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:
- Given \(F(x)=a\)
- Taking log: \(\ln F(x) = ln(a)\)
- Differentiate: \(\dfrac{d\ln F(x)}{dx}=\dfrac{d\ln a}{dx}=\dfrac{d F(x) / dx)}{F(x)}\)
- Replace the \(F(x)\) in the differentiated formula: \(\dfrac{d \boxed{F(x)}}{dx}=\boxed{F(x)}\times \dfrac{d\ln a}{dx}\)
- i.e. \(\dfrac{d \boxed{F(x)}}{dx}=\boxed{F(x)}(\dfrac{d\ln \boxed{F(x)}}{dx})\)
- Similarly, \(\dfrac{d \boxed{f(x)}}{dx}=\boxed{f(x)}(\dfrac{d\ln \boxed{f(x)}}{dx})\)
The pdf \(f(x)\) must satisfy:
- \(f(x)\ge0\text{ for all }x\)
- \(\int_{-\infty}^{\infty}{f(x)dx}=1\)
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:
- \(p(n)=cp(n-1)\)
- The mode is \(c\) when \(c<1\)
Or, by differentiating the PDF, the mode is calculated by:
- 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
| Distribution | CDF | \(E(X)\) | \(Var(X)\) | \(M_X(t)\) | Special Property | |
|
Discrete Uniform |
For \(0\leq x=1,2,…,n\), \(Pr(X=k)=\dfrac{1}{n}\) |
– | \(\dfrac{a+b}{2}\) | \(\dfrac{(b-a+1)^2-1}{12}\) | – | – |
|
Binomial |
For \(k=0,1,…n\), \(Pr(X=k)=\left( \begin{matrix} |
– | \(np\) | \(npq\) | \((pe^t+q)^n\) | – |
|
Hypergeometric |
For \(d=0,1,…D\), \(Pr(X=d)=\dfrac{\left( \begin{matrix} |
– | \(\dfrac{nD}{N}\) | \(n(\dfrac{D}{N})(\dfrac{N-D}{N})(\dfrac{N-n}{N-1})\) | – |
For sampling with replacement, \(X\sim Bin(n,p=\dfrac{D}{N}):
|
|
Trinomial |
For \(k_1+k_2+k_3=n\), \(Pr(X_1=k_1,X_2=k_2,X_3=k_3)\) \(=\dfrac{n!}{k_1!k_2!k_3!}p_1^{k_1}p_2^{k_2}p_3^{k_3}\) |
– | – | – | – | \(Cov(X_i,X_j)=-np_ip_j\) |
|
Negative Binomial |
For \(n=0,1,…\), \(Pr(N=r)=\left( \begin{matrix} |
– | \(\dfrac{r(1-p)}{p}\) | \(r(\dfrac{1-p}{p^2})\) | \((\dfrac{pe^t}{1-(1-p)e^t})^r\) | \(NB(r=1,p)\sim Geo(p)\) |
|
Geometric |
For \(n=0,1,…\), \(Pr(N=n)=p(1-p)^{n}\) |
\(S_N(n)\) \(=(1-p)^n\) |
\(\dfrac{1-p}{p}\) | \(\dfrac{1-p}{p^2}\) | \(\dfrac{pe^t}{1-(1-p)e^t}\) |
Geometric Series: \(\sum\limits_{n=1}^{\infty}{np^n}=\dfrac{p}{(1-p)^2}\) |
|
Poisson |
For \(n\ge 0\), \(Pr(N=n)=e^{-\lambda}\dfrac{\lambda^n}{n!}\) |
– | \(\lambda\) | \(\lambda\) | \(e^{\lambda(e^t-1)}\) | Sum of Poissons \(\sim Poisson(\lambda=\sum_{i=1}^{n}\lambda_i)\) |
|
Continuous Uniform |
For \(0\leq x\leq\theta\), \(f(x)=\dfrac{1}{\theta}\) |
\(F(x)\) \(=\dfrac{x}{\theta}\) |
\(\dfrac{\theta}{2}\) | \(\dfrac{\theta^{2}}{12}\) | – | – |
|
Exponential |
For \(x\ge 0\), \(f(x)=\dfrac{e^{-x/\theta}}{\theta}\) |
For \(x\ge 0\), \(F(x)\) \(=1-e^{-x/\theta}\) |
\(\theta\) | \(\theta^2\) | – |
1. \(E[X]=\theta\) => \(\int_{0}^{\infty}{x\dfrac{1}{\theta}e^{-x/\theta}}=\theta\) => \(\int_{0}^{\infty}{xe^{-x/\theta}}=\theta^2\) => \(\int_{0}^{\infty}{xe^{-ax}}=a^{-2}\) 2. \(E[X^2]=2\theta^2\) => \(\int_{0}^{\infty}{x^2\dfrac{1}{\theta}e^{-x/\theta}}=2\theta^2\) => \(\int_{0}^{\infty}{x^2e^{-x/\theta}}=2\theta^3\) => \(\int_{0}^{\infty}{x^2e^{-ax}}=2a^{-3}\) 3. Memoryless Property: \(X-x|X\ge x\) is exponential with the same distribution as X. 4. Sum of Exponential \(\sim Gamma(n,\theta)\), where \(f(x)=\dfrac{1}{\Gamma(n)\theta^n}x^{n-1}e^{-x/\theta}\) |
|
Normal |
\(f(x)=\phi(x)=\dfrac{e^{-(x-\mu)^2/2\sigma^2}}{\sigma \sqrt{2\pi}}\) | \(F_X(x)=\Phi(\dfrac{x-\mu}{\sigma})\) | \(\mu\) | \(\sigma^2\) | – |
1. \(\phi(-x)=\phi(x)\) => \(\Phi(x)=1-\Phi(-x)\) 2. \(\xi_p=\mu+\sigma z_p\) 3. If \(X\) is normal, then \(aX+b\) 4. If \(X\) and \(Y\) are independent and normal, then \(aX+bY+c\) 5. Linear Interpolation: \(f(x)=(\dfrac{x-x_1}{x_2-x_1})y_2+(\dfrac{x_2-x}{x_2-x_1})y_1\), \(x_1<x<x_2\) |
|
Bivariate Normal |
\(X\) and \(Y\) are bivariate normal if:
|
\(E[X|Y]=\mu_X+\dfrac{\rho\sigma_X}{\sigma_Y}(Y-\mu_Y)\) | \(Var(X|Y)=\sigma_X^2(1-\rho^2)\) | – |
1. If \(\sigma_{XY}=0\), X and Y a re independent. |
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.
-
- Obtain \(F_X(x)\)
- \(f_Y(y)=\dfrac{3!}{(2-1)!1!(3-1)!}F_X(x)^{2-1}f_X(x)S_X(x)^{3-2}\), 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
- The moment generating function generates moments. The nth derivative of the moment generating function evaluated at 0 is the nth raw moment:
\(M_X^{(n)}(t=0)=E[X^{n}]\) - 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
- It generates factorial moments; the nth derivative of the probability generating function evaluated at 1 is the nth factorial moment.:
\(\mu_{(n)}=P^{(n)}(1)=E[X(X-1)…(X-n+1)]\) - 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:
\(F_Y(y)=Pr(X\ge g^{-1}(y))=1-F_X(g^{-1}(y))\)
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:
- Add another target variable. For example, let \(W=X\)
- Determine the inverse transformation by expressing \(X\) and \(Y\) as functions of \(W\) and \(Z\).
- \(X=h_1(W,Z)\)
- \(Y=h_2(W,Z)\)
- Calculate the Jacobian of the inverse transformation.
Jacobian: \(J=\left| \begin{matrix}
\dfrac{\partial x}{\partial z} & \dfrac{\partial x}{\partial w} \\
\dfrac{\partial y}{\partial z} & \dfrac{\partial y}{\partial w} \\
\end{matrix} \right|\) - The joint density function of the transformed variables is:
\(f_{Z,W}(z, w)=f_{X,Y}(h_1(z, w), h_2(z, w))|J|\) - 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 \(z\).
- 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 \(x\):
-
-
- \(E[Y]=\int_{d}^{\infty}{(x-d)f_X(x)dx}=\int_{d}^{\infty}{(x-d)\theta^{-1}e^{-\theta^{-1}x}dx}=\int_{0}^{\infty}{x\theta^{-1}e^{-\theta^{-1}(x+d)}dx}\)
\(E[Y]=e^{-\theta^{-1}d}\int_{0}^{\infty}{x\theta^{-1}e^{-\theta^{-1}x}dx}=e^{-\theta^{-1}d}(\theta^{-1})\) - \(E[Y^2]=\int_{d}^{\infty}{(x-d)^2 f_X(x)dx}=\int_{d}^{\infty}{(x-d)^2 \theta^{-1}e^{-\theta^{-1}x}dx}=\int_{0}^{\infty}{x^2\theta^{-1}e^{-\theta^{-1}(x+d)}dx}\)
\(E[Y^2]=e^{-\theta^{-1}d}\int_{0}^{\infty}{x^2\theta^{-1}e^{-\theta^{-1}x}dx}=e^{-\theta^{-1}d}(2\theta^{-2})\)
- \(E[Y]=\int_{d}^{\infty}{(x-d)f_X(x)dx}=\int_{d}^{\infty}{(x-d)\theta^{-1}e^{-\theta^{-1}x}dx}=\int_{0}^{\infty}{x\theta^{-1}e^{-\theta^{-1}(x+d)}dx}\)
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