• • • \(E[Y]=\int_{d}^{\infty}{(x-d)f_X(x)dx}\)
    • \(E[Y^2]=\int_{d}^{\infty}{(x-d)^2 f_X(x)dx}\)

For exponentially distributed loss amount \(x\):

• • • \(E[Y]=\int_{d}^{\infty}{(x-d)f_X(x)dx}=\int_{d}^{\infty}{(x-d)\theta^{-1}e^{-\theta^{-1}x}dx}=\int_{0}^{\infty}{x\theta^{-1}e^{-\theta^{-1}(x+d)}dx}\)
      \(E[Y]=e^{-\theta^{-1}d}\int_{0}^{\infty}{x\theta^{-1}e^{-\theta^{-1}x}dx}=e^{-\theta^{-1}d}(\theta^{-1})\)
    • \(E[Y^2]=\int_{d}^{\infty}{(x-d)^2 f_X(x)dx}=\int_{d}^{\infty}{(x-d)^2 \theta^{-1}e^{-\theta^{-1}x}dx}=\int_{0}^{\infty}{x^2\theta^{-1}e^{-\theta^{-1}(x+d)}dx}\)
      \(E[Y^2]=e^{-\theta^{-1}d}\int_{0}^{\infty}{x^2\theta^{-1}e^{-\theta^{-1}x}dx}=e^{-\theta^{-1}d}(2\theta^{-2})\)