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CExamCredibility

Actuary.CExamCredibility History

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October 10, 2011, at 07:19 PM by 38.106.150.109 -
Added line 94:
->We can sometimes avoid evaluating the integral in b) by looking at the numerator in c), and see if it's a known distribution; then b) must be the appropriate constant that makes c) a density function.
October 10, 2011, at 07:15 PM by 38.106.150.109 -
Changed line 88 from:
->a)  Find the joint density function:
to:
->a)  Find the joint density function, by plugging in the observed data {$x_j$}:
October 09, 2011, at 08:31 PM by 38.106.150.109 -
Changed line 118 from:
->{$k$} is called the ''Buhlmann credibility factor''.
to:
->{$Z$} is called the ''Buhlmann credibility factor''.
October 05, 2011, at 08:35 PM by 38.106.150.109 -
Changed lines 131-132 from:
->Then {$k\=v/a$}, {$Z=m/m+k$}, and the credibility estimate for {$X_{n+1}$} (# of claims in year {$n+1$} by one insured) is {$Z\overline{X} + (1-Z)\mu$}, so the number of claims in year {$n+1$} is that multiplied by {$m_{n+1}$}.
to:
->Then {$k=v/a$}, {$Z=m/m+k$}, and the credibility estimate for {$X_{n+1}$} (# of claims in year {$n+1$} by one insured) is {$Z\overline{X} + (1-Z)\mu$}, so the number of claims in year {$n+1$} is that multiplied by {$m_{n+1}$}.
October 05, 2011, at 08:32 PM by 38.106.150.109 -
Changed lines 125-132 from:
to:
->The questions seem to come in this flavor:  Suppose in year {$j$} there are {$N_j$} claims from {$m_j$} policies, for {$j = 1, \ldots, n$}.  The individual policy has the Poisson distribution with parameter {$\Theta$}, and the parameter itself has the gamma distribution with parameters {$\alpha$} and {$\beta$}.  Estimate the number of claims in year {$n+1$} if there will be {$m_{n+1}$} policies.

->Here we let {$X_j=N_j/m_j$}.  The {$N_j$} claims from {$m_j$} policies are given in the data of the problem, so we can compute {$X_j=N_j/m_j$} and {$\overline{X}$} by average the {$X_j$}'s. 
->The individual policy {$X_j$} has Poisson distribution, so {$E(X_j | \Theta) = \Theta$}, and {$Var(X_j |\Theta = \Theta$}, and
{$$\mu = E(\Theta) = \alpha\beta, \qquad v=E(\Theta) =\alpha\beta, \qquad a = Var(\Theta) = \alpha\beta^2.$$}
->[Expectation and variance are taken with respect to the Gamma distribution.]
->Then {$k\=v/a$}, {$Z=m/m+k$}, and the credibility estimate for {$X_{n+1}$} (# of claims in year {$n+1$} by one insured) is {$Z\overline{X} + (1-Z)\mu$}, so the number of claims in year {$n+1$} is that multiplied by {$m_{n+1}$}.

October 04, 2011, at 09:52 PM by 38.106.150.109 -
October 04, 2011, at 09:45 PM by 38.106.150.109 -
Changed line 144 from:
-->3) Then {$\mu$} can be estimated by {$\mu = 1/r \sum_{i=1}^r \overline(X}_{i}.$}
to:
-->3) Then {$\mu$} can be estimated by {$\mu = 1/r \sum_{i=1}^r \overline{X}_{i}.$}
Changed line 150 from:
-->so that {$a=Var(\overline{X)_i)-v/n$}.
to:
-->so that {$a=Var(\overline{X}_i)-v/n$}.
October 04, 2011, at 09:45 PM by 38.106.150.109 -
Changed lines 143-144 from:
-->2) Then the old quantity {$\mu(\theta)$} can be estimated by {$\overline{X)_i = 1/n \sum_{j=1}^n X_{i,j}.$}
-->3) Then {$\mu$} can be estimated by {$\mu = \1/r \sum_{i=1}^r \overline{X}_{i}.$}
to:
-->2) Then the old quantity {$\mu(\theta)$} can be estimated by {$\overline(X)_i = 1/n \sum_{j=1}^n X_{i,j}.$}
-->3) Then {$\mu$} can be estimated by {$\mu = 1/r \sum_{i=1}^r \overline(X}_{i}.$}
Changed line 146 from:
-->5) So {$v$} is estimated by  {$v = \1/r \sum_{i=1}^r v_{i}.$}
to:
-->5) So {$v$} is estimated by  {$v = 1/r \sum_{i=1}^r v_{i}.$}
Changed lines 149-150 from:
{$$Var(\overline{X}_i) = Var(E(\overline{X}_i)|\Theta) + E(Var(\overline(X)_i|\Theta)) = \cdots = a + \frac{v}{n}, $$}
so that {$a=Var(\overline{X)_i)-v/n$}.
to:
{$$Var(\overline{X}_i) = Var(E(\overline{X}_i)|\Theta) + E(Var(\overline{X}_i|\Theta)) = \cdots = a + \frac{v}{n}, $$}
-->so that {$a=Var(\overline{X)_i)-v/n$}.
October 04, 2011, at 09:43 PM by 38.106.150.109 -
Changed line 140 from:
->a) Nonparametric estimation:  Goal is to esitimate  {$\mu, v, a$} based on observations.
to:
->a) Nonparametric estimation:  Goal is to estiimate  {$\mu, v, a$} based on observations.
Changed line 149 from:
{$$Var(\overline{X}_i) = Var(E(\overline{X}_i)|\Theta) + E(Var(\overlin(X)_i|\Theta)) = \cdots = a + \frac{v}{n}, $$}
to:
{$$Var(\overline{X}_i) = Var(E(\overline{X}_i)|\Theta) + E(Var(\overline(X)_i|\Theta)) = \cdots = a + \frac{v}{n}, $$}
Changed line 151 from:
-->8) Then {$k=v/a$}, {$Z = n/(n+k)$}, and the credibility premium is {$Z\overline{X}_i + (1-Z)\mu$}, for each policy holder {$i = 1, \ldodts, r$}.  All these quantities are estimates.
to:
-->8) Then {$k=v/a$}, {$Z = n/(n+k)$}, and the credibility premium is {$Z\overline{X}_i + (1-Z)\mu$}, for each policy holder {$i = 1, \ldots, r$}.  All these quantities are estimates.
October 04, 2011, at 09:42 PM by 38.106.150.109 -
October 04, 2011, at 09:42 PM by 38.106.150.109 -
Changed lines 142-143 from:
-->1) Observations are given for {$r$} policy holders.  The losses of each policy holder are described by {$\vec{X}_{i}=X_{i,1}, \cdots, X_{i,n_i}$}, for {$i = 1,\ldots, r$}.
-->2)
to:
-->1) Observations are given for {$r$} policy holders.  The losses of each policy holder are described by {$\vec{X}_{i}=X_{i,1}, \cdots, X_{i,n_i}$}, for {$i = 1,\ldots, r$}.  Suppose for simplicity that all {$n_i$}s are the same and equal {$n$}.
-->2) Then the old quantity {$\mu(\theta)$} can be estimated by {$\overline{X)_i = 1/n \sum_{j=1}^n X_{i,j}.$}
-->3) Then {$\mu$} can be estimated by {$\mu = \1/r \sum_{i=1}^r \overline{X}_{i}.$}
-->4) Also {$v(\theta)$} can be estimated by {$v_i = 1/(n-1) sum_{j=1}^n \left(X_{i,j}-\overline{X}_i \right)^2$}.
-->5) So {$v$} is estimated by  {$v = \1/r \sum_{i=1}^r v_{i}.$}
-->6) For {$a$}, we first estimate {$Var(\overline{X}_i)$} by {$ 1/(r-1) sum_{i=1}^r \left(X_{i}-\mu \right)^2$}.
-->7) Then note that
{$$Var(\overline{X}_i) = Var(E(\overline{X}_i)|\Theta) + E(Var(\overlin(X)_i|\Theta)) = \cdots = a + \frac{v}{n}, $$}
so that {$a=Var(\overline{X)_i)-v/n$}.
-->8) Then {$k=v/a$}, {$Z = n/(n+k)$}, and the credibility premium is {$Z\overline{X}_i + (1-Z)\mu$}, for each policy holder {$i = 1, \ldodts, r$}.  All these quantities are estimates.
October 04, 2011, at 08:42 PM by 38.106.150.109 -
Added lines 140-143:
->a) Nonparametric estimation:  Goal is to esitimate  {$\mu, v, a$} based on observations.
->steps:
-->1) Observations are given for {$r$} policy holders.  The losses of each policy holder are described by {$\vec{X}_{i}=X_{i,1}, \cdots, X_{i,n_i}$}, for {$i = 1,\ldots, r$}.
-->2)
October 03, 2011, at 08:15 PM by 38.106.150.109 -
October 02, 2011, at 08:41 PM by 38.106.150.109 -
Changed lines 136-139 from:
->In practice, we might not have a given pdfs for {$\pi$} and {$f_{X_j}$}'s, so we'd have to estimate those.
to:
->In practice, we might not have a given pdfs for {$\pi$} and {$f_{X_j}$}'s, so we'd have to estimate those.  Two kinds of situations:
->a) Nonparametric:  where {$\pi$} and {$f_{X_j}$}'s are unspecified- like in Buhlmann models where only the first two moments are needed.
->b) Semiparametric:  where {$f_{X_j}$}'s  are in parametric form, but not  {$\pi$}.

October 02, 2011, at 06:13 PM by 38.106.150.109 -
Changed line 131 from:
->Then we can show that the posterior distribution is {$\sim Gamma(\alpha+ \sum_j x_j, \beta + n$}, so the expected value of the posterior distribution is
to:
->Then we can show that the posterior distribution is {$\sim Gamma(\alpha+ \sum_j x_j, \beta + n)$}, so the expected value of the posterior distribution is
Changed line 136 from:
->In practice, we might not have a given pdfs for {$\pi$} and {$f_{X_j}$}'s, so we'd have to estimate those.
to:
->In practice, we might not have a given pdfs for {$\pi$} and {$f_{X_j}$}'s, so we'd have to estimate those.
October 02, 2011, at 03:41 PM by 38.106.150.109 -
Changed line 129 from:
->Prior distribution {$\Theta \sim Gamma(\alpha,\beta'), with {$\beta' = 1/\theta$} where the {$\theta$} is the one in the appendix.
to:
->Prior distribution {$\Theta \sim Gamma(\alpha,\beta')$}, with {$\beta' = 1/\theta$} where the {$\theta$} is the one in the appendix.
Changed lines 135-136 from:
5. Apply empirical Bayesian methods in the nonparametric and semiparametric cases.
to:
5. Apply empirical Bayesian methods in the nonparametric and semiparametric cases.
->In practice, we might not have a given pdfs for {$\pi$} and {$f_{X_j}$}'s, so we'd have to estimate those
.
Changed line 106 from:
->If the Bayesian premium is difficult to evaluate, an alternative is to alculate the ''credibility premium'' instead, that is, find {$\alpha_0, \ldots, \alpha_n$} such that the sum ${Y=\alpha_0+\sum_{j=1}^n \alpha_j X_j$} minimizes the squared error loss:
to:
->If the Bayesian premium is difficult to evaluate, an alternative is to alculate the ''credibility premium'' instead, that is, find {$\alpha_0, \ldots, \alpha_n$} such that the sum {$Y=\alpha_0+\sum_{j=1}^n \alpha_j X_j$} minimizes the squared error loss:
Changed line 132 from:
{$$\frac{\alpha+sum_j x_j}{\beta'+n}.$$}
to:
{$$\frac{\alpha+\sum_j x_j}{\beta'+n}.$$}
Changed line 129 from:
->Prior distribution {$\Theta \sim Gamma(\alpha,\beta)$}.
to:
->Prior distribution {$\Theta \sim Gamma(\alpha,\beta'), with {$\beta' = 1/\theta$} where the {$\theta$} is the one in the appendix.
Changed lines 131-132 from:
->Then we can show that the posterior distribution is {$\sim Gamma(\alpha+ \sum_j x_j, 1/\beta+n)$}.
to:
->Then we can show that the posterior distribution is {$\sim Gamma(\alpha+ \sum_j x_j, \beta + n$}, so the expected value of the posterior distribution is
{$$\frac{\alpha+sum_j x_j}{\beta'+n}.$$}
->In the situation I.a., this is ({$\alpha$} + # of claims over {$n$} years) divided by ({$\beta'$} + # of {$n$} years that we've been observing)
.
Changed lines 131-132 from:
->Then we can show that the posterior distribution is {$\sim Gamma(\alpha, \beta)$}.
to:
->Then we can show that the posterior distribution is {$\sim Gamma(\alpha+ \sum_j x_j, 1/\beta+n)$}.
Changed lines 130-131 from:
->The observation
to:
->The observations {$X_j \sim Poisson(\Theta).$}
->Then we can show that the posterior distribution is {$\sim Gamma(\alpha, \beta)$}.

Changed line 113 from:
{$$\mu(\theta) = E( X_j |\theta), \qquad v(theta) = VAR (X_j | \theta).$$}
to:
{$$\mu(\theta) = E( X_j |\theta), \qquad v(\theta) = VAR (X_j | \theta).$$}
Changed line 115 from:
{$$\mu = E(\mu(\theta)), \qquad v= E(v(theta)), \qquad a = VAR (\mu( \theta)).$$}
to:
{$$\mu = E(\mu(\theta)), \qquad v= E(v(\theta)), \qquad a = VAR (\mu( \theta)).$$}
Changed line 120 from:
{$$Var(X_j|theta) = \frac{v(\theta)}{m_j};$$} 
to:
{$$Var(X_j|\theta) = \frac{v(\theta)}{m_j};$$} 
Changed line 116 from:
Then we can show that the sum {$Y$} above can be written as {$Z\overline{X} + (1-Z)\mu$}, where
to:
Then we can show that the sum {$Y$} (the credibility premium) above can be written as {$Z\overline{X} + (1-Z)\mu$}, where
Changed lines 118-119 from:
->{$k$} is called the ''Buhlmann credibility factor''
to:
->{$k$} is called the ''Buhlmann credibility factor''.
->b)The Buhlmann-Straub model:  Same except now the variance looks like
{$$Var(X_j|theta) = \frac{v(\theta)}{m_j};$$} 
the factors {$m_j$} takes into account things that changes every year, like the # of individuals in the group in year {$j$}, or the premium income for the policy in year {$j$}.  Let {$m=sum_j m_j$}.
->Define {$\mu, v, a$} exactly as above (not involving the {$m_j$}'s).  Then the credibility premium becomes  {$Z\overline{X} + (1-Z)\mu$} where
{$$Z=\frac{m}{m+k}, \qquad k = \frac{v}{a} = \frac{\text{EPV}}{\text{VHM}}, \qquad \overline{X} = \sum_{j=1}^n \frac{m_j}{m} X_j.$$}


Changed lines 127-128 from:
If a prior distribution gives rise to a posterior distribution that's in the same family (maybe a different parameter), it's called a ''conjugate prior distribution''.  An example is the Poisson-gamma model.
to:
If a prior distribution gives rise to a posterior distribution that's in the same family (maybe a different parameter), it's called a ''conjugate prior distribution''.  An example is the Poisson-Gamma model.
->Setup:
->Prior distribution {$\Theta \sim Gamma(\alpha,\beta)$}.
->The observation

Added lines 109-110:
->[Side note:  the same solution of {$\alpha_j$}'s also minimizes {$E\left[ (E(X_{n+1}|\vec{X}) - Y )^2  \right]$} and  {$E\left[ (X_{n+1} - Y )^2  \right]$}.]
Changed line 113 from:
{$$\mu(\theta) = E( X_j |\theta), \qquad \v(theta) = VAR (X_j | \theta).$$}
to:
{$$\mu(\theta) = E( X_j |\theta), \qquad v(theta) = VAR (X_j | \theta).$$}
Changed line 115 from:
{$$\mu = E(\mu(\theta)), \qquad \v= E(v(theta)), \qquad a = VAR (\mu( \theta)).$$}
to:
{$$\mu = E(\mu(\theta)), \qquad v= E(v(theta)), \qquad a = VAR (\mu( \theta)).$$}
Changed lines 118-119 from:
to:
->{$k$} is called the ''Buhlmann credibility factor''
Changed line 107 from:
{$$Q = E\left[ \mu_{n+1}(\Theta) - Y    \right],$$}
to:
{$$Q = E\left[ (\mu_{n+1}(\Theta) - Y )^2  \right],$$}
Changed lines 110-111 from:
->a)The B\"uhlmann model:
to:
->a)The Buhlmann model:  All the {$X_j$}s have the same mean and variance and are iid conditional on {$\Theta$}.  Define the ''hypothetical mean'' and ''process variance'' as follows, respectively:
{$$\mu(\theta) = E( X_j |\theta), \qquad \v(theta) = VAR (X_j | \theta).$$}
Let {$\mu, v, a$} be the ''expected value of the hypothetical means'' (EHM), ''expected value of the process variance'' (EPV), and ''variance of the hypothetical means'' (VHM):
{$$\mu = E(\mu(\theta)), \qquad \v= E(v(theta)), \qquad a = VAR (\mu( \theta)).$$}
Then we can show that the sum {$Y$} above can be written as {$Z\overline{X} + (1-Z)\mu$}, where
{$$Z=\frac{n}{n+k}, \qquad k = \frac{v}{a} = \frac{\text{EPV}}{\text{VHM}}.$$}

Changed lines 106-109 from:
->If the Bayesian premium is difficult to evaluate, an alternative is to alculate the ''credibility premium'' instead, that is, find {$\alpha_0, \ldots, \alpha_n$} and form ${Y=\alpha_0+\sum_{j=1}^n \alpha_j X_j$}, and minimize the squared error loss:


to:
->If the Bayesian premium is difficult to evaluate, an alternative is to alculate the ''credibility premium'' instead, that is, find {$\alpha_0, \ldots, \alpha_n$} such that the sum ${Y=\alpha_0+\sum_{j=1}^n \alpha_j X_j$} minimizes the squared error loss:
{$$Q = E\left[ \mu_{n+1}(\Theta) - Y    \right],$$}
->where the expectation is over all possible joint values of {$\theta$} and {$X_j$}. 
->It's messy if the means and variances of {$X_j$} are all different, but here are two simplified situations:
->a)The B\"uhlmann model:

Changed lines 106-109 from:
to:
->If the Bayesian premium is difficult to evaluate, an alternative is to alculate the ''credibility premium'' instead, that is, find {$\alpha_0, \ldots, \alpha_n$} and form ${Y=\alpha_0+\sum_{j=1}^n \alpha_j X_j$}, and minimize the squared error loss:


October 01, 2011, at 10:08 PM by 38.106.150.109 -
Changed line 100 from:
The ''pure premium'' {$\mu$} (which does not depend on prior observations {$\vec{x}$}, or the risk parameter {$\theta$}) is the mean of the hypothetical means:
to:
->The ''pure premium'' {$\mu$} (which does not depend on prior observations {$\vec{x}$}, or the risk parameter {$\theta$}) is the mean of the hypothetical means:
October 01, 2011, at 10:01 PM by 38.106.150.109 -
Changed lines 95-96 from:
{$$f_{X_{n+1}|\vec{X}}(x_{n+1}|\vec{x}) = \frac{f_{X_{n+1},\vec{X}}(x_{n+1},\vec{x})}{f_{\vec{X}}(\vec{x})} = \frac{\int \prod_{j=1}^{n+1} f_{X_j|\theta}(x_j|\theta) \pi(\theta)  }{f_{\vec{X}}(\vec{x})} = \int f_{X_{n+1}|\Theta}(x_{n+1}|\theta) \pi_{\Theta|\vec{X}}(\theta|\vec{x}) \,d\theta.$$}
to:
{$$f_{X_{n+1}|\vec{X}}(x_{n+1}|\vec{x}) = \frac{f_{X_{n+1},\vec{X}}(x_{n+1},\vec{x})}{f_{\vec{X}}(\vec{x})} = \frac{\int \prod_{j=1}^{n+1} f_{X_j|\theta}(x_j|\theta) \pi(\theta)  }{f_{\vec{X}}(\vec{x})} = \int f_{X_{n+1}|\Theta}(x_{n+1}|\theta) \pi_{\Theta|\vec{X}}(\theta|\vec{x}) \,d\theta. $$}
Changed lines 103-104 from:
{$E(X_{n+1}|\vec{X}=\vec{x}) =  \int x_{n+1} f_{X_{n+1}|\vec{X}}(x_{n+1}|\vec{x}) \,dx_{n+1} =\ldots = \int \mu(\theta) \pi_{\Theta|\vec{X}}(\theta|\vec{x})\,d\theta.$$}
to:
{$$E(X_{n+1}|\vec{X}=\vec{x}) =  \int x_{n+1} f_{X_{n+1}|\vec{X}}(x_{n+1}|\vec{x}) \,dx_{n+1} =\ldots = \int \mu(\theta) \pi_{\Theta|\vec{X}}(\theta|\vec{x})\,d\theta.$$}
October 01, 2011, at 09:59 PM by 38.106.150.109 -
Changed line 99 from:
{$$E(X_{n+1}|\Theta=\theta) = \int x_{n+1} f_{X_{n+1}|\theta}(x_{n+1}|\theta) \,dx_{n+1}.$$}
to:
{$$\mu(\theta)=E(X_{n+1}|\Theta=\theta) = \int x_{n+1} f_{X_{n+1}|\theta}(x_{n+1}|\theta) \,dx_{n+1}.$$}
Changed lines 103-104 from:
{{$E(X_{n+1}|\vec{X}=\vec{x}) =  \int x_{n+1} f_{X_{n+1}|\vec{X}}(x_{n+1}|\vec{x}) \,dx_{n+1} =\ldots = \int \mu(\theta) \pi_{\Theta|\vec{X}}(\theta|\vec{x})\,d\theta.$$}
to:
{$E(X_{n+1}|\vec{X}=\vec{x}) =  \int x_{n+1} f_{X_{n+1}|\vec{X}}(x_{n+1}|\vec{x}) \,dx_{n+1} =\ldots = \int \mu(\theta) \pi_{\Theta|\vec{X}}(\theta|\vec{x})\,d\theta.$$}
October 01, 2011, at 09:58 PM by 38.106.150.109 -
Changed lines 97-99 from:


to:
->The next step is to find the expected values of {$X_{n+1}$}.  We can do this two ways:
->a)  Find the ''hypothetical mean'' {$\mu(\theta)$}, i.e.,
{$$E(X_{n+1}|\Theta=\theta) = \int x_{n+1} f_{X_{n+1}|\theta}(x_{n+1}|\theta) \,dx_{n+1}.$$}
The ''pure premium'' {$\mu$} (which does not depend on prior observations {$\vec{x}$}, or the risk parameter {$\theta$}) is the mean of the hypothetical means:
{$$\mu = E(E(X_{n+1}|\Theta)) = E(\mu(\Theta)) = \int \mu(\theta) \pi(\theta)\,d\theta.$$}
->b) Find the ''Bayesian premium'', which is the mean of the predictive distribution:
{{$E(X_{n+1}|\vec{X}=\vec{x}) =  \int x_{n+1} f_{X_{n+1}|\vec{X}}(x_{n+1}|\vec{x}) \,dx_{n+1} =\ldots = \int \mu(\theta) \pi_{\Theta|\vec{X}}(\theta|\vec{x})\,d\theta.$$}

October 01, 2011, at 08:56 PM by 38.106.150.109 -
Changed line 86 from:
->{$\theta$} represents a ''risk parameter'' with a pdf {$\pi(\theta)$} that describes the risk characteristics within a population.
to:
->{$\theta$} represents a ''risk parameter'' with a pdf {$\pi(\theta)$} that describes the risk characteristics within a population.  Call {$\pi(\theta)$} the ''prior distribution''.
Added lines 97-99:


Changed lines 103-104 from:
to:
If a prior distribution gives rise to a posterior distribution that's in the same family (maybe a different parameter), it's called a ''conjugate prior distribution''.  An example is the Poisson-gamma model.
October 01, 2011, at 08:50 PM by 38.106.150.109 -
Changed line 94 from:
->d)  Then the conditional density of {$X_{n+1}$}, given {$\vec{X}$}, is
to:
->d)  Then the conditional density of {$X_{n+1}$}, given {$\vec{X}$} (the ''predicative distribution''), is
October 01, 2011, at 08:46 PM by 38.106.150.109 -
Changed lines 95-96 from:
{$$f_{X_{n+1}|\vec{X}}(x_{n+1}|\vec{x}) = \frac{f_{X_{n+1},\vec{X}}(x_{n+1},\vec{x})}{f_{\vec{X}}(\vec{x})} = \frac{\int \prod_{j=1}^{n+1} f_{X_j|\theta}(x_j|\theta) \pi(\theta)  }{f_{\vec{X}}(\vec{x})} = \int f_{X_{n+1}|\Theta}(x_{n+1}|\theta) \pi_{\Theta|\vec{X}}(\theta|\vec{x}} \,d\theta.$$}
to:
{$$f_{X_{n+1}|\vec{X}}(x_{n+1}|\vec{x}) = \frac{f_{X_{n+1},\vec{X}}(x_{n+1},\vec{x})}{f_{\vec{X}}(\vec{x})} = \frac{\int \prod_{j=1}^{n+1} f_{X_j|\theta}(x_j|\theta) \pi(\theta)  }{f_{\vec{X}}(\vec{x})} = \int f_{X_{n+1}|\Theta}(x_{n+1}|\theta) \pi_{\Theta|\vec{X}}(\theta|\vec{x}) \,d\theta.$$}
October 01, 2011, at 08:46 PM by 38.106.150.109 -
Changed line 92 from:
->c)  Find the ''posterior density of {$\Theta$} given {$\vec{X}$}:
to:
->c)  Find the ''posterior density'' of {$\Theta$} given {$\vec{X}$}:
Changed lines 95-96 from:
{$$f_{X_{n+1}|\vec{X}}(x_{n+1}|\vec{x}) = \frac{f_{X_{n+1},\vec{X}}(x_{n+1},\vec{x})}{f_{\vec{X}}(\vec{x})}$$}
to:
{$$f_{X_{n+1}|\vec{X}}(x_{n+1}|\vec{x}) = \frac{f_{X_{n+1},\vec{X}}(x_{n+1},\vec{x})}{f_{\vec{X}}(\vec{x})} = \frac{\int \prod_{j=1}^{n+1} f_{X_j|\theta}(x_j|\theta) \pi(\theta)  }{f_{\vec{X}}(\vec{x})} = \int f_{X_{n+1}|\Theta}(x_{n+1}|\theta) \pi_{\Theta|\vec{X}}(\theta|\vec{x}} \,d\theta.$$}
October 01, 2011, at 08:41 PM by 38.106.150.109 -
Changed lines 95-96 from:
{$$f_{X_{n+1}|\vec{X}}(x_{n+1}|\vec{x} = \frac{f_{X_{n+1},\vec{X}}(x_{n+1},\vec{x}}{f_{\vec{X}}(\vec{x}}$$}
to:
{$$f_{X_{n+1}|\vec{X}}(x_{n+1}|\vec{x}) = \frac{f_{X_{n+1},\vec{X}}(x_{n+1},\vec{x})}{f_{\vec{X}}(\vec{x})}$$}
October 01, 2011, at 03:50 PM by 38.106.150.109 -
Changed line 88 from:
->Find the joint density function:
to:
->a)  Find the joint density function:
Changed line 90 from:
->Find the marginal distribution by integrating out {$\theta$}:
to:
->b)  Find the marginal distribution by integrating out {$\theta$}:
Changed lines 92-93 from:

to:
->c)  Find the ''posterior density of {$\Theta$} given {$\vec{X}$}:
{$$\pi_{\Theta|\vec{X}}(\theta|\vec{x}) = \frac{f_{\vec{X},\theta}(\vec{x},\theta)}{f_{\vec{X}}(\vec{x})}.$$}
->d)  Then the conditional density of {$X_{n+1}$}, given {$\vec{X}$}, is
{$$f_{X_{n+1}|\vec{X}}(x_{n+1}|\vec{x} = \frac{f_{X_{n+1},\vec{X}}(x_{n+1},\vec{x}}{f_{\vec{X}}(\vec{x}}$$}

October 01, 2011, at 03:19 PM by 38.106.150.109 -
Changed lines 91-93 from:
{$$f_{\vec{X}(\vec{x})= \int_\theta f_{\vec{X},\theta}(\vec{x},\theta)\,d\theta.$$}

to:
{$$f_{\vec{X}}(\vec{x})= \int_\theta f_{\vec{X},\theta}(\vec{x},\theta)\,d\theta.$$}

October 01, 2011, at 03:19 PM by 38.106.150.109 -
October 01, 2011, at 03:17 PM by 38.106.150.109 -
Changed line 63 from:
{$$\text{If }X \text{ and }Y {\text{ are independent, then } f_{X,Y}(x,y) = f_X(x)f_Y(y);$$}
to:
{$$\text{If }X \text{ and }Y \text{ are independent, then } f_{X,Y}(x,y) = f_X(x)f_Y(y);$$}
Changed lines 61-93 from:
to:
->Know the following basic formulas about conditional expectation:  Let {$f_{X,Y}(x,y)$} be the joing pdf for two RVs {$X$} and {$Y$}, with marginal pdfs ($f_X(x)$} and {$f_Y(y)$}.  [The whole discussion applies to discrete RVs as well.]  Then
{$$\text{conditional pdf of }X \text{ given }Y = f_{X|Y}(x|y) =\frac{f_{X,Y}(x,y)}{f_Y(y)}.$$}
{$$\text{If }X \text{ and }Y {\text{ are independent, then } f_{X,Y}(x,y) = f_X(x)f_Y(y);$$}
->so in the case of independence, the conditional and marginal distributions are the same.
->Switching the definitions and formulas around a bit, we have
{$$f_X(x) = \int f_{X|Y}(x|y)f_Y(y)\,dy.$$}
->As the result of this formula we can deduce that
-->If {$X|Y \sim Poisson(Y)$} and {$Y\sim \Gamma(\alpha, \beta)$}, then {$X\sim NegBinomial(\alpha,\beta)$}.
-->If {$X|Y \sim Normal(Y,\sigma_1^2)$} and {$Y\sim Normal(\mu,\sigma_2^2 )$}, then {$X\sim Normal(\mu,\sigma_1^2+\sigma_2^2)$}.
From the formula above we can exchange the roles of {$X$} and {$Y$} to get
{$$f_{X,Y}(x,y) = f_{X|Y}(x|y)f_Y(y) = f_{Y|X}(y|x)f_X(x),$$}
->yielding ''Bayes's Theorem'':
{$$f_{X|Y}(x|y) = \frac{ f_{Y|X}(y|x)f_X(x)}{f_Y(y)}.$$}
->We can also define the ''conditional expection'' (which is a random variable that is a function of {$Y$}:
{$$E(X|Y) = \int x f_{X|Y}(x|y) \, dx.$$}
->It can be shown that
{$$E(E(X|Y)) = \int E(X|Y=y) f_Y(y) \, dy = E(X).$$}
->This can be generalized to any functions of {$X, Y$}:
{$$E(E(h(X,Y)|Y)) = \int \int h(x,y) f_{X|Y}(x|y)\,dx f_Y(y) \, dy = \int\int h(x,y) f_{X,Y}(x,y)=E(h(X,Y)).$$}
->Getting into variances,
{$$Var(X|Y) = E( (X-E(X|Y))^2 | Y) = E(X^2Y) - E(X|Y)^2.$$}
->Quite a bit of derivation later we can show that
{$$Var(X) = E(Var(X|Y)) + Var(E(X|Y)).$$}

->Now the Baysian anaysis.  Setup:
->{$\theta$} represents a ''risk parameter'' with a pdf {$\pi(\theta)$} that describes the risk characteristics within a population.
->{$\vec{X}=(X_1, \ldots, X_n)$} represent the ''claim amount'' that already occurred in the past; {$X_{n+1}$} is the next one we want to predict.  Each has a conditional pdf {$f_{X_j|\theta}(x_j|\theta)$} (they may be all the same if {$X_j$}'s are identically distributed).  Assume the {$X_j$}s are independent conditional on {$\theta$}.  Then
->Find the joint density function:
{$$f_{\vec{X},\theta}(\vec{x},\theta)= \prod_{j=1}^n f_{X_j|\theta}(x_j|\theta) \pi(\theta).$$}
->Find the marginal distribution by integrating out {$\theta$}:
{$$f_{\vec{X}(\vec{x})= \int_\theta f_{\vec{X},\theta}(\vec{x},\theta)\,d\theta.$$}

Changed lines 58-59 from:
{$$Z = \frac{\xi\sqrt{n}}{\sigma\sqrt{\lambda_0}}=\sqrt{frac{n}{\lambda_0\frac{\sigma^2}{\xi^2}}} = \text{Sq root of the ratio of the actual count to the count required for full credibility}.$$}
to:
{$$Z = \frac{\xi\sqrt{n}}{\sigma\sqrt{\lambda_0}}=\sqrt{\frac{n}{\lambda_0\frac{\sigma^2}{\xi^2}}} = \text{Sq root of the ratio of the actual count to the count required for full credibility}.$$}
Changed lines 58-59 from:
{$$Z = \frac{\xi\sqrt{n}}{\sigma\sqrt{\lamdbda_0}}=\sqrt{frac{n}{\lambda_0\frac{\sigma^2}{\xi^2}}} = \text{Sq root of the ratio of the actual count to the count required for full credibility}.$$}
to:
{$$Z = \frac{\xi\sqrt{n}}{\sigma\sqrt{\lambda_0}}=\sqrt{frac{n}{\lambda_0\frac{\sigma^2}{\xi^2}}} = \text{Sq root of the ratio of the actual count to the count required for full credibility}.$$}
Changed line 54 from:
{$$n \geq \lambda_0 \left(\frac{\sigma}{\xi}\right)^2 \Leftrightarrow  Var(\overline{X} = \frac{\sigma^2}{n} \leq \frac{\xi^2}{\lambda_0},$$}
to:
{$$n \geq \lambda_0 \left(\frac{\sigma}{\xi}\right)^2 \Leftrightarrow  Var(\overline{X}) = \frac{\sigma^2}{n} \leq \frac{\xi^2}{\lambda_0},$$}
Changed lines 58-59 from:
{$$Z = \frac{\xi\sqrt{n}}{\sigma{\sqrt{\lamdbda_0}}=\sqrt{frac{n}{\lambda_0\frac{\sigma^2}{\xi^2}}} = \text{Sq root of the ratio of the actual count to the count required for full credibility}.$$}
to:
{$$Z = \frac{\xi\sqrt{n}}{\sigma\sqrt{\lamdbda_0}}=\sqrt{frac{n}{\lambda_0\frac{\sigma^2}{\xi^2}}} = \text{Sq root of the ratio of the actual count to the count required for full credibility}.$$}
Changed line 54 from:
{$$n \geq \lambda_0 \left(\frac{\sigma}{\xi}\right)^2 \Leftrightarrow  Var(\overline{X} = \frac{\sigma^2?{n} \leq \frac{\xi^2}{\lambda_0},$$}
to:
{$$n \geq \lambda_0 \left(\frac{\sigma}{\xi}\right)^2 \Leftrightarrow  Var(\overline{X} = \frac{\sigma^2}{n} \leq \frac{\xi^2}{\lambda_0},$$}
Changed lines 58-59 from:
{$$Z = \frac{\xi\sqrt{n}}{\sigma{\sqrt{\lamdbda_0}}=\sqrt{frac{n}{\lambda_0\frac{\sigma^2}{\xi^2}}} = {\text{Sq root of the ratio of the actual count to the count required for full credibility}.$$}
to:
{$$Z = \frac{\xi\sqrt{n}}{\sigma{\sqrt{\lamdbda_0}}=\sqrt{frac{n}{\lambda_0\frac{\sigma^2}{\xi^2}}} = \text{Sq root of the ratio of the actual count to the count required for full credibility}.$$}
Changed lines 54-55 from:
{$$n \geq \lambda_0 \left(\frac{\sigma}{\xi}\right)^2 \Leftrightdoublearrow$$}
to:
{$$n \geq \lambda_0 \left(\frac{\sigma}{\xi}\right)^2 \Leftrightarrow  Var(\overline{X} = \frac{\sigma^2?{n} \leq \frac{\xi^2}{\lambda_0},$$}
->so we can set the variance of {$P_c$} to be the same as {$\xi^2/\lambda_0$}:
{$$\frac{\xi^2}{\lambda_0} = Var(P_c) = Z^2 Var(\overline{X}) = Z^2 \frac{\sigma^2}{n}.$$}
->So
{$$Z = \frac{\xi\sqrt{n}}{\sigma{\sqrt{\lamdbda_0}}=\sqrt{frac{n}{\lambda_0\frac{\sigma^2}{\xi^2}}} = {\text{Sq root of the ratio of the actual count to the count required for full credibility}.
$$}
Changed lines 53-55 from:
->Partial Credibility:  Let {$M$} is the manual premium, and {$P_c = Z\overline{X} + (1-Z) M$} be the ''credibility premium''
to:
->Partial Credibility:  Let {$M$} is the manual premium, and {$P_c = Z\overline{X} + (1-Z) M$} be the ''credibility premium''.  Note that in the standard for full credibility for frequency,
{$$n \geq \lambda_0 \left(\frac{\sigma}{\xi}\right)^2 \Leftrightdoublearrow$$}

Added line 53:
->Partial Credibility:  Let {$M$} is the manual premium, and {$P_c = Z\overline{X} + (1-Z) M$} be the ''credibility premium''
Changed lines 31-32 from:
->II.a.  Suppose we have more info in the case of I.a.:  each {$X_j$} is compund Poisson distributed, i.e., {$X_j = Y_{j,1}+\cdots+Y_{j,N_j}$}, where each {$N_j$} is Poisson with parameter {$\lambda$} and all the {$Y$}'s represent the claim distributions, and have mean {$\theta_Y$} and variance {$\sigma_Y^2$}.
to:
->II.  Suppose we have more info in the case of I.a.:  each {$X_j$} is compund Poisson distributed, i.e., {$X_j = Y_{j,1}+\cdots+Y_{j,N_j}$}, where each {$N_j$} is Poisson with parameter {$\lambda$} and all the {$Y$}'s represent the claim distributions, and have mean {$\theta_Y$} and variance {$\sigma_Y^2$}.
September 30, 2011, at 07:58 PM by 38.106.150.109 -
Changed line 21 from:
{$$n \geq \lambda_0 \left(\frac{\sigma}{\xi}\right)^2,\qquad \text{in terms of # of exposures (policies),}$$}
to:
{$$n \geq \lambda_0 \left(\frac{\sigma}{\xi}\right)^2,\qquad \text{in terms of # of exposures/policies (I.a), or in terms of # of claims (I.b),}$$}
Changed lines 27-28 from:
->The quantity {$\lambda_0 \left(\frac{\sigma}{\xi}\right)^2$} is what the notes call ''Standard for Full Credibility for Severity''.
to:
->The quantity {$\lambda_0 \left(\frac{\sigma}{\xi}\right)^2$} is what the notes call ''Standard for Full Credibility for Severity''. (In the situation I.b.)
September 30, 2011, at 07:53 PM by 38.106.150.109 -
Changed lines 51-52 from:
->The quantity {$\lambda_0 \left(\theta_Y+\frac{\sigma_Y^2}{\theta_Y}\right)$} is what the notes call ''Standard for Full Credibility for Pure Premium''.
to:
->The middle quantity {$\lambda_0 \left(1+\left(\frac{\sigma_Y}{\theta_Y}\right)^2\right)$} is what the notes call ''Standard for Full Credibility for Pure Premium''.
September 30, 2011, at 07:52 PM by 38.106.150.109 -
Changed lines 29-30 from:
->In practice, we have to estimate {$\xi$} and {$\sigma$} with the data we're presented.  ( {$\sigma$} is approximated with the "n-1" unbiased estimator.)
to:
->In practice, we have to estimate {$\xi$} and {$\sigma$} with the data we're presented.  ({$\sigma$} is approximated with the "n-1" unbiased estimator.)
Changed lines 51-52 from:

to:
->The quantity {$\lambda_0 \left(\theta_Y+\frac{\sigma_Y^2}{\theta_Y}\right)$} is what the notes call ''Standard for Full Credibility for Pure Premium''.
September 30, 2011, at 07:13 PM by 38.106.150.109 -
Changed lines 29-30 from:
->In practice, we have to estimate {$\xi$} and {$\sigma$} with the data we're presented.  ({$\simga$} is approximated with the "n-1" unbiased estimator.)
to:
->In practice, we have to estimate {$\xi$} and {$\sigma$} with the data we're presented.  ( {$\sigma$} is approximated with the "n-1" unbiased estimator.)
Changed lines 41-43 from:
->In practice, we estimate {$\lambda$} by taking # of claims in the data divided by # of policies, and {$\theta_Y$} and {$\sigma_Y$} by the given data.  ({$\simga_Y$} is approximated with the "n-1" unbiased estimator.)

to:
->In practice, we estimate {$\lambda$} by taking # of claims in the data divided by # of policies, and {$\theta_Y$} and {$\sigma_Y$} by the given data.  ({$\sigma_Y$} is approximated with the "n-1" unbiased estimator.)

September 30, 2011, at 07:11 PM by 38.106.150.109 -
Added lines 27-28:
->The quantity {$\lambda_0 \left(\frac{\sigma}{\xi}\right)^2$} is what the notes call ''Standard for Full Credibility for Severity''.
Changed lines 39-40 from:
to:
->The quantity {$\lambda_0$} is what the notes call ''Standard for Full Credibility for Frequency''.
September 30, 2011, at 06:55 PM by 38.106.150.109 -
September 30, 2011, at 06:50 PM by 38.106.150.109 -
September 30, 2011, at 06:50 PM by 38.106.150.109 -
Changed line 21 from:
{$$n \geq \lambda_0 \left(\frac{\sigma}{\xi}\right)^2,\qquad \text{in terms of # of exposures,}$$}
to:
{$$n \geq \lambda_0 \left(\frac{\sigma}{\xi}\right)^2,\qquad \text{in terms of # of exposures (policies),}$$}
Changed lines 27-28 from:
->In practice, we have to estimate {$\
to:
->In practice, we have to estimate {$\xi$} and {$\sigma$} with the data we're presented.  ({$\simga$} is approximated with the "n-1" unbiased estimator.)
Added lines 38-40:
->In practice, we estimate {$\lambda$} by taking # of claims in the data divided by # of policies, and {$\theta_Y$} and {$\sigma_Y$} by the given data.  ({$\simga_Y$} is approximated with the "n-1" unbiased estimator.)

September 30, 2011, at 03:35 PM by 38.106.150.109 -
Changed line 14 from:
->Full Credibility:  Choose {$r$} close to 0, {$p$} close to 1, and assign full credibility to the (average) # of claims/losses if
to:
->Full Credibility:  Choose {$r$} close to 0, {$p$} close to 1.  We would like the RV {$\overline{X}$} to be "stable", i.e., that the difference between {$\overline{X}$} and {$\xi$} is small, most of the time.  More precisely,
Changed lines 16-19 from:
->It turns out this is equivalent to making {$n$} very large, i.e.,
{$$n \geq \lambda_0 \left(\frac{\sigma}{\xi}\right)^2,\qquad \text{in terms of # of policies,}$$}
->where {$\lambda_0 = (y_p/r)^2$}, and {$y_p$} is such that {$\Phi(y_p) = (p+1)/2$}. {$\Phi$ is the cdf for the standard normal distribution.)
to:
->Rewrite as
{$$Prob( \left| \frac
{\overline{X}-\xi}{\sigma/\sqrt{n}}  \right| \leq \frac{r\xi\sqrt{n}}{\sigma}) \geq p.$$}
->This holds whenever {$ \frac{r\xi\sqrt{n
}}{\sigma} \geq y_p$}, where
{$$Prob( \left| \frac{\overline{X}-\xi}{\sigma/\sqrt{n}}  \right| \leq y_p) = p;$$}
->we call this the ''full credibility standard'' for the given {$r$} and {$p$} values:
{$$n \geq \lambda_0 \left(\frac{\sigma}{\xi}\right
)^2,\qquad \text{in terms of # of exposures,}$$}
->where {$\lambda_0 = (y_p/r)^2$}.
->We actually find {$y_p$} by assuming {$overline{X}$} is approximately normal with mean {$\xi$} and variance {$\sigma^2/n$}; then
{$$p = Prob( \left| \frac{\overline{X}-\xi}{\sigma/\sqrt{n}}  \right| \leq y_p) \approx  Prob( \left| Z \right| \leq y_p) = 2\Phi(y_p)-1, $$}
-> where {$\Phi$} is the cdf for the standard normal distribution.  Then {$y_p$} is such that {$\Phi(y_p) = (p+1)/2 $}.

->In practice, we have to estimate {$\

September 30, 2011, at 02:47 PM by 38.106.150.109 -
Changed line 14 from:
->Full Credibility:  Choose {$r$} close to 0, {$p$} close to 1, and assign full credibility to the # of claims/losses if
to:
->Full Credibility:  Choose {$r$} close to 0, {$p$} close to 1, and assign full credibility to the (average) # of claims/losses if
Changed line 17 from:
{$$n \geq \lambda_0 \left(\frac{\sigma}{\xi}\right)^2,$$}
to:
{$$n \geq \lambda_0 \left(\frac{\sigma}{\xi}\right)^2,\qquad \text{in terms of # of policies,}$$}
September 30, 2011, at 11:48 AM by 38.106.150.109 -
Changed line 33 from:
{$$n \geq \lambda_0 \frac{\lambda(\theta_Y^2+\sigma_Y^2}{ \lambda^2\theta_Y^2}= \lambda_0/\lambda \left(1+\left(\frac{\sigma_Y}{\theta_Y}\right)^2\right), \qquad \text{in terms of # of policies,}$$}
to:
{$$n \geq \lambda_0 \frac{\lambda(\theta_Y^2+\sigma_Y^2)}{ \lambda^2\theta_Y^2}= \frac{\lambda_0}{\lambda} \left(1+\left(\frac{\sigma_Y}{\theta_Y}\right)^2\right), \qquad \text{in terms of # of policies,}$$}
September 30, 2011, at 10:06 AM by 38.106.150.109 -
Changed lines 20-21 from:
-->II.a.  Suppose we have more info in the case of I.a.:  each {$X_j$} is compund Poisson distributed, i.e., {$X_j = Y_{j,1}+\cdots+Y_{j,N_j}$}, where each {$N_j$} is Poisson with parameter {$\lambda$} and all the {$Y$}'s represent the claim distributions, and have mean {$\theta_Y$} and variance {$\sigma_Y^2$}.
to:
->II.a.  Suppose we have more info in the case of I.a.:  each {$X_j$} is compund Poisson distributed, i.e., {$X_j = Y_{j,1}+\cdots+Y_{j,N_j}$}, where each {$N_j$} is Poisson with parameter {$\lambda$} and all the {$Y$}'s represent the claim distributions, and have mean {$\theta_Y$} and variance {$\sigma_Y^2$}.
September 30, 2011, at 10:06 AM by 38.106.150.109 -
Changed lines 3-5 from:
->Setup:  several kinds of problems:
-->I.a.  Past periods are labeled {$1,2,\ldots, n$}; there are {$X_j$} claims (or losses) in each period {$j$}.
-->I.b.  Different policies are labeled {$1,2,\ldots, n$} and each {$X_j$} is the claim or loss for each policy {$j$}.
to:
Setup:  several kinds of problems:
->I.a.  Past periods are labeled {$1,2,\ldots, n$}; there are {$X_j$} claims (or losses) in each period {$j$}.
->I.b.  Different policies are labeled {$1,2,\ldots, n$} and each {$X_j$} is the claim or loss for each policy {$j$}.
Changed line 16 from:
It turns out this is equivalent to making {$n$} very large, i.e.,
to:
->It turns out this is equivalent to making {$n$} very large, i.e.,
Changed lines 18-19 from:
where {$\lambda_0 = (y_p/r)^2$}, and {$y_p$} is such that {$\Phi(y_p) = (p+1)/2$}. {$\Phi$ is the cdf for the standard normal distribution.)
to:
->where {$\lambda_0 = (y_p/r)^2$}, and {$y_p$} is such that {$\Phi(y_p) = (p+1)/2$}. {$\Phi$ is the cdf for the standard normal distribution.)
September 30, 2011, at 10:05 AM by 38.106.150.109 -
Changed lines 22-23 from:
->Full Credibility:
-->Use {$N_j$}'s instead of {$X_j$}'s and get
to:
->Full Credibility on the average number of claims:
-->The RV to consider here is {$N_j$}:
Changed line 25 from:
-->so full credibility for the # of claims/losses is
to:
-->so full credibility for the average number of claims is
Added lines 29-37:
->Full Credibility on the average total payment:
-->The RV to consider here is {$X_j$},
{$$ \xi = E(X_j) = \lambda\theta_Y, \qquad \sigma^2 = Var(X_j) = \lambda(\theta_Y^2+\sigma_Y^2,$$}
-->so full credibility for the average total payment is
{$$n \geq \lambda_0 \frac{\lambda(\theta_Y^2+\sigma_Y^2}{ \lambda^2\theta_Y^2}= \lambda_0/\lambda \left(1+\left(\frac{\sigma_Y}{\theta_Y}\right)^2\right), \qquad \text{in terms of # of policies,}$$}
{$$n\lambda \geq \lambda_0 \left(1+\left(\frac{\sigma_Y}{\theta_Y}\right)^2\right), \qquad \text{in terms of # of expected claims,}$$}
{$$n\lambda\theta_Y \geq \lambda_0 \left(\theta_Y+\frac{\sigma_Y^2}{\theta_Y}\right), \qquad \text{in terms of # of expected total dollars of claims.}$$}

September 29, 2011, at 09:42 PM by 38.106.150.109 -
Changed lines 20-21 from:
-->II.a.  Suppose we have more info in the case of I.a.:  each {$X_j$} is compund Poisson distributed, i.e., {$X_j = Y_{j,1}+\cdots+Y_{j,N_j}$}, where {$N_j$} is Poisson with parameter {$\lambda$} and all the {$Y$}'s have mean {$\theta_Y$} and variance {$\sigma_Y^2$}.
to:
-->II.a.  Suppose we have more info in the case of I.a.:  each {$X_j$} is compund Poisson distributed, i.e., {$X_j = Y_{j,1}+\cdots+Y_{j,N_j}$}, where each {$N_j$} is Poisson with parameter {$\lambda$} and all the {$Y$}'s represent the claim distributions, and have mean {$\theta_Y$} and variance {$\sigma_Y^2$}.
Changed lines 23-25 from:
-->If we're talking about # of claims/losses, then

to:
-->Use {$N_j$}'s instead of {$X_j$}'s and get
{$$ \xi = E(N_j) = \lambda, \qquad \sigma^2 = Var(N_j) = \lambda,$$}
-->so full credibility for the # of claims/losses is
{$$n \geq \lambda_0 \left(\frac{\sqrt{\lambda}}{\lambda}\right)^2= \lambda_0/\lambda, \qquad \text{in terms of # of policies,}$$}
{$$n\lambda \geq \lambda_0, \qquad \text{in terms of # of expected claims.}$$}

September 29, 2011, at 09:28 PM by 38.106.150.109 -
Changed lines 3-25 from:
to:
->Setup:  several kinds of problems:
-->I.a.  Past periods are labeled {$1,2,\ldots, n$}; there are {$X_j$} claims (or losses) in each period {$j$}.
-->I.b.  Different policies are labeled {$1,2,\ldots, n$} and each {$X_j$} is the claim or loss for each policy {$j$}.
->Assume that
-->{$E(X_j) = \xi $} for every {$j$}.  (Stable mean)
-->{$Var(X_j = \sigma^2$} for every {$j$}. (Stable variance)
-->Past experience is  {$\overline{X} = (X_1+ \cdots + X_n)/n$}.
-->Then {$E(\overline{X})= \xi, Var(\overline{X}) = \sigma^2/n$}. 
-->Define ''manual premium'' to be {$M$}, some value for the mean that's determined from other past experiences.
-->Goal:  Determine what the new premium should be (between {$M$} and {$\xi$}.

->Full Credibility:  Choose {$r$} close to 0, {$p$} close to 1, and assign full credibility to the # of claims/losses if
{$$Prob( -r\xi \leq \overline{X}-\xi \leq r\xi) \geq p.$$}
It turns out this is equivalent to making {$n$} very large, i.e.,
{$$n \geq \lambda_0 \left(\frac{\sigma}{\xi}\right)^2,$$}
where {$\lambda_0 = (y_p/r)^2$}, and {$y_p$} is such that {$\Phi(y_p) = (p+1)/2$}. {$\Phi$ is the cdf for the standard normal distribution.)

-->II.a.  Suppose we have more info in the case of I.a.:  each {$X_j$} is compund Poisson distributed, i.e., {$X_j = Y_{j,1}+\cdots+Y_{j,N_j}$}, where {$N_j$} is Poisson with parameter {$\lambda$} and all the {$Y$}'s have mean {$\theta_Y$} and variance {$\sigma_Y^2$}.

->Full Credibility:
-->If we're talking about # of claims/losses, then

Added lines 1-10:
!!Credibility (20-25%)
1. Apply limited fluctuation (classical) credibility including criteria for both full and partial credibility.

2. Perform Bayesian analysis using both discrete and continuous models.

3. Apply Bühlmann and Bühlmann-Straub models and understand the relationship of these to the Bayesian model.

4. Apply conjugate priors in Bayesian analysis and in particular the Poisson-gamma model.

5. Apply empirical Bayesian methods in the nonparametric and semiparametric cases.
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Page last modified on October 10, 2011, at 07:19 PM